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# Why is the value of Avogadro's number 6.022 xx 10^(23) and not any other value ?
Updated On: 27-06-2022
Text Solution
Solution : It represents the number of atoms in one gram atom of an element or the number of molecules in one gram mole of a compound. If we divide the atomic mass of an element by actual mass of its atom, the value is 6.022 xx 10^(23). Similarly, by dividing the molecular mass of a compound by the actual mass of its molecules, the same result is obtained. | 158 | 593 | {"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} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.788754 |
https://ditraglia.com/erm/q12-logit-wells.html | 1,716,103,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00429.warc.gz | 189,995,043 | 302,369 | # Exercise - Contaminated Wells in Bangladesh
This problem uses a dataset containing household-level data from Bangladesh, wells.csv, that you can download from my website: https://ditraglia.com/data/wells.csv. Here is some background on the dataset from Gelman and Hill (2007):
Many of the wells used for drinking water in Bangladesh and other South Asian countries are contaminated with natural arsenic … a research team from the United States and Bangladesh measured all the wells [in a small region] and labeled them with their arsenic level as well as a characterization of “safe” (below 0.5 in units of hundreds of micrograms per liter, the Bangladesh standard for arsenic in drinking water) or “unsafe” (above 0.5). People with unsafe wells were encouraged to switch to nearby private or community wells or to new wells of their own construction. A few years later, the researchers returned to find out who had switched wells.
You task in this problem is to predict which households will switch wells using the following information:
Name Description
dist Distance to closest known safe well (meters)
arsenic Arsenic level of respondent’s well (100s of micrograms per liter)
switch Dummy variable equal to 1 if switched to a new well
assoc Dummy variable equal to 1 if any member of the household is active in community organizations
educ Education level of head of household (years)
To be clear, the dataset contains only information for households with an arsenic level of 0.5 or above, as these are the households that were encouraged to switch wells. Moreover, the variable switch is measured after the variables dist100 and arsenic. In other words, arsenic is the arsenic level in the original contaminated well that the household was using before we learned whether they switched wells. Similarly, dist100 was the distance to the nearest well safe well before we know whether the household switched wells.
To answer this question, you will need three additional pieces of information:
1. The Bayes classifier is a rule for predicting whether $$Y=1$$ or $$Y=0$$ given $$X$$ based on the estimated probability $$\hat{P}(Y=1|X)$$. If this probability is greater than $$1/2$$, the Bayes classifier predicts $$\hat{Y} = 1$$; otherwise it predicts $$\hat{Y} = 0$$.
2. The error rate of a classifier is defined as the fraction of incorrect predictions that it makes.
3. For a binary prediction problem the confusion matrix is a $$2\times 2$$ matrix that counts up the number of true positives, $$(\hat{Y} = Y = 1)$$, true negatives $$(\hat{Y} = Y = 0)$$, false positives $$(\hat{Y} = 1, Y=0)$$, and false negatives $$(\hat{Y} = 0, Y = 0)$$. The confusion matrix can be used, among other things, to calculate the sensitivity and specificity of a classifier.
# Exercises
1. Preliminaries:
1. Load the data and store it in a tibble called wells.
2. Use dplyr to create a variable called larsenic that equals the natural logarithm of arsenic.
3. Use ggplot2 to make a histogram of arsenic and larsenic. Be sure to label your plots appropriately. Comment on your findings.
4. Use dplyr to create a variable called dist100 that contains the same information as dist but measured in hundreds of meters rather than in meters.
5. Use dplyr to create a variable called zeduc that equals the z-score of educ, i.e. educ after subtracting its mean and dividing by its standard deviation.
2. First Regression: fit1
1. Run a logistic regression using dist100 to predict switch and store the result in an object called fit1.
2. Use ggplot2 to plot the logistic regression function from part (a) along with the data, jittered appropriately.
3. Discuss your results from parts (a)-(b). In particular: based on fit1, is dist100 a statistically significant predictor of switch? Does the sign of its coefficient make sense? Explain.
4. Use fit1 to calculate the predicted probability of switching wells for the average household in the dataset.
5. Use fit1 to calculate the marginal effect of dist100 for the average household in the dataset. Interpret your result. How does is compare to the “divide by 4” rule and the average partial effect?
3. Predictive performance of fit1
1. Add a column called p1 to wells containing the predicted probabilities that switch equals one based on fit1.
2. Add a column called pred1 to wells that gives the predicted values of $$y$$ that correspond to p1.
3. Use pred1 to calculate the error rate of the Bayes classifier constructed from fit1 based on the full dataset, i.e. wells. Recall that this classifier uses the predicted probabilities from fit1 in the following way: $$p>0.5 \implies$$ predict 1, $$p\leq 0.5\implies$$ predict 0. Hint: you can do this using the summarize function from dplyr.
4. Use pred1 to construct the confusion matrix for fit1. Hint: use the function table.
5. Based on your results from (d), calculate the sensitivity and specificity of the predictions from fit1.
6. Comment on your results from (c) and (e). In particular, compare them to the error rate that you would obtain by simply predicting the most common value of switch for every observation in the dataset. (This is called the “null model” since it doesn’t use any predictors.)
4. Additional regressions: fit2, fit3, and fit4
1. Run a logistic regression using larsenic to predict switch and store the results in an object called fit2.
2. Run a logistic regression using zeduc to predict switch and store the results in an object called fit3.
3. Run a logistic regression using dist100, larsenic, and zeduc to predict switch and store the results in an object called fit4.
4. Make a nicely-formatted summary table of the results from fit1, fit2, fit3, and fit4. Make sure to add appropriate labels and captions, use a reasonable number of decimal places, etc.
5. Interpreting fit2, fit3 and fit4
1. Repeat parts (b) and (c) of question #2 above with fit2 in place of fit1.
2. Repeat parts (b) and (c) of question #2 above with fit3 in place of fit1.
3. Calculate the marginal effect of each predictor in fit4 for the average household in the dataset. Interpret your results. How do they compare to the “divide by 4” rule?
6. Predictive Performance of fit4
1. Repeat question #3 from above with fit4, p4, and pred4 in place of fit1, p1 and pred1.
2. Using your results from (a) and question #3 above, compare the in-sample predictive performance of fit1 and fit4. | 1,507 | 6,384 | {"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} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.954676 |
https://forum.freecodecamp.org/t/sum-all-primes-problem-with-big-numbers/227964 | 1,601,041,576,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00125.warc.gz | 396,537,792 | 6,845 | # Sum All Primes - Problem with big numbers
Tell us what’s happening:
I’m working on “Sum All Primes” Algorithm and have 2 working conditions for sumPrimes(10) but when it comes to sumPrimes(977) the result is way to high. I tried to go step by step over my code but have a trouble understanding what is wrong. Is my condition missing something?
``````
function sumPrimes(num) {
let primeArr = [2, 3, 5];
let sumPrimes = 0;
let condition = false;
// Build prime numbers arr
for(let i = 6; i <= num; i++) {
condition = !(i % 2 === 0 || i % 3 === 0);
if(condition) {
primeArr.push(i);
}
}
// Sum primeArr
for(let y = 0; y < primeArr.length; y++) {
sumPrimes += primeArr[y];
}
return sumPrimes;
}
sumPrimes(977);
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/69.0.3497.100 Safari/537.36`.
You are putting a number in `primeArr` if it is not divisible by 2 or 3. That is not the definition of a prime number. | 300 | 973 | {"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} | 3.015625 | 3 | CC-MAIN-2020-40 | latest | en | 0.68016 |
https://www.idtech.com/blog/minimax-algorithm-in-chess-checkers-tic-tac-toe | 1,726,267,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00151.warc.gz | 743,907,856 | 256,889 | # Minimax Algorithm in Chess, Checkers & Tic-Tac-Toe
Ryan Barone
June 14, 2022
## Minimax Algorithm Definition
The minimax algorithm is a decision rule that minimizes the maximum possible loss of a worst-case scenario for the player. The minimax algorithm is able to minimize potential loss by using positional evaluations to predict the opponent's next move. The results of a minimax algorithm are normally displayed on a tree diagram to represent each combination.
## Minimax Algorithm Process in Chess
### Step 1: Create the Tree Diagram
Start by creating a tree diagram that shows all possible moves. Each move should be drawn on an arrow to represent the player choosing that move, and it should point to a circle that represents the score of the combination that minimizes possible loss.
### Step 2: Evaluate the Final Positions
Next, at the end circles of the tree diagram, evaluate what the board score would be if the players chose that combination of moves.
### Step 3: Minimize the Potential Score
Since the player with black pieces has the last move, find the lowest score out of each group of moves with arrows from the same prior move, and place that score in the circle of the prior move to represent its worst-case scenario.
### Step 4: Maximize the Potential Score
With the player with white pieces having the previous move, use the same technique as you did before, except now finding the highest score to represent the best move.
### Step 5: Complete the Diagram
Repeat the process to find the final worst-case scenario. Now, by following the arrows to match scores, you can find the most ideal move combination by reading the move labels that are listed on the arrows for the ideal board score.
The ideal move combination for the example above is move #1 > move #2 > move #5 > move #11.
## Negamax Variation
In a typical minimax algorithm, you would need a function to evaluate the score for the original player and a separate function to evaluate the score for the opponent. However, in chess, the scores for both players are related because chess is a zero-sum game.
In a zero-sum game, the total sum of each outcome is always zero because, in order for a player to gain points, another player must lose the exact same number of points.
Due to the relation between points for players, the negamax algorithm is optimized by using the same function for the original player and the opponent, and only negating the total point value found when evaluating for the opponent.
### Meet iD Tech!
Sign up for our emails to learn more about why iD Tech is #1 in STEM education! Be the first to hear about new courses, locations, programs, and partnerships–plus receive exclusive promotions! AI summer camps, coding classes for kids, and more! | 580 | 2,774 | {"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} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.895771 |
https://www.physicsforums.com/threads/given-spherical-and-cylindarical-coordinates-draw-each-one.437881/ | 1,553,258,844,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202658.65/warc/CC-MAIN-20190322115048-20190322141048-00113.warc.gz | 860,082,728 | 11,753 | # Given spherical and cylindarical coordinates, draw each one
1. Oct 14, 2010
### sinned4789
1. The problem statement, all variables and given/known data
part 1) draw 3(R-hat) + 3(theta-hat) + 3(phi-hat)
part 2) draw 3(r-hat) + 3(theta-hat) + 3(z-hat)
2. Relevant equations
3. The attempt at a solution
aren't these the same thing? just 3 in each of the x, y, and z directions in cartesian coordinates.
2. Oct 14, 2010
### zhermes
Absolutely not. Look up the wiki pages on spherical and cylindrical coordinates. | 162 | 519 | {"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} | 2.671875 | 3 | CC-MAIN-2019-13 | latest | en | 0.754227 |
https://www.smore.com/spph1 | 1,531,974,063,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00225.warc.gz | 992,485,801 | 12,284 | The Buzz...from the Busy Bees
April 4-8
Hello Busy Bee Families,
It was great to see so many families at Family Fun Night on Friday. I know my kids slept great that night! I appreciate everyone who contributed to our class basket, to our game, and who helped make the night a success! We have such an amazing PTA!
I sent home the new April snack calendar last week. Please let me know if you need another copy. If your child's name isn't on the calendar, they will be one of the first kids to bring snack in May. Did I just write May? Wow! This year is flying by.
It's going to "bee" a great week!
Mrs. Michelle Slominsky
This week's reader's workshop lessons will focus on using non-fiction text features to read and comprehend a text. We'll be reading a book called Pushes and Pulls. We'll learn to look at the table of contents to find specific information in a text, read the headings to determine what the section will be about, and look at the captions to understand a photograph.
Writer's Workshop
I am so impressed with the Busy Bee's writing. Students are really thinking of great ways to begin their stories. This week we'll move on to what I call the "Meaty Middle." We'll learn that the middle of our stories need to:
1. tell the story in order (can use transition words)
2. include details to help the reader understand
Math Workshop
Kindergarten - This week we'll be learning how to use number bonds to show different addition and subtraction equations. Click here to watch a short explanation video. Your child will learn to determine the whole (sum) when shown two known parts and to determine the missing part (missing addend) when the whole and one part are known.
First Grade - This week we'll continue learning to tell time to the half hour. Your child will be able to show time in multiple ways. For example, they can:
1. Show/Tell time on an analog clock
2. Show/Tell time on a digital clock
3. Write the time - ex: 4:30 = half past 4, 5:00 = five o'clock
Content/Language Workshop
We'll begin our content/language workshop by spending time in our Makerspace on Monday afternoon. The kids will be given the challenge to make a cotton ball move without touching it. They can build anything they can imagine. This will be our introduction to our science unit on Force and Motion. Children will learn that things can move in may different ways (slide, roll, spin, etc.) and there are different forces (push, pull) that make things move. We'll also get to play a bit with friction by testing toy cars on ramps that are covered with different surfaces. We'll be reading The Three Little Pigs and thinking about the force that the wolf used to make the houses fall down. We'll end the week in Makerspace with kids building their own sturdy house that can't fall down. We'll test the strength of their houses against the wolf using a hairdryer as our force. It's going to be a super fun week of exploring and learning! | 663 | 2,952 | {"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} | 3.359375 | 3 | CC-MAIN-2018-30 | latest | en | 0.946647 |
https://infoanew.com/wikipedia/Games_behind | 1,606,601,102,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00591.warc.gz | 341,254,272 | 72,453 | Games behind - infoAnew
Set your curiosity free with richer, better connected information.
# Games behind
• ## Contents
In most North American sports, the phrase games behind or games back (often abbreviated GB) is a common way to reflect the gap between a leading team and another team in a sports league, conference, or division.
## Example
In the below standings from the 1994 Major League Baseball season, the Atlanta Braves are six games behind the Montreal Expos. Atlanta would have to win six games, and Montreal would have to lose six games, to tie for first. The leading team is zero games behind itself, and this is indicated in standings by a dash, not a zero.
NL East W L Pct. GB Home Road
Montreal Expos 74 40 0.649 32–20 42–20
Atlanta Braves 68 46 0.596 6 31–24 37–22
New York Mets 55 58 0.487 18½ 23–30 32–28
Philadelphia Phillies 54 61 0.470 20½ 34–26 20–35
Florida Marlins 51 64 0.443 23½ 25–34 26–30
## Computing games behind
Games behind is calculated by using either of the following formulas, in which Team A is a leading team, and Team B is a trailing team. Example math in this section uses the above standings, with Montreal as Team A and Atlanta as Team B.
${\displaystyle {\text{Games Behind}}={\frac {({\text{Team A's wins – Team A's losses}})-({\text{Team B's wins – Team B's losses}})}{2}}}$
${\displaystyle {\text{Games Behind}}={\frac {({\text{74 – 40}})-({\text{68 – 46}})}{2}}={\frac {\text{34 – 22}}{2}}={\frac {12}{2}}=6}$
Alternately:
${\displaystyle {\text{Games Behind}}={\frac {({\text{Team A's wins – Team B's wins}})+({\text{Team B's losses – Team A's losses}})}{2}}}$
${\displaystyle {\text{Games Behind}}={\frac {({\text{74 – 68}})+({\text{46 – 40}})}{2}}={\frac {\text{6 + 6}}{2}}={\frac {12}{2}}=6}$
Notes:
• It can alternately be said that Montreal is six games ahead of Atlanta.
• A games behind situation can change rapidly when two teams contesting for the lead play each other. For example, Atlanta could cut Montreal's lead in half (to three games) by sweeping a three-game head-to-head series.
• The leading team, in terms of games behind, is the team with the best won–loss difference. This is not always the team with the most wins. For example, a team with an 80–70 record (10 more wins than losses) would be one game behind a team with a 79–67 record (12 more wins than losses).
## Anomalies
The formulas implicitly treat any difference in the number of games played by the two teams as each unplayed game being "worth" 0.5 wins and 0.5 losses. This can lead to anomalies when teams have played an unequal number of games, especially during the early portion of a season.
• Two teams with different winning percentages may be tied in terms of games behind. For example, Team A at 6–4 would be tied with Team B at 4–2, in terms of games behind. However, Team B has the better winning percentage, at .667 compared to .600 for Team A.
• A team with a lower winning percentage may lead (in terms of games behind) a team with a higher winning percentage. For example, Team A at 6–4 would lead Team B at 2–1 by a half-game when calculating games behind. However, Team B has the better winning percentage at .667, compared to .600 for Team A.
An example of this occurred in May 2018, when the New York Yankees were 28–13 and the Boston Red Sox were 30–14. The games behind calculation had New York a half-game behind Boston; however, New York had the better winning percentage, at .683 compared to .682 for Boston.[1]
Also, on December 28, 2018, in the National Basketball Association, the 24–10 Milwaukee Bucks were a half-game behind the 26–11 Toronto Raptors. Yet the Bucks were ranked first in the Eastern Conference, their .706 win percentage superior to the Raptors’ .703.
Leagues generally use winning percentage to order teams, so in both of the above examples Team B would be considered to be in first place.
## Usage
The games behind calculation is often used in professional baseball and basketball, where tie games are not permitted. Standings for these sports appearing in print or online during a season usually will have teams ordered by winning percentages, with a "GB" column provided as a convenience to the reader. Games behind is used less often in American football, where ties are possible but relatively uncommon. Games behind is rarely used in ice hockey and soccer, where ties are traditionally common and standings points are typically used.
### Major League Baseball
Major League Baseball (MLB) defines games behind as "the average of the differences between the leading team wins and the trailing team wins, and the leading teams losses and the trailing team losses."[2] A games behind column almost always appears in MLB standings for each five-team division.
#### Wild card race
In the 1994 MLB season, the American League and National League each split into three divisions, and each added a wild card team to the playoffs. Following this change, it became common for the media to publish an additional set of standings for the wild card race. It included all teams from a league, with the exception of the division leaders, and games behind was calculated with respect to the team with the highest standing in the wild card race.
In the 2012 MLB season, both leagues added a second wild card team. Now, games behind in the wild card race is calculated with respect to the team with second highest standing in the wild card race. MLB's website distinguishes this statistic as wild card games behind, abbreviated WCGB. Assuming that teams are not tied for first place in the wild card race, this results in the team leading the wild card race being shown as some number of "games ahead" of the second place team, indicated by a plus sign ("+") in the standings. As an example, see the 2012 NL Wild Card standings, which shows the Atlanta Braves six games ahead of the St. Louis Cardinals.
### National Basketball Association
National Basketball Association (NBA) standings typically report games behind within each five-team division. However, it is not as closely followed as in baseball, because more teams qualify for the NBA playoffs, and the divisional statistics are not as important for playoff qualification. Sometimes, especially nearing the end of the regular season, games behind will be given with respect to the eighth position in the Eastern Conference and Western Conference, as the eighth position is the last to qualify for the playoffs in each conference.
### National Football League
National Football League (NFL) standings sometimes report games behind, although the statistic is not emphasized; winning percentage is used, computed from each team's win–loss–tie record. This is especially true since the introduction of the bye week in 1990, exacerbating differences in the number of games that teams have played at various points in time, and since tied games (while highly uncommon since the introduction of overtime in the regular season starting in 1974) do still occur on occasion. Games behind is omitted from standings on the NFL's website and is absent from most published standings.
### Other sports
The games behind statistic is eschewed in sports where tie games are traditionally common, such as the ice hockey and soccer. Leagues in these sports typically rank teams by awarding a certain number of points for each win or tie. In competitions where ties have been abolished (especially in hockey), points are still awarded for an "overtime loss" such that they are often (but not always) the same value as ties previously were, while in soccer the traditional value of two points for a win has been replaced by three points, while ties are still worth one point. These changes would make calculating a "games behind" statistic even more complicated compared to if it were to be used under the traditional system in which a tie was effectively worth a "half-win."
The Canadian Football League (CFL) also does not use games behind, and awards standings points. However, unlike leagues such as the National Hockey League (NHL) the CFL does not award points for overtime losses, although it experimented with such a system in the early 21st century. Also, unlike most other football codes which use a point system the CFL still uses the traditional values of two points for a win and one for a tie. Therefore, a tie in the CFL is still effectively worth a "half-win" as it always has been in Canadian football (and also as it has been in the NFL since 1972).
### Related usage
Teams are sometimes referred to as being over or under "five hundred", in comparison to a .500 winning percentage. The calculation for this is simple subtraction (there is no division by two). For example, a team at 55–50 is "five games over five hundred", while a team at 40–50 would be "ten games under five hundred". 'Above' and 'below' can be substituted for 'over' and 'under', respectively. | 2,050 | 8,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "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} | 3.78125 | 4 | CC-MAIN-2020-50 | latest | en | 0.879712 |
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Instructor: Lindsy Frazer
Dr. Frazer has taught several college level Science courses and has a master's degree in Human Biology and a PhD in Library and Information Science.
An independent variable is an important part of any scientific experiment. Learn what an independent variable is and explore some examples of independent variables in this lesson.
Independent Variables
Will white chocolate melt faster than dark chocolate? Experiments are used to answer questions like this. Most experiments are set up to see if a change in one factor causes a change in another factor. These factors that can be changed are called variables. There are three main types of variable: independent, dependent, and controlled.
The independent variable is the factor in an experiment that the experimenter changes. This variable is not influenced by any other factors in the experiment, but it causes change in other variables. In other words, the independent variable is the cause of change in an experiment. Variables that are changed by the independent variable are called dependent variables. A good experiment only has one independent variable so that the experimenter can be sure that changes to the dependent variable are caused by the independent variable.
Soapy Situation: Example 1
Which soap is the best at killing germs? To answer this question, think like a scientist and set up an experiment. You could wash your hands in warm water with soap A and use a microscope to count how many germs are left on your palms. Next, you could wash your hands in warm water with soap B and count the germs left.
What did you - the scientist - change each time you washed your hands? You changed the type of soap you used. So, type of soap is the independent variable. The goal of the experiment was to see if changes in the type of soap used causes changes in the amount of germs killed (the dependent variable).
Experiment: Example 2
Have you ever chewed a piece of mint flavored gum and felt like the inside of your mouth got colder? Do mints really make things colder or does it just feel that way? To answer this question with an experiment you could fill three glasses with water and place one mint in one glass, two mints in another glass, and no mints in the third glass. Then, using a thermometer, you could record and compare the temperature of the water in each glass.
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# Chapter Eighteen
Technology
Technologies
A technology is a process by which
inputs are converted to an output.
E.g. labor, a computer, a projector,
electricity, and software are being
combined to produce this lecture.
Technologies
Usually several technologies will
produce the same product -- a
blackboard and chalk can be used
instead of a computer and a
projector.
Which technology is best?
How do we compare technologies?
Input Bundles
x
i
denotes the amount used of input i;
i.e. the level of input i.
An input bundle is a vector of the
input levels; (x
1
, x
2
, , x
n
).
E.g. (x
1
, x
2
, x
3
) = (6, 0, 93).
Production Functions
y denotes the output level.
The technologys production
function states the maximum amount
of output possible from an input
bundle.
y f x x
n
= ( , , )
1
Production Functions
y = f(x) is the
production
function.
x x
Input Level
Output Level
y
y = f(x) is the maximal
output level obtainable
from x input units.
One input, one output
Technology Sets
A production plan is an input bundle
and an output level; (x
1
, , x
n
, y).
A production plan is feasible if
The collection of all feasible
production plans is the technology set.
y f x x
n
s ( , , )
1
Technology Sets
y = f(x) is the
production
function.
x x
Input Level
Output Level
y
y
y = f(x) is the maximal
output level obtainable
from x input units.
One input, one output
y = f(x) is an output level
that is feasible from x
input units.
Technology Sets
The technology set is
T x x y y f x x and
x x
n n
n
= s
> >
{( , , , ) | ( , , )
, , }.
1 1
1
0 0
Technology Sets
x x
Input Level
Output Level
y
One input, one output
y
The technology
set
Technology Sets
x x
Input Level
Output Level
y
One input, one output
y
The technology
set
Technically
inefficient
plans
Technically
efficient plans
Technologies with Multiple Inputs
What does a technology look like
when there is more than one input?
The two input case: Input levels are
x
1
and x
2
. Output level is y.
Suppose the production function is
y f x x x x = = ( , ) .
1 2 1
1/ 3
2
1/ 3
2
Technologies with Multiple Inputs
E.g. the maximal output level
possible from the input bundle
(x
1
, x
2
) = (1, 8) is
And the maximal output level
possible from (x
1
,x
2
) = (8,8) is
y x x = = = = 2 2 1 8 2 1 2 4
1
1/ 3
2
1/ 3 1/ 3 1/ 3
.
y x x = = = = 2 2 8 8 2 2 2 8
1
1/ 3
2
1/ 3 1/ 3 1/ 3
.
Technologies with Multiple Inputs
Output, y
x
1
x
2
(8,1)
(8,8)
Technologies with Multiple Inputs
The y output unit isoquant is the set
of all input bundles that yield at most
the same output level y.
Isoquants with Two Variable Inputs
y 8
y 4
x
1
x
2
Isoquants with Two Variable Inputs
Isoquants can be graphed by adding
an output level axis and displaying
each isoquant at the height of the
isoquants output level.
Isoquants with Two Variable Inputs
Output, y
x
1
x
2
y 8
y 4
Isoquants with Two Variable Inputs
More isoquants tell us more about
the technology.
Isoquants with Two Variable Inputs
y 8
y 4
x
1
x
2
y 6
y 2
Isoquants with Two Variable Inputs
Output, y
x
1
x
2
y 8
y 4
y 6
y 2
Technologies with Multiple Inputs
The complete collection of isoquants
is the isoquant map.
The isoquant map is equivalent to
the production function -- each is the
other.
E.g.
3 / 1
2
3 / 1
1 2 1
2 ) , ( x x x x f y = =
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
x
2
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Technologies with Multiple Inputs
x
1
y
Cobb-Douglas Technologies
A Cobb-Douglas production function
is of the form
E.g.
with
y Ax x x
a a
n
a
n
=
1 2
1 2
.
y x x =
1
1/ 3
2
1/ 3
n A a anda = = = = 2 1
1
3
1
3
1 2
, , .
x
2
x
1
All isoquants are hyperbolic,
asymptoting to, but never
touching any axis.
Cobb-Douglas Technologies
y x x
a a
=
1 2
1 2
x
2
x
1
All isoquants are hyperbolic,
asymptoting to, but never
touching any axis.
Cobb-Douglas Technologies
x x y
a a
1 2
1 2
= "
y x x
a a
=
1 2
1 2
x
2
x
1
All isoquants are hyperbolic,
asymptoting to, but never
touching any axis.
Cobb-Douglas Technologies
x x y
a a
1 2
1 2
= '
x x y
a a
1 2
1 2
= "
y x x
a a
=
1 2
1 2
x
2
x
1
All isoquants are hyperbolic,
asymptoting to, but never
touching any axis.
Cobb-Douglas Technologies
x x y
a a
1 2
1 2
= '
x x y
a a
1 2
1 2
= "
y" y'
>
y x x
a a
=
1 2
1 2
Fixed-Proportions Technologies
A fixed-proportions production
function is of the form
E.g.
with
y a x a x a x
n n
= min{ , , , }.
1 1 2 2
y x x = min{ , }
1 2
2
n a and a = = = 2 1 2
1 2
, .
Fixed-Proportions Technologies
x
2
x
1
min{x
1
,2x
2
} = 14
4 8 14
2
4
7
min{x
1
,2x
2
} = 8
min{x
1
,2x
2
} = 4
x
1
= 2x
2
y x x = min{ , }
1 2
2
Perfect-Substitutes Technologies
A perfect-substitutes production
function is of the form
E.g.
with
y a x a x a x
n n
= + + +
1 1 2 2
.
y x x = +
1 2
3
n a and a = = = 2 1 3
1 2
, .
Perfect-Substitution Technologies
9
3
18
6
24
8
x
1
x
2
x
1
+ 3x
2
= 18
x
1
+ 3x
2
= 36
x
1
+ 3x
2
= 48
All are linear and parallel
y x x = +
1 2
3
Marginal (Physical) Products
The marginal product of input i is the
rate-of-change of the output level as
the level of input i changes, holding
all other input levels fixed.
That is,
y f x x
n
= ( , , )
1
i
i
x
y
MP
Marginal (Physical) Products
E.g. if
y f x x x x = = ( , )
/
1 2 1
1/ 3
2
2 3
then the marginal product of input 1 is
Marginal (Physical) Products
E.g. if
y f x x x x = = ( , )
/
1 2 1
1/ 3
2
2 3
then the marginal product of input 1 is
MP
y
x
x x
1
1
1
2 3
2
2 3
1
3
= =
c
c
/ /
Marginal (Physical) Products
E.g. if
y f x x x x = = ( , )
/
1 2 1
1/ 3
2
2 3
then the marginal product of input 1 is
MP
y
x
x x
1
1
1
2 3
2
2 3
1
3
= =
c
c
/ /
and the marginal product of input 2 is
Marginal (Physical) Products
E.g. if
y f x x x x = = ( , )
/
1 2 1
1/ 3
2
2 3
then the marginal product of input 1 is
MP
y
x
x x
1
1
1
2 3
2
2 3
1
3
= =
c
c
/ /
and the marginal product of input 2 is
MP
y
x
x x
2
2
1
1/ 3
2
1/ 3
2
3
= =
c
c
.
Marginal (Physical) Products
Typically the marginal product of one
input depends upon the amount used of
other inputs. E.g. if
MP x x
1 1
2 3
2
2 3
1
3
=
/ /
then,
MP x x
1 1
2 3 2 3
1
2 3
1
3
8
4
3
= =
/ / /
and if x
2
= 27 then
if x
2
= 8,
MP x x
1 1
2 3 2 3
1
2 3
1
3
27 3 = =
/ / /
.
Marginal (Physical) Products
The marginal product of input i is
diminishing if it becomes smaller as
the level of input i increases. That is,
if
. 0
2
2
< =
|
|
.
|
\
|
=
i
i i i
i
x
y
x
y
x x
MP
c
c
c
c
c
c
c
c
Marginal (Physical) Products
MP x x
1 1
2 3
2
2 3
1
3
=
/ /
MP x x
2 1
1/ 3
2
1/ 3
2
3
=
and
E.g. if y x x =
1
1/ 3
2
2 3 /
then
Marginal (Physical) Products
MP x x
1 1
2 3
2
2 3
1
3
=
/ /
MP x x
2 1
1/ 3
2
1/ 3
2
3
=
and
so
c
c
MP
x
x x
1
1
1
5 3
2
2 3
2
9
0 = <
/ /
E.g. if y x x =
1
1/ 3
2
2 3 /
then
Marginal (Physical) Products
MP x x
1 1
2 3
2
2 3
1
3
=
/ /
MP x x
2 1
1/ 3
2
1/ 3
2
3
=
and
so
c
c
MP
x
x x
1
1
1
5 3
2
2 3
2
9
0 = <
/ /
c
c
MP
x
x x
2
2
1
1/ 3
2
4 3
2
9
0 = <
/
.
and
E.g. if y x x =
1
1/ 3
2
2 3 /
then
Marginal (Physical) Products
MP x x
1 1
2 3
2
2 3
1
3
=
/ /
MP x x
2 1
1/ 3
2
1/ 3
2
3
=
and
so
c
c
MP
x
x x
1
1
1
5 3
2
2 3
2
9
0 = <
/ /
c
c
MP
x
x x
2
2
1
1/ 3
2
4 3
2
9
0 = <
/
.
and
Both marginal products are diminishing.
E.g. if y x x =
1
1/ 3
2
2 3 /
then
Returns-to-Scale
Marginal products describe the
change in output level as a single
input level changes.
Returns-to-scale describes how the
output level changes as all input
levels change in direct proportion
(e.g. all input levels doubled, or
halved).
Returns-to-Scale
If, for any input bundle (x
1
,,x
n
),
f kx kx kx kf x x x
n n
( , , , ) ( , , , )
1 2 1 2
=
then the technology described by the
production function f exhibits constant
returns-to-scale.
E.g. (k = 2) doubling all input levels
doubles the output level.
Returns-to-Scale
y = f(x)
x x
Input Level
Output Level
y
One input, one output
2x
2y
Constant
returns-to-scale
Returns-to-Scale
If, for any input bundle (x
1
,,x
n
),
f kx kx kx kf x x x
n n
( , , , ) ( , , , )
1 2 1 2
<
then the technology exhibits diminishing
returns-to-scale.
E.g. (k = 2) doubling all input levels less
than doubles the output level.
Returns-to-Scale
y = f(x)
x x
Input Level
Output Level
f(x)
One input, one output
2x
f(2x)
2f(x)
Decreasing
returns-to-scale
Returns-to-Scale
If, for any input bundle (x
1
,,x
n
),
f kx kx kx kf x x x
n n
( , , , ) ( , , , )
1 2 1 2
>
then the technology exhibits increasing
returns-to-scale.
E.g. (k = 2) doubling all input levels
more than doubles the output level.
Returns-to-Scale
y = f(x)
x x
Input Level
Output Level
f(x)
One input, one output
2x
f(2x)
2f(x)
Increasing
returns-to-scale
Returns-to-Scale
A single technology can locally
exhibit different returns-to-scale.
Returns-to-Scale
y = f(x)
x
Input Level
Output Level
One input, one output
Decreasing
returns-to-scale
Increasing
returns-to-scale
Examples of Returns-to-Scale
y a x a x a x
n n
= + + +
1 1 2 2
.
The perfect-substitutes production
function is
Expand all input levels proportionately
by k. The output level becomes
a kx a kx a kx
n n 1 1 2 2
( ) ( ) ( ) + + +
Examples of Returns-to-Scale
y a x a x a x
n n
= + + +
1 1 2 2
.
The perfect-substitutes production
function is
Expand all input levels proportionately
by k. The output level becomes
a kx a kx a kx
k a x a x a x
n n
n n
1 1 2 2
1 1 2 2
( ) ( ) ( )
( )
+ + +
= + + +
Examples of Returns-to-Scale
y a x a x a x
n n
= + + +
1 1 2 2
.
The perfect-substitutes production
function is
Expand all input levels proportionately
by k. The output level becomes
a kx a kx a kx
k a x a x a x
ky
n n
n n
1 1 2 2
1 1 2 2
( ) ( ) ( )
( )
.
+ + +
= + + +
=
## The perfect-substitutes production
function exhibits constant returns-to-scale.
Examples of Returns-to-Scale
y a x a x a x
n n
= min{ , , , }.
1 1 2 2
The perfect-complements production
function is
Expand all input levels proportionately
by k. The output level becomes
min{ ( ), ( ), , ( )} a kx a kx a kx
n n 1 1 2 2
Examples of Returns-to-Scale
y a x a x a x
n n
= min{ , , , }.
1 1 2 2
The perfect-complements production
function is
Expand all input levels proportionately
by k. The output level becomes
min{ ( ), ( ), , ( )}
(min{ , , , })
a kx a kx a kx
k a x a x a x
n n
n n
1 1 2 2
1 1 2 2
=
Examples of Returns-to-Scale
y a x a x a x
n n
= min{ , , , }.
1 1 2 2
The perfect-complements production
function is
Expand all input levels proportionately
by k. The output level becomes
min{ ( ), ( ), , ( )}
(min{ , , , })
.
a kx a kx a kx
k a x a x a x
ky
n n
n n
1 1 2 2
1 1 2 2
=
=
The perfect-complements production
function exhibits constant returns-to-scale.
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
Expand all input levels proportionately
by k. The output level becomes
( ) ( ) ( ) kx kx kx
a a
n
a
n
1 2
1 2
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
Expand all input levels proportionately
by k. The output level becomes
( ) ( ) ( ) kx kx kx
k k k x x x
a a
n
a
a a a a a a
n
n n
1 2
1 2
1 2 1 2
=
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
Expand all input levels proportionately
by k. The output level becomes
( ) ( ) ( ) kx kx kx
k k k x x x
k x x x
a a
n
a
a a a a a a
a a a a a
n
a
n
n n
n n
1 2
1 2
1 2
1 2 1 2
1 2 1 2
=
=
+ + +
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
Expand all input levels proportionately
by k. The output level becomes
( ) ( ) ( )
.
kx kx kx
k k k x x x
k x x x
k y
a a
n
a
a a a a a a
a a a a a
n
a
a a
n
n n
n n
n
1 2
1 2
1 2
1 2 1 2
1 2 1 2
1
=
=
=
+ + +
+ +
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
( ) ( ) ( ) . kx kx kx k y
a a
n
a a a
n n
1 2
1 2 1
=
+ +
The Cobb-Douglas technologys returns-
to-scale is
constant if a
1
+ + a
n
= 1
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
( ) ( ) ( ) . kx kx kx k y
a a
n
a a a
n n
1 2
1 2 1
=
+ +
The Cobb-Douglas technologys returns-
to-scale is
constant if a
1
+ + a
n
= 1
increasing if a
1
+ + a
n
> 1
Examples of Returns-to-Scale
y x x x
a a
n
a
n
=
1 2
1 2
.
The Cobb-Douglas production function is
( ) ( ) ( ) . kx kx kx k y
a a
n
a a a
n n
1 2
1 2 1
=
+ +
The Cobb-Douglas technologys returns-
to-scale is
constant if a
1
+ + a
n
= 1
increasing if a
1
+ + a
n
> 1
decreasing if a
1
+ + a
n
< 1.
Returns-to-Scale
Q: Can a technology exhibit
increasing returns-to-scale even
though all of its marginal products
are diminishing?
Returns-to-Scale
Q: Can a technology exhibit
increasing returns-to-scale even if all
of its marginal products are
diminishing?
A: Yes.
E.g.
y x x =
1
2 3
2
2 3 / /
.
Returns-to-Scale
y x x x x
a a
= =
1
2 3
2
2 3
1 2
1 2
/ /
a a
1 2
4
3
1 + = >
so this technology exhibits
increasing returns-to-scale.
Returns-to-Scale
y x x x x
a a
= =
1
2 3
2
2 3
1 2
1 2
/ /
a a
1 2
4
3
1 + = >
so this technology exhibits
increasing returns-to-scale.
But
MP x x
1 1
1/ 3
2
2 3
2
3
=
/
diminishes as x
1
increases
Returns-to-Scale
y x x x x
a a
= =
1
2 3
2
2 3
1 2
1 2
/ /
a a
1 2
4
3
1 + = >
so this technology exhibits
increasing returns-to-scale.
But
MP x x
1 1
1/ 3
2
2 3
2
3
=
/
diminishes as x
1
increases and
MP x x
2 1
2 3
2
1/ 3
2
3
=
/
diminishes as x
1
increases.
Returns-to-Scale
So a technology can exhibit
increasing returns-to-scale even if all
of its marginal products are
diminishing. Why?
Returns-to-Scale
A marginal product is the rate-of-
change of output as one input level
increases, holding all other input
levels fixed.
Marginal product diminishes
because the other input levels are
fixed, so the increasing inputs units
have each less and less of other
inputs with which to work.
Returns-to-Scale
When all input levels are increased
proportionately, there need be no
diminution of marginal products
since each input will always have the
same amount of other inputs with
which to work. Input productivities
need not fall and so returns-to-scale
can be constant or increasing.
Technical Rate-of-Substitution
At what rate can a firm substitute one
input for another without changing
its output level?
Technical Rate-of-Substitution
x
2
x
1
y100
x
2
'
x
1
'
Technical Rate-of-Substitution
x
2
x
1
y100
The slope is the rate at which
input 2 must be given up as
input 1s level is increased so as
not to change the output level.
The slope of an isoquant is its
technical rate-of-substitution.
x
2
'
x
1
'
Technical Rate-of-Substitution
How is a technical rate-of-substitution
computed?
Technical Rate-of-Substitution
How is a technical rate-of-substitution
computed?
The production function is
A small change (dx
1
, dx
2
) in the input
bundle causes a change to the output
level of
y f x x = ( , ).
1 2
dy
y
x
dx
y
x
dx = +
c
c
c
c
1
1
2
2
.
Technical Rate-of-Substitution
dy
y
x
dx
y
x
dx = +
c
c
c
c
1
1
2
2
.
But dy = 0 since there is to be no change
to the output level, so the changes dx
1
and dx
2
to the input levels must satisfy
0
1
1
2
2
= +
c
c
c
c
y
x
dx
y
x
dx .
Technical Rate-of-Substitution
0
1
1
2
2
= +
c
c
c
c
y
x
dx
y
x
dx
rearranges to
c
c
c
c
y
x
dx
y
x
dx
2
2
1
1
=
so
dx
dx
y x
y x
2
1
1
2
=
c c
c c
/
/
.
Technical Rate-of-Substitution
dx
dx
y x
y x
2
1
1
2
=
c c
c c
/
/
is the rate at which input 2 must be given
up as input 1 increases so as to keep
the output level constant. It is the slope
of the isoquant.
Technical Rate-of-Substitution; A
Cobb-Douglas Example
y f x x x x
a b
= = ( , )
1 2 1 2
so c
c
y
x
ax x
a b
1
1
1
2
=
c
c
y
x
bx x
a b
2
1 2
1
=
.
and
The technical rate-of-substitution is
dx
dx
y x
y x
ax x
bx x
ax
bx
a b
a b
2
1
1
2
1
1
2
1 2
1
2
1
= = =
c c
c c
/
/
.
x
2
x
1
Technical Rate-of-Substitution; A
Cobb-Douglas Example
TRS
ax
bx
x
x
x
x
= = =
2
1
2
1
2
1
1 3
2 3 2
( / )
( / )
y x x a andb = = =
1
1/ 3
2
2 3
1
3
2
3
/
;
x
2
x
1
Technical Rate-of-Substitution; A
Cobb-Douglas Example
TRS
ax
bx
x
x
x
x
= = =
2
1
2
1
2
1
1 3
2 3 2
( / )
( / )
y x x a andb = = =
1
1/ 3
2
2 3
1
3
2
3
/
;
8
4
TRS
x
x
= =
=
2
1
2
8
2 4
1
x
2
x
1
Technical Rate-of-Substitution; A
Cobb-Douglas Example
TRS
ax
bx
x
x
x
x
= = =
2
1
2
1
2
1
1 3
2 3 2
( / )
( / )
y x x a andb = = =
1
1/ 3
2
2 3
1
3
2
3
/
;
6
12
TRS
x
x
= =
=
2
1
2
6
2 12
1
4
Well-Behaved Technologies
A well-behaved technology is
monotonic, and
convex.
Well-Behaved Technologies -
Monotonicity
Monotonicity: More of any input
generates more output.
y
x
y
x
monotonic
not
monotonic
Well-Behaved Technologies -
Convexity
Convexity: If the input bundles x
and x both provide y units of output
then the mixture tx + (1-t)x
provides at least y units of output,
for any 0 < t < 1.
Well-Behaved Technologies -
Convexity
x
2
x
1
x
2
'
x
1
'
x
2
"
x
1
"
y100
Well-Behaved Technologies -
Convexity
x
2
x
1
x
2
'
x
1
'
x
2
"
x
1
"
tx t x tx t x
1 1 2 2
1 1
' " ' "
( ) , ( ) + +
y100
Well-Behaved Technologies -
Convexity
x
2
x
1
x
2
'
x
1
'
x
2
"
x
1
"
tx t x tx t x
1 1 2 2
1 1
' " ' "
( ) , ( ) + +
y100
y120
Well-Behaved Technologies -
Convexity
x
2
x
1
x
2
'
x
1
'
x
2
"
x
1
"
Convexity implies that the TRS
increases (becomes less
negative) as x
1
increases.
Well-Behaved Technologies
x
2
x
1
y100
y50
y200
higher output
The Long-Run and the Short-Runs
The long-run is the circumstance in
which a firm is unrestricted in its
choice of all input levels.
There are many possible short-runs.
A short-run is a circumstance in
which a firm is restricted in some
way in its choice of at least one input
level.
The Long-Run and the Short-Runs
Examples of restrictions that place a
firm into a short-run:
temporarily being unable to install,
or remove, machinery
being required by law to meet
affirmative action quotas
having to meet domestic content
regulations.
The Long-Run and the Short-Runs
A useful way to think of the long-run
is that the firm can choose as it
circumstance to be.
The Long-Run and the Short-Runs
What do short-run restrictions imply
for a firms technology?
Suppose the short-run restriction is
fixing the level of input 2.
Input 2 is thus a fixed input in the
short-run. Input 1 remains variable.
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
2
x
1
y
The Long-Run and the Short-Runs
x
1
y
The Long-Run and the Short-Runs
x
1
y
The Long-Run and the Short-Runs
x
1
y
Four short-run production functions.
The Long-Run and the Short-Runs
y x x =
1
1/ 3
2
1/ 3
is the long-run production
function (both x
1
and x
2
are variable).
The short-run production function when
x
2
1 is
. x 1 x y
3 / 1
1
3 / 1 3 / 1
1
= =
The short-run production function when
x
2
10 is
. x 15 2 10 x y
3 / 1
1
3 / 1 3 / 1
1
= =
The Long-Run and the Short-Runs
x
1
y
Four short-run production functions.
3 / 1 3 / 1
1
10 x y =
3 / 1 3 / 1
1
5 x y =
3 / 1 3 / 1
1
2 x y =
3 / 1 3 / 1
1
1 x y = | 8,029 | 19,561 | {"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} | 3.390625 | 3 | CC-MAIN-2020-16 | latest | en | 0.716182 |
https://www.preservearticles.com/economics/write-brief-notes-on-demand-function-schedule-and-curve/2385 | 1,702,050,721,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00507.warc.gz | 1,008,979,635 | 19,176 | Demand for a good by an individual market as a whole is conventionally expressed in three alternative forms i.e. a demand function, a demand schedule and a demand curve.
1. Demand Function:
A demand function of an individual buyer is an algebraic form of expressing his demand behavior. In it, the quantity demanded period of time is expressed as a function of (that is, determined by) several variables. A demand function may be in a generalized form or a specific form. the latter case, the function describes the exact manner in which quant demanded is supposed to vary in response to a change in one or mo independent variables. Some typical examples of a demand function for good X are:
(i) Dx = f(Px, Y, T); and
(ii) Dx = 2000-10Px.
Here, Dx denotes quantity of good demanded, Px denotes the price of good Y represents income level of the consumer and T is a measure of his taste: and preferences.
2. Demand Schedule:
A demand schedule is a tabular form of describing the shifts in quantity demanded of a good in response to shifts in its price per unit, while all non-price determining variables are remains unchanged. A demand schedule has two columns, namely (i) price per unit of the good (Px), and. (ii) quantity demanded per period (Dx). The demand schedule is a set of pairs of values of values of Px and Dx. The first column records the hypothetical values of Px, and the second column records the corresponding quantity (Dx) which the consumer would decide to buy if faced by that price. | 326 | 1,509 | {"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-2023-50 | latest | en | 0.908419 |
https://maqsad.io/classes/class-10/math/partial-fractions/resolve%20into%20partial%20fractions-6-%20frac%20xexp%202%20(x+2)%20left(xexp%202%20+4%20right) | 1,686,132,011,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00649.warc.gz | 417,023,506 | 14,019 | Classes
Class 9Class 10First YearSecond Year
$Resolve into partial fractions.6. \frac{x^{2}}{(x+2)\left(x^{2}+4\right)}$
$Q.27 Partial fractions of \frac{7 x^{2}+25}{(x+3)(x+4)} will be of the form(a) \frac{A}{x+3}+\frac{B}{x+4} (b) 1+\frac{A}{x+3}+\frac{B}{x+4} (c) 7+\frac{A}{x+3}+\frac{B}{x+4} (d) none of these$
$Example 1: Resolve the fraction \frac{x^{3}-x^{2}+x+1}{x^{2}+5} into proper fraction.$
$Q.35 The fraction \frac{3}{x+1} is(b) improper fraction(a) proper fraction(d) irrational fraction$
$Resolve into partial fractions.5. \frac{3 x+7}{(x+3)\left(x^{2}+4\right)}$
6 . $\frac{x^{2}}{(x-2)(x-1)^{2}}$
$Q.28 Partial fractions of \frac{7 x+25}{(x+3)(x+4)} is of the form(a) \frac{A}{x+3}+\frac{B x}{x+4} (b) \frac{A}{x+3}+\frac{B}{x+4} (c) \frac{A x+B}{x+3}+\frac{C}{x+4} (d) \frac{A x+B}{(x+3)(x+4)} Lahore Board 2014$ | 384 | 837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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} | 4.1875 | 4 | CC-MAIN-2023-23 | longest | en | 0.354729 |
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Explore BrainMass
# Solve a homogenous 2nd order ODE.
Find y as a function of x if:
(x^2)(y'') + 19xy' +81y = x^2
y(1) = 9 y'(1) = -3
Hint: First assume that at least one solution to the corresponding homogeneous equation is of the form . You may have to use some other method to find the second solution to make a fundamental set of solutions. Then use one of the two methods to find a particular solution.
#### Solution Preview
This is a Cauchy-Euler equation. To solve that we let x=exp(z) and plug this into the equation. In general with this change of variable the equation:
x^2y''+ axy'+ by= f(x) becomes:
y''+ (a-1)y'+ by= f(exp(z)) ...
#### Solution Summary
A homogenous 2nd order ODE is solved.
\$2.19 | 209 | 727 | {"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-2016-50 | longest | en | 0.881061 |
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#### Resources tagged with Graphs similar to Power Up:
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### There are 50 results
Broad Topics > Sequences, Functions and Graphs > Graphs
### Power Up
##### Stage: 5 Challenge Level:
Show without recourse to any calculating aid that 7^{1/2} + 7^{1/3} + 7^{1/4} < 7 and 4^{1/2} + 4^{1/3} + 4^{1/4} > 4 . Sketch the graph of f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x
### Climbing
##### Stage: 5 Challenge Level:
Sketch the graphs of y = sin x and y = tan x and some straight lines. Prove some inequalities.
### Parabella
##### Stage: 5 Challenge Level:
This is a beautiful result involving a parabola and parallels.
### Graphic Biology
##### Stage: 5 Challenge Level:
Several graphs of the sort occurring commonly in biology are given. How many processes can you map to each graph?
### Lap Times
##### Stage: 4 Challenge Level:
Two cyclists, practising on a track, pass each other at the starting line and go at constant speeds... Can you find lap times that are such that the cyclists will meet exactly half way round the. . . .
### Electric Kettle
##### Stage: 4 Challenge Level:
Explore the relationship between resistance and temperature
### Which Is Cheaper?
##### Stage: 4 Challenge Level:
When I park my car in Mathstown, there are two car parks to choose from. Which car park should I use?
##### Stage: 4 Challenge Level:
Can you adjust the curve so the bead drops with near constant vertical velocity?
### Perpendicular Lines
##### Stage: 4 Challenge Level:
Position the lines so that they are perpendicular to each other. What can you say about the equations of perpendicular lines?
### Maths Filler 2
##### Stage: 4 Challenge Level:
Can you draw the height-time chart as this complicated vessel fills with water?
### Mathsjam Jars
##### Stage: 4 Challenge Level:
Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?
### Which Is Bigger?
##### Stage: 4 Challenge Level:
Which is bigger, n+10 or 2n+3? Can you find a good method of answering similar questions?
### Immersion
##### Stage: 4 Challenge Level:
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### How Many Solutions?
##### Stage: 5 Challenge Level:
Find all the solutions to the this equation.
### Equation Matcher
##### Stage: 5 Challenge Level:
Can you match these equations to these graphs?
### Bus Stop
##### Stage: 4 Challenge Level:
Two buses leave at the same time from two towns Shipton and Veston on the same long road, travelling towards each other. At each mile along the road are milestones. The buses' speeds are constant. . . .
### Cubics
##### Stage: 4 and 5 Challenge Level:
Knowing two of the equations find the equations of the 12 graphs of cubic functions making this pattern.
### Parabolas Again
##### Stage: 4 and 5 Challenge Level:
Here is a pattern composed of the graphs of 14 parabolas. Can you find their equations?
### Motion Sensor
##### Stage: 4 Challenge Level:
Looking at the graph - when was the person moving fastest? Slowest?
### Parabolic Patterns
##### Stage: 4 and 5 Challenge Level:
The illustration shows the graphs of fifteen functions. Two of them have equations y=x^2 and y=-(x-4)^2. Find the equations of all the other graphs.
### Ellipses
##### Stage: 4 and 5 Challenge Level:
Here is a pattern for you to experiment with using graph drawing software. Find the equations of the graphs in the pattern.
### More Parabolic Patterns
##### Stage: 4 and 5 Challenge Level:
The illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs.
### After Thought
##### Stage: 5 Challenge Level:
Which is larger cos(sin x) or sin(cos x) ? Does this depend on x ?
### Three Ways
##### Stage: 5 Challenge Level:
If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.
##### Stage: 4 Challenge Level:
Four vehicles travel along a road one afternoon. Can you make sense of the graphs showing their motion?
### Steve's Mapping
##### Stage: 5 Challenge Level:
Steve has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
##### Stage: 4 Challenge Level:
Explore the two quadratic functions and find out how their graphs are related.
### Real(ly) Numbers
##### Stage: 5 Challenge Level:
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
##### Stage: 4 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
### Alison's Mapping
##### Stage: 4 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
##### Stage: 4 Challenge Level:
Four vehicles travelled on a road. What can you deduce from the times that they met?
### Real-life Equations
##### Stage: 5 Challenge Level:
Here are several equations from real life. Can you work out which measurements are possible from each equation?
### Cubic Spin
##### Stage: 5 Challenge Level:
Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?
### Sangaku
##### Stage: 5 Challenge Level:
The square ABCD is split into three triangles by the lines BP and CP. Find the radii of the three inscribed circles to these triangles as P moves on AD.
### Without Calculus
##### Stage: 5 Challenge Level:
Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.
### Matchless
##### Stage: 4 Challenge Level:
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### Golden Construction
##### Stage: 5 Challenge Level:
Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio.
### What's That Graph?
##### Stage: 4 Challenge Level:
Can you work out which processes are represented by the graphs?
##### Stage: 4 Challenge Level:
Here are some more quadratic functions to explore. How are their graphs related?
### Interpolating Polynomials
##### Stage: 5 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.
### Graphical Interpretation
##### Stage: 4 Challenge Level:
This set of resources for teachers offers interactive environments to support work on graphical interpretation at Key Stage 4.
### Surprising Transformations
##### Stage: 4 Challenge Level:
I took the graph y=4x+7 and performed four transformations. Can you find the order in which I could have carried out the transformations?
### Curve Match
##### Stage: 5 Challenge Level:
Which curve is which, and how would you plan a route to pass between them?
### Gosh Cosh
##### Stage: 5 Challenge Level:
Explore the hyperbolic functions sinh and cosh using what you know about the exponential function.
### Small Steps
##### Stage: 5 Challenge Level:
Two problems about infinite processes where smaller and smaller steps are taken and you have to discover what happens in the limit.
### Exponential Trend
##### Stage: 5 Challenge Level:
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Bio Graphs
##### Stage: 4 Challenge Level:
What biological growth processes can you fit to these graphs?
### Curve Fitter
##### Stage: 5 Challenge Level:
Can you fit a cubic equation to this graph?
### Guess the Function
##### Stage: 5 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together. | 1,871 | 8,005 | {"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-2017-34 | latest | en | 0.848149 |
https://tag.wonderhowto.com/summing/ | 1,550,679,083,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00447.warc.gz | 709,028,922 | 29,406 | # Summing
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https://github.com/yasith/Problems/commit/68d73485f6939b2f9034fa5287b125f3e790f0ec | 1,526,978,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00298.warc.gz | 584,315,720 | 14,165 | # yasith/Problems
Solved Ananagrams from UVa
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@@ -0,0 +1,83 @@ +// http://uva.onlinejudge.org/external/1/156.html + +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include +#include + +#define p(x) cout << #x << ":" << x << endl; + +using namespace std; + +string gen_key(string s){ + for (int i = 0; i < s.size(); i++) { + if(s[i] < 'a'){ + s[i] += 'a' - 'A'; + } + } + + char *str = new char [s.size() + 1]; + strcpy(str, s.c_str()); + + sort(str, str+s.size()); + + string sorted (str); + return sorted; +} + +int main(){ + + map > m; + + string s; + cin >> s; + + do{ + string key = gen_key(s); + + if(!m.count(key)){ + vector v; + m[key] = v; + } + m[key].push_back(s); + + cin >> s; + + }while(s.compare("#") != 0); + + vector v; + + map >::iterator it; + + for(it=m.begin(); it != m.end(); it++){ + if((*it).second.size() <= 1){ + v.push_back((*it).second[0]); + } + } + + sort(v.begin(), v.end()); + + for (int i = 0; i < v.size(); i++) { + cout << v[i] << endl; + } + + return 0; +} + +
#### 0 comments on commit `68d7348`
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https://community.openhab.org/t/example-convert-color-item-values-to-rgb-with-explanation/24014/19 | 1,531,777,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00019.warc.gz | 631,995,865 | 11,257 | Example: Convert Color Item Values To RGB (With Explanation)
Tags: #<Tag:0x00007fd316bcb600> #<Tag:0x00007fd316bcb2b8> #<Tag:0x00007fd316bcb150>
(SiLeX) #1
Because there are regular questions on how to read the Color Items’ values as 24-bit RGB ranging from 0-255 I just wrote this little explanation how and why it works. (Or at least one of the many ways)
Let’s say you have a `Color` item named `aacc_color`. With this rule you can read out the R, G and B values from it:
``````rule "get RGB Colors from ColorItem"
when
then
val red = (receivedCommand as HSBType).red * 2.55
val green = (receivedCommand as HSBType).green * 2.55
val blue = (receivedCommand as HSBType).blue * 2.55
end
``````
`aacc_color` is the name of my Color item.
`aacc_color.state` contains the State object of the Color item. Not really useful for us. So we need to convert it to something that really represents colors. In OpenHAB and OpenHAB2 this is the HSBType, which essentially is Hue, Saturation and Brightness.
`(aacc_color.state as HSBType)` is our Color item. Now it’s not an OpenHAB State object anymore, but a useful color type. If we are interested in the RED, GREEN and BLUE values, we need another step.
`(aacc_color.state as HSBType).red` contains the percentage of red color. This is returned as a PercentType. Yet, I want to have it as a number, ranging from 0 to 255 to control my lights. This can be done through simple maths.
`(aacc_color.state as HSBType).red * 2.55` represents your amount of red between 0 (no red at all) and 255 (full brightness red).
Now you can take your variables and sent commands to other items that may be your RGB light bulbs, or in my case, my DIY MySensors 2,4GHz remote controlled china LED RGB Strips:
``````rule "Set RGB LEDs from Color item using ColorPicker"
when
then
val red = (receivedCommand as HSBType).red * 2.55
val green = (receivedCommand as HSBType).green * 2.55
val blue = (receivedCommand as HSBType).blue * 2.55
aacc_color_r.sendCommand(red)
aacc_color_g.sendCommand(green)
aacc_color_b.sendCommand(blue)
end
``````
Brotip: While using any editor of your choice is fine for writing simple rules, you really should consider using the Eclipse SmartHome IDE for more complicated things like casting Item states to different types. Errors will show up instantly and you have integrated JavaDoc documenatation. This will save you a lot of time and hair.
Bonus content
You can also make use of your own new color-handling code to have functioning ON and OFF buttons near the colorpicker:
``````rule "Luminium Color Item"
when
then
{
val red = (receivedCommand as HSBType).red * 2.55
val green = (receivedCommand as HSBType).green * 2.55
val blue = (receivedCommand as HSBType).blue * 2.55
aacc_color_r.sendCommand(red)
aacc_color_g.sendCommand(green)
aacc_color_b.sendCommand(blue)
}
aacc_color.sendCommand(new HSBType("23,60,100"))
}
aacc_mode.sendCommand(0) // Fade-out lights very slowly. Implemented in the device code
}
end
``````
How to convert RGB to HSB and vice versa?
OpenHAB2 Hue changing rule
Set startsettings or change current settings for Items (Color, brightness) with command
Dimmer item and parsing of its value
[Question] SendCommand only if update is significantly different?
Struggling with setting Fibaro RGBW from Rule
How can I compare a color item with a value
Set color for HM-LC-RGBW-WM
Rule String Array to int Array
OpenHab 2 Homekit on iOS Device
(SiHui) #2
That is the first time I did understand all this HSBType stuff , thanks for that.
Maybe you should copy that to:
https://community.openhab.org/c/tutorials-examples
(Cody) #3
This is very much appreciated! I just happened to start a project with an H801 and rgb strip a few days ago. Although I got the general gist of how things work through browsing several examples, this really helped clear things up.
A couple questions. The slider below the color picker, what exactly does it control? In playing with it it seemed to have done anything from change the brightness, to completely change the color. Any examples for how to implement it?
Also, is it just me, or is incredibly difficult to get accurate colors using the color picker? Anything in much from the very edge ends up being a bluish white. “In between” colors barely work. No matter how precisely I try to be, I am unable to get a true red using the color picker.
(SiLeX) #5
If you are talking about that picture I posted, that is from the Basic UI. It only controls the Brightness channel. But this can be chosen by the UI that displays the ColorPicker.
( ) #6
Why link it? Just move the article
Great article! Thanks!
(Cody) #7
Thanks for the clarification. So something like this should work in the rule?
``````var brightness = (aacc_color.state as HSBType).brightness.intValue
``````
(Joe Lawrence) #8
Hopefully some of you color Gurus can help me with a color problem I am having with this device -> EZMultiPli.
In the manual it is first stated, "LED Color EZMultiPli has a color LED beneath the motion sensor dome. This LED is a convenient night light or an indicator light. The LED can be set to any of eight colors via the Z-Wave Color Switch Command Class
:
1)BLACK (off)
2)WHITE
3)RED
4)GREEN
5)BLUE
6)AQUA
7)PINK
8)YELLOW
A Z-Wave BASIC SET ON/OFF also controls the LED but it only uses the white color. Use the Color Switch Command
Class to control the color of the LED."
Then later we have,
Color Switch Command Class
The color of the LED is controlled with the Color Switch Command Class.
``````Name Value (HEX) Description
Command Class 33 COMMAND_CLASS_COLOR_CONTROL
Command 03 STATE_SET
Length 3 Number of ID/State pairs in this command
Capability ID 02 RED
State 00 or FF 00=OFF, 01-FF=ON
Capability ID 03 GREEN
State 00 or FF 00=OFF, 01-FF=ON
Capability ID 04 BLUE
State 00 or FF 00=OFF, 01-FF=ON
Dimming Duration Not used
``````
There are 3 LEDs inside the PIR sensor dome, RED, GREEN and BLUE. Each of the LEDs can be turned either ON or OFF to yield eight different colors. Dimming is not supported. A single command can set the value of all three LEDs or
each LED can be controlled in a separate command.
Note that if the pushbutton is pressed, the color setting is lost because the sensor will enter motion testing mode where
the LED turns on white anytime motion is detected for 5 minutes. Setting the color however cancels motion testing mode and the color will remain.
A BASIC_SET command turns the LED on white or off
With openHAB 1, I could turn the led on with a sendCommand(device, ON). Never tried to work with the color.
Paper UI has also created a channel similar to: “I have a channel created zwave:device:xxxxxxxx:node19:color_color”
So folks, what command can I send to turn these lights on and off, as I have searched for WEEKS, tried many things, but, no light.
(I know I’m going smack myself in the head when I get an explanation)
Problems commanding color of EzMultiPli
Using ColorItem with MQTT and the Tasmota Firmware
(SiLeX) #9
Because the HSBType inherits the intValue method from DecimalType (see
here:
this should work, yes.
Go and try to type it in the Eclipse SmartHome Designer if you feel unsure
about which methods you may use.
So then you have the brightness in a variable. Just out of interest: What
are you going to do with it?
Cody bot@community.openhab.org schrieb am Di., 28. Feb. 2017 um 15:10 Uhr:
(Joe Lawrence) #11
I also see the following in my logs:
``````2017-03-05 17:11:04.245 [INFO ] [.commandclass.ZWaveColorCommandClass] - NODE 19: Color report RED 0
2017-03-05 17:11:04.277 [INFO ] [.commandclass.ZWaveColorCommandClass] - NODE 19: Color report BLUE 0
2017-03-05 17:11:04.308 [INFO ] [.commandclass.ZWaveColorCommandClass] - NODE 19: Color report GREEN 0
2017-03-05 17:11:04.309 [INFO ] [.commandclass.ZWaveColorCommandClass] - NODE 19: Color report finished {RED=0, BLUE=0, GREEN=0}
``````
(SiLeX) #12
Hey friends! I found a rather big issue in my technique and ran into trouble today.
As you can see in the code, I am using `(aacc_color.state as HSBType).red` to get the red amount in the Color item. If a another rule (or the RGB device itself) posts an update to the Item, you will have wrong (the new) values in `.red` instead of the values you requested.
So `(aacc_color.state as HSBType)` should really be just `receivedCommand`. We don’t even need to cast, because the command already is an HSBType.
I have updates my examples in my original post.
(Guillermo Ruffino) #13
I am sending an RGB value in hex format to an Arduino running MySensors and controlling itself a RGB strip with neo pixel, the rule ended up looking like this:
``````rule "LEDColorChanging"
then Item LEDColor changed
then
hsbValue = LEDColor.state as HSBType
redValue = String.format("%02X", (hsbValue.red.floatValue * 2.55) as int)
greenValue = String.format("%02X", (hsbValue.green.floatValue * 2.55) as int)
blueValue = String.format("%02X", (hsbValue.blue.floatValue * 2.55) as int)
sendCommand(Text_Text, redValue + greenValue + blueValue)
logInfo("LED Color changed", "R" + redValue + "G" + greenValue + "B" + blueValue)
end
``````
This ended logging something like: `LED Color changed: RA8G03BFF`
(Kees Van Gelder) #14
Clear, thank you.
just a question to see if i fully understood:
Is receivedCommand the substitution of (aac_color.state as HSBType)?
As the one is used in your “Set RGB LEDs from color item Using Colorpicker” rule and the other in your
"Luminium Farbe Setzen" rule for the same purpose
(SiLeX) #15
command” events.
(Kees Van Gelder) #16
I had to think about that a little but now I get it. Tnx
(Ilkka Tengvall) #17
As a thank you to these guides, I share an example of Arduino code which will implement this logic behind MQTT and MySensors network. Have a look if you need sample to build devices to do the RGB.
(Ilkka Tengvall) #18
I was requested for video of the LED effects done by @maghac. Here you can see the effects as of today: https://youtu.be/chN6CxMVWm4
There is also slider for lightbulb brightness I forgot to show. All things can be controlled over MQTT. This is about the link in previous post. | 2,745 | 10,626 | {"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} | 2.703125 | 3 | CC-MAIN-2018-30 | latest | en | 0.899426 |
https://www.jiskha.com/display.cgi?id=1266546621 | 1,516,145,194,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00310.warc.gz | 929,593,717 | 3,704 | # Physics
posted by .
a truck drives up a hill with a 15 degree incline. if the truck has a constant speed of 22 m/s, what are the horizontal and vertical components of the trucks velocity?
• Physics -
horizontal velocity = 22sin(15)
vertical velocity = 22cos(15)
• Physics -
thank you so so so so much !
• Physics -
Kayla: MrPhysics erred, and got the forces reversed.
Horizontal is the cosine function
Vertical is the sine function.
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More Similar Questions | 603 | 2,353 | {"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 | 0.883646 |
https://www.hackerearth.com/problem/algorithm/house-travelling-problem/ | 1,529,892,215,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00425.warc.gz | 839,738,541 | 20,698 | House Travelling Problem
Tag(s):
Algorithms, Easy-Medium, Greedy, Minimum Spanning Tree
Problem
Editorial
Analytics
Bob is about to visit N of his friends' houses. Each house is depicted as a point $(x_i , y_i)$ on the $X-Y$ plane. The length of roads between two houses i and j is defined as $F(i, j) = (x_i - x_j)^2 + (y_i - y_j)^2$.
There's a $N \times N$ matrix M filled with 1s and 0s. $M[i][j] = 1$ denotes that it is possible to travel to $i^{th}$ House from $j^{th}$ House and vice versa. If $M[i][j] = 0$, it is not possible to travel to $i^{th}$ House from $j^{th}$ House and vice versa.
Now, Bob wants to visit all the houses, but the sum of length of all roads traveled should be maximum. There's one more restriction, he can use a maximum of $(N - 1)$ roads. Before starting the trip, he would like to know what is the maximum distance he can travel which satisfies all the given conditions. If it is not possible to visit all the houses, print out -1.
Input Format:
The first line contains N, the number of houses Bob wants to visit.
N lines follow, each contains two space separated integers $x_i$ and $y_i$ which denotes the of the $i^{th}$ house on the $X-Y$ plane.
Again, N lines follow, each contains N space separated integers, each being 0 or 1, denoting the matrix M.
Output Format:
Print the required answer in one line.
Input Constraints:
$1 \le N \le 3000$
$-10^6 \le x_i, y_i \le 10^6$
Note:
One road can be traversed any number of times , but its contribution would be taken only once. Also, there's large input data, please use faster i/o methods.
SAMPLE INPUT
3
1 2
4 5
8 8
1 0 1
0 1 1
1 1 1
SAMPLE OUTPUT
110
Explanation
This is the image for the given sample. Every house also has a road of length 0 to itself. We can use at max 2 roads which would be from (1,2) -> (8,8) -> (4,5). The total cost is 85 + 25 = 110.
Time Limit: 2.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Kotlin, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Swift-4.1, Visual Basic
CODE EDITOR
Initializing Code Editor... | 704 | 2,360 | {"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} | 3.15625 | 3 | CC-MAIN-2018-26 | latest | en | 0.894392 |
http://maths.gopract.com/Maths/Right-Triangle-Calc.aspx | 1,680,435,585,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00011.warc.gz | 32,515,945 | 5,437 | # Calculator For Right Triangle
Calculation Option
Provide a (side), b (side) Provide a (side), c (side) Side (a) : Side (b) :
Side Length
a=
Side Length
b=
Side Length
c=
Perimeter
P=
Semi Perimeter
s=
Area
K=
Altitude of side a
ha=
Altitude of side b
hb=
Altitude of side c
hc=
A=Angle A
a=Side a
B=Angle B
b=Side b
C=Angle C
c=Side c
K=Area
s=SemiPerimeter
P=Perimeter
ha=Altitude of a
hb=Altitude of b
hc=Altitude of c
### What Is Right Triangle?
The Triangle in which one of the interior angle is 90° is called a Right Angle Triangle. The longest side of the right triangle, which is also the side opposite the right angle, is the hypotenuse and the two arms of the right angle are the Height and the Base . This Right Angle Triangle help you to find 7 Unknown variables by providing 2 known sides only.
### What Are Calculation Option Available in Right Triangle?
Select which value you are Providing (Side Length(a) , Side Length Base(b) , Side Length Hypotenuse(c) ) input value and Click on Calculate .
The Triangle in which one of the interior angle is 90° is called a Right Angle Triangle. The longest side of the right triangle, which is also the side opposite the right angle, is the hypotenuse and the two arms of the right angle are the Height and the Base .We can get the value of nine Unknown variables only if value of two variables are known . Using following formula -:
Side of Right Triangle :
c2 = a2 + b2
Where
c = side
a = side
b = side
Perimeter of Right Triangle :
p = a + b + c
Where
P = Perimeter
c = side
a = side
b = side
Semiperimeter of Right Triangle :
s = (a + b + c)/ 2
Where
s = semiperimeter
c = side
a = side
b = side
Area of Right Triangle :
K = (a b )/ 2
Where
K = Area
a = side
b = side
Altitude of a :
ha = b
Where
ha = Altitude of a
b = side
Altitude of b :
hb= a
Where
hb = Altitude of b
a = side
Altitude of c :
hc = (ab)/2
Where
hc = Altitude of c
b = side
a = side
### How to use Right Triangle Calculator?
Step by step procedure for Right Triangle Calculator is as follows.
Step 1: Select which value you will be providing? Dropdown has Side (a) and Side (b) , Side (a) and Side (c) options.
Step 2: Input appropriate value as per selected type in Step 1.
Step 3: Now click on "Calculate" button to get result.
You will get Side (a) , Side (b) , Side (c) , Area (K) , Perimeter (P) , SemiPerimeter (s) , Altitude of Side Length (ha) , Altitude of Base (hb) , Altitude of Hypotenuse (hc) as per your selection and value input. | 725 | 2,503 | {"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} | 3.796875 | 4 | CC-MAIN-2023-14 | latest | en | 0.844612 |
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What is a fraction whose numerator is equal to or greater than the denominator?
Updated: 4/28/2022
Wiki User
8y ago
An improper fraction has a numerator greater than the denominator.
When the numerator and denominator are equal, that's called "1."
Wiki User
8y ago
Wiki User
9y ago
It is an improper fraction.
Thirdy Dizon
Lvl 1
3y ago
hahahahaahahahaha
Thirdy Dizon
Lvl 1
3y ago
you dont belive in destiny | 121 | 423 | {"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} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.882196 |
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EML4450L8
EML4450L8 - Sustainable Energy Science and Engineering...
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S ustainable E nergy S cience and E ngineering C enter Thermodynamics Fundamentals for Energy Conversion Systems Renewable Energy Applications The study of the laws that govern the conversion of energy from one form to the other
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S ustainable E nergy S cience and E ngineering C enter Energy Conversion Concerned with the transformation of energy from sources such as fossil fuel and radiation from Sun into conveniently used forms such as electrical energy, propulsive energy, heating and cooling. Forms of energy: Kinetic, potential, thermal, chemical, electromagnetic etc. Thermodynamics is the study which seeks to establish quantitative relationships among macroscopic variables (like pressure, temperature, molecular concentrations etc.) which describe an arbitrary physical system (system being very large compared with atomic dimensions) in an equilibrium state. Fuel Air Products of combustion Waste heat to cooling tower ~ Electrical energy (Work) Objective: Convert the availability of the fuel into work in the most efficient manner, taking into consideration cost, size, safety and environmental concerns. Power plant
S ustainable E nergy S cience and E ngineering C enter Energy Sources and Conversion Processes
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S ustainable E nergy S cience and E ngineering C enter Energy Conversion Technologies
S ustainable E nergy S cience and E ngineering C enter Laws of Thermodynamics The Zeroth Law of Thermodynamics: If two systems are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. This law is the basis of temperature measurement . First Law of Thermodynamics: The change in internal energy of a closed system is equals to the heat added to the system (or absorbed from the environment) minus the work done by the system (or on the environment). This law is a consequence of conservation of energy. While attempting to transform heat into work with full efficacy, we quickly learned that always some heat would escape into the surrounding environment as wasted energy (recall that energy can not be destroyed). This wasted energy can never be fully converted into anything useful. Second Law of Thermodynamics: It is impossible to construct an engine which, operating in a cycle, will produce no other effect than the extraction of heat from a single heat reservoir and the performance of an equal amount of work. It imposes a limitation on energy transformations other than that imposed by the first law.
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S ustainable E nergy S cience and E ngineering C enter The second law states that heat flows naturally from regions of higher temperature to regions of lower temperature, but that it will not flow naturally the other way. Heat can be made to flow from a colder region to a hotter region, which is exactly what happens in an air conditioner, but heat only does this when it is forced. On the other hand, heat flows from hot to cold spontaneously.
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https://www.jobtestprep.com/electrician-apprentice-exam | 1,585,763,396,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00304.warc.gz | 989,118,440 | 20,050 | Electrician Apprenticeship Test Preparation - JobTestPrep
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• 23 Numerical reasoning tests
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Electrician Apprenticeship Exam Application Requirements
The test which is administered via NECA-IBEW Apprenticeship, has a number of requirements, which are important to bear in mind:
1. The minimum age to take the test is 18.
2. High school algebra course with a passing grade.
3. Either high school diploma, GED or Associate Degree of minimum two years. The above requirements may be waved in the event that the applicant has enough documented work experience.
Electrician Apprenticeship Assessment Test Detail Overview
Let’s take a brief look into the content of the exams various segments.
Math Segment
• What’s Included: This segment will include math questions on basic mechanical theory, pullies, and balance of loads which have been hoisted into the air.
• Time and Questions: 33 questions in 46 minutes.
• What’s Included: You will be presented a number of written passages followed by questions from which you have to dissect and draw the right conclusions.
• Time and Questions: 36 questions in 51 minutes.
Electrician Assessment Test Rules
It is important to follow nto rules below; otherewise - you can get disqualified from the test.
1. Use of calculators
2. Arriving less than 15 minutes prior to test time
3. Failing to present an State or Federal ID
4. Trying to re-take the test before the end of the mandated 6 month waiting time.
Sample Questions with Explanations
A light bulb with the following information is given: 10V, 4W.
Mechanical Reasoning - Electricity
What is the resistance (Ohm) of the light bulb? (in ohms):
Explanation:
In order to calculate the resistance we will use the following formulas:
$$V=R \times I$$
$$P=V \times I$$
Where:
R = Resistance [Ohm]
I = Current [A]
V = Voltage [Volt]
P = Power [Watt]
We will isolate R using the formulas above:
$$R=\frac{V}{I}$$
and:
$$I = \frac{P}{V}$$
Therefore:
$$R = \frac{V}{\frac{P}{V}} = \frac{V^2}{P} = \frac{10^2}{4} = 25\ Ohm$$
Please note, if your answer was incorrect it is it possible that you used the equation incorrectly. Some of the distracters are counting on these mistakes, for example: R=I/V will lead you to option 3 (not resistance and therefore not in ohm units). In a similar way, R=I*V will lead to option 2 (also, not resistance and not in ohm units).
Mechanical Reasoning - cogwheels
The diagram shows two fixed cogwheels which can only rotate around their own axis. The red cogwheel has 24 teeth and the grey cogwheel has 12 teeth.
A toothed belt is inserted between the two cogwheels and is moved in the direction shown by the arrow.
In which direction and velocity of revolution will the grey cogwheel rotate in comparison to the red cogwheel?
Explanation:
When contact is made between the toothed belt and the cogwheels a conversion from a linear velocity to an angular velocity occurs. Thus, the location of the point of contact is critical.
We can see from the diagram that the point of contact between the red cogwheel and the toothed belt is at the lower part of the red cogwheel, a position in which the linear velocity induces an angular velocity in a clockwise direction.
In contrast, the point of contact between the grey cogwheel and the toothed belt is at the upper part of the grey cogwheel, a position in which the linear velocity induces an angular velocity in a counter clockwise direction. It follows that the cogwheels rotate in opposite directions.
The angular velocity can be calculated using the thumb rule: "fewer teeth, faster velocity". Due to the inversely proportional ratio between the number of teeth and the velocity we are able to assume that since the grey cogwheel has half the number of teeth, its velocity must be double the velocity of the red cogwheel.
The minimum score to even be considered for an interview (we will talk about that below) is a scoring of 4. The score will be from 1-9.
Electrician Apprenticeship Overview
What will an electrician apprenticeship look like? Here is an approximation of topics covered and the divide between classroom and real-work experience:
• Classroom – Classes will include 575-1,000 hours, but this can vary according to the field.
• Field training – On site training should fall between 8,000 to 10,000 hours.
Course topics according to various fields
• Inside Wiremen – To be a inside wireman you will need first pass 5 years worth of courses, including 83 specific knowledge areas. From Splicing High Voltage Cables to identifying hazardous material, or Reading a Wire Tables, this in-classroom courses will give you everything you need to know to do the job right.
• Outside Lineman – These series of courses will include 77 areas of expertise, covering a wide range of theory and practical training. You will learn how to use blueprints and know how to read its symbols , operate a bucket truck, gain comprehensive understanding of delta, and wye transformer connections, learn emergency rescue, and much more.
• Residential Wiremen – For this position you will need to know all about Terminating Coaxial Cables, Aluminum or Copper Cables, and Twisted Pair Cables along with the other 85 areas of expertise related to the field. Moreover, the courses will arm you will all relevant safety rules and guidelines to keep you safe during your work.
• Sound and Communication Technicians – The comprehensive sequence of courses will equip you well for the job. This will include, Structured Wiring, Basics of Telephony, understanding of LAN, Principles of Grounding, , Video, Voice, and Data Systems among 124 other areas expertise needed to excel in the field.
Panel Interview
The panel interview will include a mix of 4-10 interviews from the Training Center, and the National Electrical Contractors Association. It should be noted that aside from testing your knowledge the interview will also include a series of Behavioral Interview Questions. These question are designed to bring to light how you past experiences will affect performance, communication and output while on the job.
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There was a problem sending your message. Please try again in a few minutes. | 1,570 | 7,104 | {"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} | 3.8125 | 4 | CC-MAIN-2020-16 | latest | en | 0.899291 |
https://www.recruitmentinboxx.com/ceed-question-papers/ | 1,680,129,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00312.warc.gz | 1,058,608,902 | 19,724 | Home / Question Paper 2022 / CEED Question Papers 2019 | CEED Previous Year Question Papers PDF
# CEED Question Papers 2019
Candidates preparing for the CEED exam must download PDF of CEED Question Papers 2019 from here!!! With the help of these Previous Year CEED Solved Question Papers links candidates may take their preparation to the next level.
Industrial Design Centre, IIT Bombay conducts Common Entrance Examination for Design (CEED) for those candidates who want to get admission in PG degree in the field of technological design.
CEED Previous Year Question Papers 2019
The main motive to conduct the CEED exam is to assess the candidate’s aptitude for design, logical skills, and communication skills. So you are suggested to practice more and more from the CEED question papers.
CEED Previous Year Question Papers will help you to gain knowledge regarding the pattern of the exam. More information CEED previous question papers with solutions pdf along with the CEED Question Papers 2019 on the below section of this page which is well planned and structured by the team of www.recruitmentinboxx.com
## CEED Question Papers
Year CEED Question Papers 2018 https://recruitmentinboxx.com/transvestite-dating-websites/ 2017 Hit Here 2016 Question Paper 1 Hit Here 2016 Question Paper 2 Hit Here 2015 Hit Here 2014 Hit Here 2013 Hit Here 2012 Hit Here 2011 Hit Here 2010 Hit Here
Question 1: Using the Relative Polar Coordinate System to add a 3 inch line that is 45 degrees from the end point of the line created above the drafter would ________.
1. type @3<45 and enter
2. type 3 back slash and enter
3. type 3 forward slash and enter
4. type 3 and try to use the Dynamic Input readout to find the end point
Question 2: When using the direct entry method to create a horizontal line exactly three inches long the drafter should ________ first.
1. pull the line from away from the starting point very carefully
2. insure that Ortho is on
3. enter 3 for the length of the line
Question 3: The UCS icon represents the intersection of the ________.
1. X axis
2. Y axis
3. Z axis
4. All of the above
Get Here Detailed: CEED Syllabus
Question 4: Understanding the Cartesian coordinate system will help the beginning drafter ________.
1. control the accuracy of the drawing
2. produce a visible grid system
3. select a different text style
4. choose various line types
Question 5: A line with a tapering width can be easily created by using the ________ tool.
1. circle
2. eclipse
3. line
4. polyline
Question 6: Rabindranath Tagore’s ‘Jana Gana Mana’ has been adopted as India’s National Anthem. How many stanzas of the said song were adopted?
1. Only the first stanza
2. The whole song
3. Third and Fourth stanza
4. First and Second stanza
Question 7: The famous Dilwara Temples are situated in
2. Rajasthan
3. Maharashtra
Question 8: Bijapur is known for its
1. severe drought condition
2. Gol Gumbaz
3. heavy rainfall
4. statue of Gomateswara
Check Previous Year – Qualifying Marks: CEED Cut Off
Question 9: Who is the author of the book ‘Nineteen Eighty Four’?
1. Thomas Hardy
2. Emile Zola
3. George Orwell
4. Walter Scott
Question 10: Who is the author of the book ‘Forbidden Verses’?
1. Salman RushDie
2. Abu Nuwas
3. Taslima Nasrin
4. H. Lawrence
Question 11: Pick out the correct statement about differential used in automotives
1. Differential Is always located on the rear axle
2. Reduces the wear of the axle
3. Allows wheels to run at different speeds
4. Improves fuel efficiency of the automotive
Question 12: Dilip Chhabria is best known as
1. Animator
2. Illustrator
3. Product designer
4. Automobile designer
Question 13: which of the following file format is lossless?
1. JPEG
2. PNG
3. BMP
4. SVG
Question 14: Who is known as “Father of Indian Cinema?”
1. Shyam Benegal
2. Satyajit Ray
3. Raj Kapoor
Question 15: The following part is made out of aluminium block of 50 x 40 x 10 mm. Which of the following is the best process to make the rounded corners and pockets inside?
1. Turning
2. Drilling
3. Milling
4. Rolling
Question 16: A smooth, long, flexible strip used to draw a smooth curve through several points on a plane is called ___?
1. Ruler
2. French curves
3. Stripe
4. Spline
Also Know Here: How to Become A Fashion Designer Without a Degree
Question 17: A farmer owns a rectangular farm that has a road on all the four sides. He starts from one corner of the field and walks north. After going 600 meters, he takes a right turn and walks another 400 meters. Again he takes a right and walks for 100 meters Estimate the minimum distance between his starting point and the current position.
1. 360 m
2. 500 m
3. 420 m
4. None of the above
Question 18: Find the odd man out
1. Wings of Fire
2. Ignited Minds
3. Letters for a Nation
4. India 2020
Question 19: Which color does not occur in standard Google logo?
1. Blue
2. Green
3. Purple
4. Yellow
Question 20: With a 50 mm lens when you change aperture from 2.8 to 8, the depth of field of the image will
1. Increase
2. Decrease
3. Remain unchanged
4. None of the above
CEED 2019
Common Entrance Examination for Design is going to be conducted on 09-01-2019 to provide admission to eligible students in M.Des. CEED 2019 will be of three hours in two parts. Part A will be a computer based test while Part B will be subjective type. Candidates will be given one hour to complete part A and two hours for the part B.
Note:
Candidates you are advised to prepare according to the CEED Question Papers and get yourself ready to appear in the exam. For any query regarding CEED exam question papers, you may write us in the below stated comment box, we will soon try to solve your queries.
Directly Get Here: CEED Admit Card
You may also follow us on Facebook and Google Plus, as we update all the information regarding CEED question papers with solutions or you can get all the updates related to CEED previous year question papers with solutions by subscribing to our web portal that is www.recruitmentinboxx.com
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June 20, 2022 | 1,890 | 7,374 | {"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} | 3.109375 | 3 | CC-MAIN-2023-14 | longest | en | 0.853478 |
https://goprep.co/q42-in-fig-10-102-qr-is-a-common-tangent-to-the-given-i-1nk9vx | 1,624,126,316,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00119.warc.gz | 263,414,622 | 32,229 | # In Fig. 10.102, Q
Given:
PT = 3.8 cm
Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
By property 1,
PQ = PT (Tangent from P)
PR = PT (Tangent from P)
Now,
QR = PQ + PR
QR = PT + PT
QR = 3.8 cm + 3.8 cm
QR = 7.6 cm
Hence, QR = 7.6 cm
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view all courses | 228 | 825 | {"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} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.876608 |
https://medium.com/@dhavatar/displaying-transit-information-using-google-maps-api-29a18cb0ac24?source=---------2------------------ | 1,560,927,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998923.96/warc/CC-MAIN-20190619063711-20190619085711-00457.warc.gz | 518,476,459 | 28,266 | # Displaying transit information using Google Maps API
One of my friends was interested in trying to map out the number of buses needed in order to get to Facebook in Seattle. He was mentioning that he had a problem trying to figure out what was water and what was land using the Google Maps API. After some discussion of possible solutions, searching online, and viewing his prototype, I figured this would be an interesting problem to try to solve. Besides, how hard would it be to 1) determine water and land on Google Maps using their API, 2) determine the number of buses needed to reach Facebook across the map, and 3) display the data in a pleasing way?
### The Goal
Make a visually pleasing map that shows the number of buses needed to travel to the Facebook office from any location in the greater Seattle area.
### Problem 1a: What is water?
I tackled the first problem: given a random GPS point in the Seattle area, can I determine that it’s over water or land? You would think the provided Map API will give enough information, but this turns out to be a hard problem using only their endpoints. The first solution would be to use their reverse geocoding to determine what the type is at that coordinate as it returns an array that says land, sea, ocean, lake, etc. However, it turns out this type is unreliable and many times, when over water, it returns an incorrect type. Trying to determine it based on elevation also didn’t work as the elevation varied too much across both water and land.
There was another suggestion of doing image processing, where you take a static map and go through the pixels to determine if it’s water based on the pixel color. At first glance, this seems like an difficult way of doing it as now you need to figure out which shades of blue is water, but Google Maps has a style API which allows you to customize the colors on the map for different features. This seemed like the way to go.
Doing some more searching, I found the exact style needed to get the result I wanted, which generated the map below (same as the link above).
Opening this map in python and checking a few pixels, this indexed PNG gave a value of 0 for the black areas (water) and some value between 1–255 for the rest, which was perfect. Now I know exactly what is water and what is land. But now I have another problem.
### Problem 1b: What is the GPS coordinate of a pixel?
Luckily for me, some other people doing research on this figured out the math to it. In this StackOverflow question, there’s a python function that did exactly what I needed for the static image I had. The max resolution of a static Google Map is 640px x 640px (for a free account), and knowing the zoom level and the center coordinate of the image, I just had to plug in the the pixel coordinate to get a GPS coordinate.
Would the distance between a pixel be granular enough for what I was trying to do? Since I know the GPS coordinates of any pixel, I did some math to determine that the distance between two pixels is about 169 ft (51.5 m), which is more detail than I thought. In the end, I chose to use a 5 pixel distance (845 ft/258 m) between each location to check.
### Problem 2: How many buses does it take to get to the FB office?
I chose a 400px x 400px area on the map to start mapping out the number of buses. Using the Directions API, I just plopped down the starting GPS coordinate and the FB office as the destination, choose transit as the option, and got a JSON result that contained a list of steps with the number of TRANSIT types and the total duration to travel there.
However, because I’m using 5px increments, that means I would need to make at most 6,400 API calls to get the number of buses needed at each of those locations. Using a free API key, the quota is 2,500 API calls per day. As I wasn’t in a rush to get this done, I wrote the program to do as much as it could, recording the results per column of pixels in a file, and doing something else as I waited for the quota to reset.
During my testing, I found out I needed to give a specific time for the directions API as the time of day influences the number buses available. Running my script at 1 AM led to several results saying it was faster to walk 2.5 miles because there were no buses running at the time (though thinking about it, walking/biking 2.5 miles wouldn’t be too bad). I set the time to December 6 at 8 AM to get a rough idea on transit to the office during work time.
### Problem 3: How do I display this data in a pleasing way?
If the results were plotted on the map, then the map would be littered with a bunch of dots that don’t say much. My first draft was to use the heatmap option, which has a bunch of issues by itself. The way Google does a heatmap is it gives each point a fixed radius in pixels and an intensity color based on a gradient. The problem with the fixed pixel size is when you zoom in/out on the map, the information changes as the radius is smaller/larger in respect to the size of the map, which will also change the colors as the amount they bleed into each other changes. To solve this, I used some JavaScript functions that calculated the Mercator projection and a formula to convert the current zoom level to a pixel radius that would be visually consistent no matter how close/far you were in the map. It wasn’t perfect, but it worked well enough.
Tweaking the radius until it looked good visually, it was a pretty good draft and gave a great overview of travel times at a glance. There was a problem with what was shown though.
If you take a look at the small section in Queen Anne, there appears to be data saying it takes 3 buses to get to Facebook, except the office is really close to there. Why is this the case? At first, I thought it was because the dots were merging together to show an incorrect result, and doing a quick spot check in Google Maps showed that it’s supposed to be 2 buses. However if you click on the transit option to give more details, you see this:
I didn’t know there were bus lines that transformed into another number when it reached one of the stations. This explains why it showed 3 buses as it considered this line as two buses. The API unfortunately doesn’t have any indication that it’s the same line and doesn’t contain the text “continue on the same vehicle,” so with a bit of brainstorming, I made a small modification. If there’s two transit steps and the arrival time of the previous matches the departure time of the next step, then it’s the same line. Generally, if you have to transfer to another bus/train, there’s a small walk step in between or the arrival/departure times will differ slightly. I reran the script and the results looked much better.
There’s also a few spots on the map where the Directions API didn’t give any results, so scattered throughout are blank spots. To make the data look cleaner, I wrote a second part that looped through the map again and compared it against the JSON output. If there was a part that was land, but didn’t appear in the JSON, I copied the previous value. A median filter would probably work better, but this was good enough. With a small change to the color gradient, I now have the final heatmap.
I also made a choropleth map version that allows you to switch between the number of transit needed and the travel time. It’s not as smooth since it uses square blocks for each point, but the colors are more distinct. It also allows you to switch between the number of buses and the travel time.
In total, I made 11,846 API calls to the Directions API, which required 4 days since I waited a day for the free 2,500 limit to reset (because I’m cheap). It was a fun project to learn about Google Map API and how to display different data on it. | 1,716 | 7,779 | {"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} | 2.859375 | 3 | CC-MAIN-2019-26 | latest | en | 0.960678 |
https://convertoctopus.com/309-8-hours-to-weeks | 1,696,318,990,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00149.warc.gz | 198,780,153 | 7,882 | ## Conversion formula
The conversion factor from hours to weeks is 0.005952380952381, which means that 1 hour is equal to 0.005952380952381 weeks:
1 hr = 0.005952380952381 wk
To convert 309.8 hours into weeks we have to multiply 309.8 by the conversion factor in order to get the time amount from hours to weeks. We can also form a simple proportion to calculate the result:
1 hr → 0.005952380952381 wk
309.8 hr → T(wk)
Solve the above proportion to obtain the time T in weeks:
T(wk) = 309.8 hr × 0.005952380952381 wk
T(wk) = 1.8440476190476 wk
The final result is:
309.8 hr → 1.8440476190476 wk
We conclude that 309.8 hours is equivalent to 1.8440476190476 weeks:
309.8 hours = 1.8440476190476 weeks
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 week is equal to 0.54228534538412 × 309.8 hours.
Another way is saying that 309.8 hours is equal to 1 ÷ 0.54228534538412 weeks.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred nine point eight hours is approximately one point eight four four weeks:
309.8 hr ≅ 1.844 wk
An alternative is also that one week is approximately zero point five four two times three hundred nine point eight hours.
## Conversion table
### hours to weeks chart
For quick reference purposes, below is the conversion table you can use to convert from hours to weeks
hours (hr) weeks (wk)
310.8 hours 1.85 weeks
311.8 hours 1.856 weeks
312.8 hours 1.862 weeks
313.8 hours 1.868 weeks
314.8 hours 1.874 weeks
315.8 hours 1.88 weeks
316.8 hours 1.886 weeks
317.8 hours 1.892 weeks
318.8 hours 1.898 weeks
319.8 hours 1.904 weeks | 505 | 1,733 | {"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} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.854364 |
https://bettersheets.co/formulas/t-dist-rt | 1,719,321,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00543.warc.gz | 103,779,515 | 8,015 | Formulas > =T.DIST.RT()
# How To Use T.DIST.RT() Function in Google Sheets
Description
Returns the right tailed Student distribution for a value x.
Common Questions about the T.DIST.RT Formula:
• What does the T.DIST.RT formula do?
• How is the T.DIST.RT formula used?
• How do I input information into the T.DIST.RT formula?
How Can the T.DIST.RT Formula be Used Appropriately:
• The formula can be used to calculate the probability of the right-tailed t-distribution.
• The formula is most commonly used to test for significance in a given data set.
• The T.DIST.RT formula can also be used in regression analysis, to illustrate the probability of a given outcome.
How Can the T.DIST.RT Formula be Commonly Mistyped:
• It is often mistyped as TDIST RT or T.DIST RT.
• The “.” between “T” and “DIST” is often left out.
What Are Some Common Ways the T.DIST.RT Formula is Used Inappropriately:
• The formula should not be used to calculate values that are not related to t-distribution probabilities.
• The formula should not be used as a substitution for a chi-squared test.
• The formula should not be used without first setting up the appropriate parameters.
What Are Some Common Pitfalls When Using the T.DIST.RT Formula:
• Inputting incorrect values into the formula.
• Forgetting to enter a negative sign for the degrees of freedom argument.
• Not including the right number of arguments in the formula.
What Are Common Mistakes When Using the T.DIST.RT Formula:
• Inputting incorrect values into the formula.
• Forgetting to include the degrees of freedom argument.
• Not including the right number of arguments in the formula.
What Are Common Misconceptions People Might Have With the T.DIST.RT Formula:
• That it can be used as a substitute for a chi-squared test.
• That it can be used to calculate any value, not just t-distribution probabilities.
• That it can be used without first setting up parameters.
How To Actually Use T.DIST.RT() in Sheets
`T.DIST.RT(x, degrees_freedom)`
Looking for a video tutorial? Members can email me anytime! Check out the blogs below for more info on this formula. Or generate your formula. Find more formulas here at BetterSheets.co/formulas
Generate a T.DIST.RT() formula for your needs with AI
Whatever you need to do in sheets, you can generate a formula. Use the Better Sheets Formula generator to create a formula for any need. Completely free for members.
Asa
Looking for more help inside sheets get the free Add-on: Asa. Ask Sheets Anything. Go ahead, ask it any problem you migth have. Bring your own APIKEY and generate formulas inside of Google Sheets.
Better Sheets | 610 | 2,638 | {"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} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.817495 |
https://community.powerbi.com/t5/Desktop/Calculating-downtime-for-shifts-when-downtime-crosses-shifts/m-p/422208/highlight/true | 1,585,733,258,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00383.warc.gz | 398,588,880 | 127,794 | cancel
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Frequent Visitor
## Calculating downtime for shifts when downtime crosses shifts
We have 4 crews(A,B,C,D), that work rotating 12 hr shifts and they are 6:00 to 18:00 and 18:00 to 6:00. We capture downtime in a table with a starttime and an endtime. We have a slicer that sets a start date of say 4/29 6:00 and end date of 5/6 6:00. We have a table with the shift schedule, where there are 2 rows for each day, like 04/29/18 6:00 and 4/29/18 18:00 with the shift letter.
Each shift's downtime is the time down in their shift, while the total downtime for that segment could span mutliple shifts and be outside the slicer limits. I'm looking for a way to get a breakdown of each shifts downtime and sum that downtime up for a total. I'm using Directquery now, but if its impossible I can switch to import. I wasn't sure how to start to set the relationships and what measures to create. I had something working between the slicers with IF statements but I had to switch to Import before it would work. However, this didn't address breaking it down by shifts. I'm more of a SQL person than a DAX person so it something needs to be prepared on the SQL side, I can do that.
I can capture screen shots of the data is thats helpful.
Any help would be appreciated!
Bart
1 ACCEPTED SOLUTION
Accepted Solutions
Frequent Visitor
## Re: Calculating downtime for shifts when downtime crosses shifts
I spent some time and I ended up figuring it out on my own.
2 REPLIES 2
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Community Support
## Re: Calculating downtime for shifts when downtime crosses shifts
HI @blynn,
Please provide some sample data or pbix file to help us clarify your scenario and coding formula.
Regards,
Xiaoxin Sheng
Community Support Team _ Xiaoxin Sheng
If this post helps, please consider Accept it as the solution to help the other members find it more quickly
Learning resources: Power BI
Frequent Visitor
## Re: Calculating downtime for shifts when downtime crosses shifts
I spent some time and I ended up figuring it out on my own.
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https://beta.geogebra.org/m/zuszcfar | 1,723,380,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00677.warc.gz | 105,189,978 | 29,304 | GeoGebra Classroom
# Estimate the value of π with a unit circle.
## Estimate the value of π with a unit circle.
1. Select the CIRCLE: CENTER & RADIUS tool. Click wherever you want the center to be, and then enter a radius of 0.5 so that the diameter will be 1. 2. Zoom in so that you can easily see the circle. 3. Select the POLYGON tool. Click on the circle's circumference in four different places to create a square with its vertices on the circumference. Click on the first vertex to finish the square. (Each time you click, make sure your mouse's arrow icon turns into a hand icon so that the points you create will be anchored to the circle.) 4. Select the MOVE tool. Drag the the vertices around to make sure they will stay on the circle. Then arrange them to make a square. 5. Select the DISTANCE OR LENGTH tool. Click somewhere inside the circle to make a perimeter measurement appear. Record its perimeter below.
## What is the perimeter of an inscribed polygon with 12 sides?
6. Repeat the instructions above, but this time create a dodecagon (12 sides). 7. Zoom in to confirm that each side is shorter than the arc of the circle beside it. 8. What is the perimeter of this 12-sided approximation of a circle?
## What is the perimeter of an inscribed polygon with many sides?
9. Repeat the instructions above with a polygon that has as many sides. (Archimedes used 96). 10. Zoom in to confirm that each side is shorter than the arc of the circle beside it. 11. How many sides does your shape have, and what is its perimeter? | 361 | 1,542 | {"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} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.901823 |
http://www.scansteel.mx/library/calculus-an-integrated-approach-to-functions-and-their-rates-of-change | 1,556,151,840,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578675477.84/warc/CC-MAIN-20190424234327-20190425020327-00455.warc.gz | 287,735,972 | 8,984 | By Robin J. Gottlieb
A massive grievance of professors instructing calculus is that scholars wouldn't have the perfect historical past to paintings in the course of the calculus direction effectively. this article is focused at once at this underprepared viewers. this can be a single-variable (2-semester) calculus textual content that comes with a conceptual re-introduction to key precalculus principles through the exposition as acceptable. this is often the correct source for these colleges facing poorly ready scholars or for colleges introducing a slower paced, built-in precalculus/calculus path.
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Extra info for Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition
Example text
The number of pounds of sugar cane it takes to support a panda for a day is a function of the weight, x, of the panda. Express this function with a formula. The formula will involve the constants S, N, and C. ii. 3 Representations of Functions 27 W , the number of weeks they are to be supported. Express this function as a formula. The formula will involve the constants S, N, C, P , and Q. SOLUTION i. Let’s use the strategy of taking the problem apart into a series of simpler questions. We’ll use unit analysis to help us.
Instead, we must draw a triangle that involves the radius of the bowl itself. This radius must emanate from the center of the sphere. 11 4h − h2. ◆ Functioning with friends. Javier goes to a pizza shop intending to order a small pizza and eat it. When he enters the shop he sees some of his friends and they decide to split a large pizza. If the radius of a large pizza is twice the radius of a small pizza, what fraction of the large pizza should be allocated to Javier to give him the amount of food he originally intended to eat?
A) C(A) (b) C(2A) (c) 2C(A) (d) C(A + 1) (e) C(A) + 1 1 x+1 , 14. If f (x) = find the following. Simplify your answer where possible. (a) f (0) (b) f (3) (f ) f (b + 3) 15. If g(x) = (a) g(0) 16. If h(x) = (a) h(0) (g) [f (7)]2 (c) f (− 41 ) (h) (d) f (b) f (b2) (i) [f (b)]2 (e) f (b − 1) √ x 2 +4 , 2 find the following. Simplify your answer where possible. √ √ (e) −g(3t) (f ) g( t − 4) (b) g(2) (c) g( 5) (d) g √1 2 x2 1−2x , find (b) h(3) (c) h(p + 1) (d) h(3p) (e) 2h(3p) (f ) 1 h(2p) 17. If j (x) = 3x 2 − 2x + 1, find the following. | 813 | 2,901 | {"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} | 3.375 | 3 | CC-MAIN-2019-18 | longest | en | 0.911307 |
https://communities.sas.com/t5/ODS-and-Base-Reporting/first-variables/td-p/146196?nobounce | 1,531,724,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00414.warc.gz | 651,931,759 | 33,355 | ## first.variables
Solved
Occasional Contributor
Posts: 12
# first.variables
My data is like:
Id year month response 100131 2014 3 0 100131 2014 3 1 100131 2014 4 0 100131 2014 4 0 100131 2014 5 0 100131 2014 5 0 100131 2014 5 0 100131 2014 5 0 100131 2014 5 0 100131 2014 5 0 100131 2014 5 1 100131 2014 5 1 100131 2014 5 1 100131 2014 5 1 100131 2014 5 1
I want to count the number of 0 response and number of 1 response for each month. i have sorted the data by id year month and response.
the output must be the following:
id year month response count 100131 2014 3 0 1 100131 2014 3 1 1 100131 2014 4 0 2 100131 2014 5 0 6 100131 2014 5 1 5
the code i have used is :
data step......
......
...
if first.month and first.response then count=0;
count+1;
if last.response then output;
but i am not getting the expected output..
please help me in fixing this.
thanks in advance..
Accepted Solutions
Solution
01-30-2015 12:11 PM
Respected Advisor
Posts: 3,852
## Re: first.variables
data response;
input id year month response;
cards;
100131
2014
3
0
100131
2014
3
1
100131
2014
4
0
100131
2014
4
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
;;;;
run;
proc print;
run;
data counts;
set response;
by id month;
array c[0:1] _temporary_;
if first.month then do; c[0]=0; c[1]=0; end;
c[response]+
1;
if last.month then do;
do response=0,1;
count = c[response];
output;
end;
end;
run;
proc print;
run;
All Replies
Respected Advisor
Posts: 3,852
## Re: first.variables
Where is variable RESULT?
Without a variable RESULT in a BY statement there can be no FIRST.RESULT or LAST.RESULT
Occasional Contributor
Posts: 12
## Re: first.variables
Posted in reply to data_null__
sorry it is the 'response' I have unknowingly written as result. may be the code goes like this;
if first.month and first.response then count=0;
count+1;
if last.response then output;
Respected Advisor
Posts: 3,852
## Re: first.variables
Leave off FIRST.MONTH from the first IF.
Occasional Contributor
Posts: 12
## Re: first.variables
Posted in reply to data_null__
the problem with my data is that, not all the id will have 0 and 1 some times the value is only 0 for all the month, so during that time the 0 will be counted despite the month change. but what I want is the response count for each month when it is 0 and 1.. so i cannot omit the first.month
Solution
01-30-2015 12:11 PM
Respected Advisor
Posts: 3,852
## Re: first.variables
data response;
input id year month response;
cards;
100131
2014
3
0
100131
2014
3
1
100131
2014
4
0
100131
2014
4
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
0
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
100131
2014
5
1
;;;;
run;
proc print;
run;
data counts;
set response;
by id month;
array c[0:1] _temporary_;
if first.month then do; c[0]=0; c[1]=0; end;
c[response]+
1;
if last.month then do;
do response=0,1;
count = c[response];
output;
end;
end;
run;
proc print;
run;
Super User
Posts: 13,498
## Re: first.variables
proc freq data=have;
tables id*year*month*response/ list nopercent nocum;
run;
if you want a dataset as well add OUT=newdatasetname after the / above.
Occasional Contributor
Posts: 12
## Re: first.variables
I got the same result in proc freq using similar code like urs. but I wanted to try in datastep if possible.
🔒 This topic is solved and locked.
Need further help from the community? Please ask a new question.
Discussion stats
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• 3 in conversation | 1,288 | 3,696 | {"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} | 2.640625 | 3 | CC-MAIN-2018-30 | latest | en | 0.566489 |
https://www.physicsforums.com/threads/population-growth.107720/ | 1,624,184,700,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487660269.75/warc/CC-MAIN-20210620084505-20210620114505-00411.warc.gz | 866,451,025 | 15,636 | # Population Growth
I am quite sure the first approach is to use partial fractions but I am
unclear how to finish this equation
1/P*dP/dt=b+aP
HallsofIvy
Homework Helper
How to finish it? You haven't started it!
$$\frac{1}{P}\frac{dP}{dt}= b+aP$$
is separable:
$$\frac{dP}{P(b+aP)}= dt$$
We can write
$$\frac{1}{P(b+ aP)}= \frac{A}{P}+ \frac{B}{b+ aP}$$
for some A and B. Multiply both sides by P(b+ aP):
$$1= A(b+ aP)+ BP$$
What do you get if P= 0? What do you get if P= -a/b? Put those values for A and B into the fractions and integrate.
Solve P=? or t=?
I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:
A=1/b and B=-a/b. And I know both a and b from a linear regression.
So substituting back I get
((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides
(1/b)lnP +(-1/b)ln(b+aP)=t +C
Now my question is here how do I solve for P= f(t).
HallsofIvy
Homework Helper
zoldman said:
I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:
A=1/b and B=-a/b. And I know both a and b from a linear regression.
So substituting back I get
((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides
(1/b)lnP +(-1/b)ln(b+aP)=t +C
Now my question is here how do I solve for P= f(t).
The rest is algebra:
$$\frac{1}{b}ln P- \frac{1}{b}ln(b+aP)= ln\left(\frac{P}{b+aP}\right)^\frac{1}{b}= t+ c$$
Take exponential of both sides:
$$\left(\frac{P}{b+aP}\right)^\frac{1}{b}= e^{t+ C}= C'e^t$$
(C= eC)
Take ath power of both sides:
$$\frac{P}{b+aP}=C"e^{at}$$
(C"= C'a)
multiply both sides by b+ aP and expand:
$$P= C"e^{at}(b+ aP)= C"be^{at}+ aC"e^{at}P$$
$$P- aC"e^{at}P= P(1- aC"e^{at})= C"be^{at}$$
$$P= \frac{C"be^{at}}{1- aC"e^{at}$$
mathwonk
Homework Helper
2020 Award
i thionk that logistic equation is more usefully written as dP/dt
= aP(1 - P/N).
of course it is separable and solving it, shows that the populaion approaches N as time goes on.
saltydog
Homework Helper
Starting with:
$$\frac{1}{b}ln(P)-\frac{1}{b}ln(b+aP)=t+c$$
multiplying by b and collecting logarithms:
$$ln\left[\frac{P}{b+aP}\right]=b(t+c)$$
Taking exponentials:
$$\frac{P}{b+aP}=e^{b(t+c)}$$
multiplying both sides by b+aP and collecting the P's:
$$P\left[1-ae^{b(t+c)}\right]=be^{b(t+c)}$$
Isolating the P and then multiplying the top and bottom of the rational expression by $e^{b(t+c)}$ to make it cleaner leaves:
$$P(t)=\frac{b}{e^{-b(t+c)}-a}$$
Now, how about completely characterizing the solutions in terms of a and b assuming some initial condition like P(0)=1. What would it look like whatever it was? (just a suggestion )
Last edited:
I am working on the same problem; however I started out a bit different. Follwing the above; now all I need to do is substitue my a and b values; my initial condition is P sub 0 = 3.9
Am I going in the right direction? | 994 | 2,827 | {"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} | 4.40625 | 4 | CC-MAIN-2021-25 | latest | en | 0.796363 |
https://bobbylaymancars.net/auto-service/how-do-you-determine-engine-rotation-direction.html | 1,656,309,996,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00674.warc.gz | 185,344,030 | 18,495 | # How do you determine engine rotation direction?
Contents
Engine rotation direction is determined by observing Flywheel rotation from the flywheel end of the engine looking forward to the pulley end of the engine.
## How do you find the correct direction of rotation of an engine?
Instruct an assistant to “bump” the starter over, to rotate the engine. Watch the rotation of the flywheel from the rear. If the flywheel turns counterclockwise, or left-handed, it denotes a standard rotation. If the flywheel turns clockwise, or right-handed, it signifies a reverse rotation engine.
## How do you determine the rotation of a marine engine?
How to Determine Boat Engine Rotation. Left Hand (LH) rotation is considered to be standard while Right hand (RH) is considered an “Opposite” rotation. You should ALWAYS view the the engine rotation while standing behind the engine and looking at the flywheel.
## How do you rotate an engine clockwise and anticlockwise?
A DC motor can be very simply rotated both ways in clockwise as well as in anticlockwise directions by flipping the supply inputs to it. However the above reversing requires flipping of both of its wire polarity with the connected supply. Therefore it cannot be done by using a single relay or switch.
## How do you tell if a motor is CW or CCW?
You can determine shaft end perspective by simply holding your motor up in front of you and pointing the shaft at you. If the shaft is pointing at you and rotates to the right, your motor is clockwise shaft end, or CWSE. If the shaft rotates to the left, your motor is counterclockwise shaft end, or CCWSE.
## How do you find the reverse rotation of a marine engine?
How to Determine Boat Engine Rotation. Left Hand (LH) rotation is considered to be standard while Right hand (RH) is considered an “Opposite” rotation. You should ALWAYS view the the engine rotation while standing behind the engine and looking at the flywheel.
## What is a reverse rotation engine?
This is done by making the angle of the drive gear on the cam and its mating gear opposite, when the cam turns the opposite direction. This makes the thrust of the gears in the opposite direction. For example the SE Chevy thrust is up and is taken by the base of the distributor housing and the drive gear.
## Do marine motors run backwards?
There is no real ‘forward’ or ‘backwards’ in a marine engine. The last couple of ships I worked on had low speed marine diesel engines that could run in either direction. These engines were hooked directly to the propeller shaft and there was no gear box installed.
## Why do motors rotate anticlockwise?
It is due to the counter clockwise motion of the motor. The motor rotates in the counter clockwise direction the blades of the fan are attached to the motor hence the blades rotates in the anti clockwise direction. So it is due to the motor rotation the fan rotates in counter clockwise direction.
IT\'S INTERESTING: Are bigger engines more to insure?
## What does counterclockwise clock mean?
: in a direction opposite to that in which the hands of a clock rotate as viewed from in front.
## How do you change the direction of an AC motor?
In order to reverse the direction of an AC motor, the magnetic fields must be altered to provoke movement in the opposite direction. Since each wire consists of a positive and negative current within the magnetic fields, the flip-flopping of main and starter wires causes the motor to run in reverse rotation. | 726 | 3,493 | {"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-2022-27 | latest | en | 0.908 |
https://en.sourcecodeaplikasi.info/implement-numpy-linalg-norm-for-matrix-rows-a-guide/ | 1,696,375,019,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00218.warc.gz | 267,364,814 | 13,150 | # Implement Numpy.Linalg.Norm for Matrix Rows: A Guide.
Posted on
If you’re a data scientist or a math enthusiast, then you’re probably familiar with matrices and their properties. One of the most commonly used properties is the norm of a matrix, which is a measure of its magnitude or size. The numpy.linalg.norm function allows you to calculate the norm of a matrix efficiently and accurately, with several options for specifying which norm to use. In this guide, we’ll focus on how to implement numpy.linalg.norm specifically for matrix rows.
Why focus on matrix rows, you might ask? Well, often times in data analysis or machine learning tasks, we want to compare or measure the distances between the rows of two or more matrices. When dealing with large datasets, it can be time-consuming and memory-intensive to loop through each row individually. Fortunately, numpy provides us with an optimized function to compute the norm of matrix rows in a single line of code.
In this article, we’ll walk you through the steps to implement numpy.linalg.norm for matrix rows, show you some examples of how to use it, and discuss some best practices and tips for working with matrix rows in numpy. Whether you’re a beginner or an advanced user of numpy, this guide will provide you with valuable insights and techniques for working with matrices and their norms.
So, if you’re ready to take your numpy skills to the next level and learn how to efficiently compute the norm of matrix rows, then read on!
“How To Apply Numpy.Linalg.Norm To Each Row Of A Matrix?” ~ bbaz
## Introduction
In the field of data science and machine learning, there are many mathematical operations that need to be conducted on matrices. One such operation is finding the norm of matrix rows. In this article, we will discuss the implementation of the Numpy.Linalg.Norm function for matrix rows.
## What is Numpy?
Numpy is a popular open-source package for numerical computation in Python. It provides various functions and tools for handling multi-dimensional arrays and matrices. Numpy also offers various linear algebra functions such as Eigenvalues, Eigenvectors, Matrix multiplication, determinant, and norm, among others.
## What is Norm?
The norm of a vector or matrix is a way of measuring the magnitude or length of a vector or matrix. It is represented by || || double vertical lines. The Euclidean norm is considered the most popular type of norm where it is the square root of the sum of squares of all elements in a vector/matrix.
## How to Implement Numpy.Linalg.Norm for Matrix Rows?
To find the norm of matrix rows using Numpy.Lingalg.Norm, we only need to provide the matrix and axis value, which should be one (1). Suppose matrix ‘X’ has i rows and j columns; then the output matrix will be a one-dimensional matrix of length i.
## Example
Let’s consider an example where we have the following matrix:
2 4 -1 -3 1 5 9 2 0
To find the norm of matrix rows using Numpy.Linalg.Norm, the code will be:
“`import numpy as npX = np.array([[2,4,-1],[-3,1,5],[9,2,0]])print(np.linalg.norm(X,axis=1))“`
The output will be:
“`array([4.58257569, 6.55743852, 9.21954446])“`
## Comparison with other Libraries
### Scipy
Scipy is another popular open-source library in Python for scientific computing. It also provides functions for linear algebra operations such as Eigenvalues, Eigenvectors, and norms, among others. To implement the norm of matrix rows using Scipy, we need to provide the matrix and axis value, which should be one (0 for columns). Here is how we can do it:
“`from scipy.linalg import normX = np.array([[2,4,-1],[-3,1,5],[9,2,0]])print(norm(X,axis=1))“`
The output will be the same as Numpy’s implementation:
“`array([4.58257569, 6.55743852, 9.21954446])“`
### Math Library
The Math library is a built-in library in Python that provides mathematical operations such as finding square roots, cosines, sines, and logarithmic values. However, it does not have built-in functions for matrix operations such as norms. We need to implement our function for finding the norm of matrix rows. Here is an example:
“`import mathX = [[2,4,-1],[-3,1,5],[9,2,0]]for row in X: sum_of_squares = sum([x**2 for x in row]) print(math.sqrt(sum_of_squares))“`
The output will be:
“` 4.58257569495584 6.557438524302 9.219544457292887 “`
## Conclusion
In conclusion, Numpy.Linalg.Norm is a useful function to find the norm of matrix rows in Python quickly. It’s simple and easy to implement. However, it’s not the only option, and other libraries such as Scipy and implementing our function using Math libraries are also valid options. The choice of library depends on the use case and personal preferences.
Thank you for taking the time to read our guide on implementing Numpy.Linalg.Norm for matrix rows. We hope that you found the content valuable and that it has provided insights that will help you in your work with matrices.
By familiarizing yourself with Numpy.Linalg.Norm, you will be able to effectively compute the magnitude of vectors and normalize data which is important in data science, statistics, and machine learning.
We encourage you to continue practicing and experimenting with Numpy.Linalg.Norm for matrix rows so that you can take full advantage of its potential. With consistent effort, you will be able to confidently apply the concepts outlined in this guide and achieve better results in your work.
Once again, thank you for visiting our blog and we hope that you will find more value in our future posts.
1. What is Numpy.Linalg.Norm?
2. Numpy.Linalg.Norm is a function in the Numpy library that calculates the matrix or vector norm.
3. What is matrix norm?
4. Matrix norm is a measure of the size of a matrix. It is calculated using a specific formula and is used to evaluate the properties of matrices.
5. What is vector norm?
6. Vector norm is a measure of the size of a vector. It is calculated using a specific formula and is used to evaluate the properties of vectors.
7. How do you calculate the norm of matrix rows using Numpy.Linalg.Norm?
8. You can use the Numpy.Linalg.Norm function to calculate the norm of matrix rows by passing in the matrix and specifying the axis parameter as 1.
9. What does the axis parameter do in Numpy.Linalg.Norm?
10. The axis parameter specifies the axis along which the norm is calculated. When axis=1, the norm of each row is calculated. When axis=0, the norm of each column is calculated. | 1,488 | 6,471 | {"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} | 3.75 | 4 | CC-MAIN-2023-40 | latest | en | 0.902785 |
https://www.clutchprep.com/chemistry/practice-problems/80400/the-speed-of-light-in-a-vacuum-is-2-998-x-108-m-s-a-what-is-its-speed-in-kilomet | 1,643,262,581,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00605.warc.gz | 752,338,141 | 21,683 | # Problem: The speed of light in a vacuum is 2.998 x 10 8 m/s. (a) What is its speed in kilometers per hour (km/h)? (b) What is its speed in miles per minute (mi/min)?
###### FREE Expert Solution
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###### FREE Expert Solution
97% (115 ratings)
###### Problem Details
The speed of light in a vacuum is 2.998 x 10 8 m/s.
(a) What is its speed in kilometers per hour (km/h)?
(b) What is its speed in miles per minute (mi/min)?
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Dimensional Analysis concept. You can view video lessons to learn Dimensional Analysis Or if you need more Dimensional Analysis practice, you can also practice Dimensional Analysis practice problems .
What is the difficulty of this problem?
Our tutors rated the difficulty of The speed of light in a vacuum is 2.998 x 10 8 m/s. (a) What... as medium difficulty.
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Our expert Chemistry tutor, Dasha took 5 minutes to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Kobko-Litskevitch's class at HUNTER. | 307 | 1,291 | {"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-2022-05 | latest | en | 0.922525 |
https://kids.kiddle.co/Algebraic_solution | 1,696,366,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00435.warc.gz | 367,086,626 | 5,164 | # Algebraic solution facts for kids
Kids Encyclopedia Facts
An algebraic solution is an algebraic expression which is the solution of an algebraic equation in terms of the coefficients of the variables. It is found only by addition, subtraction, multiplication, division, and the extraction of roots (square roots, cube roots, etc.).
The most well-known example is the solution of the general quadratic equation.
$x=\frac{-b \pm \sqrt {b^2-4ac\ }}{2a},$
$ax^2 + bx + c =0\,$
(where a ≠ 0).
There is a more complicated solution for the general cubic equation and quartic equation. The Abel-Ruffini theorem states that the general quintic equation does not have an algebraic solution. This means that the general polynomial equation of degree n, for n ≥ 5, cannot be solved by using algebra. However, under certain conditions, we can get algebraic solutions; for example, the equation $x^{10} = a$ can be solved as $x=a^{1/10}.$ | 223 | 932 | {"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": 4, "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-2023-40 | latest | en | 0.907432 |
https://www.huffingtonpost.com/phil-davis/how-to-get-rich-slowly_b_8126532.html?ec_carp=1074688246398894320 | 1,548,231,240,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00123.warc.gz | 834,605,231 | 50,327 | # How to Get Rich Slowly
09/12/2015 06:40 am ET Updated Dec 06, 2017
Do you want to be a Millionaire?
Sure you do, why not. \$1,000,000 is a nice amount of money -- even at 5 percent interest in retirement, it's enough to drop \$50,000 into your bank account each year to supplement whatever else you may have coming in. So let's agree that it's nice to have a Million Dollars.
Now I'll say something you may not agree with -- if you are not on a path to have \$1M in the bank by the time you retire -- it's probably your own fault.
Yes, I'm sorry but it's true. It's really not as hard as you think to save \$1M and I'm going to teach you how and, as long as you are under 50, I can show you how to get on the path to turning \$50,000 into \$1M in 25 years. If you are under 40 -- it's going to be a piece of cake to save \$1M by the time you are 65 -- as long as you can follow our plan.
First I need to convince you of the value of SAVING YOUR MONEY -- this is not something most people are good at -- especially young people. Unless you accept the VALUE of saving your money, you will not be able to make the wise choices you need to make to get on the path to saving \$1M.
This is a compound rate table. It's a mathematical fact. If you start with just \$10,000, even at 10 percent, you will have more than \$450,000 in 40 years. I'm 51 now and I can tell you that 30 years ago, I could have bought a cheaper car after college and saved \$10,000. I could have gone to 20 less concerts and saved \$2,000 and 20 less fancy dinner dates for \$2,000 and taken one less ski trip for \$2,000 -- you get the idea. I was young and I was successful but I didn't know at the time that EACH \$10,000 I spent in my 30s would cost me \$450,000 in my 70s.
That was my excuse then, what's your excuse now? Now I'm 51 and when I'm 91 I'm sure I'll appreciate having an extra \$450,000. I'll appreciate it so much, in fact, that I'm motivated to put away \$50,000, so I'll have \$2.25M. The point is, you need to accept the reality of these numbers and understand that EACH \$5 cup of coffee you buy at Starbucks is costing your future self \$225. Flying first class for \$1,500 instead of coach for \$300 is costing your future self \$54,000 -- are you really that rich?
Now, keep in mind that we're using the conservative 10 percent rate. At Philstockworld, we have 4 virtual portfolios for our Members that are averaging over 20 percent this year. At 20 percent, each dollar you save compounds into \$237 in just 30 years. While we can't ALWAYS have huge winning years, we only need to AVERAGE better than 10 percent returns and we will be in fantastic shape -- if we begin by deciding to save our money.
Using our "Be the House -- Not the Gambler" strategy, which we have discussed in our previous videos like "How To Buy a Stock for a 15-20 percent Discount" and "The Secret to Consistent 20-40 percent Annual Returns" we can put ourselves on a path to growing our savings at rates we desire to build a great retirement account. This is what we teach our Members every day at Philstockworld.
But it all has to start with you. YOU need to decide to forgo that \$5 cup of coffee, that \$500 day at the ballgame, that \$1,000 weekend trip because you fell stressed -- think how much less stressed you will feel in 40 years when that \$1,000 is an extra \$45,000! Then there are the big decisions you make along the way to retirement: Do you need that new car? Do you NEED that big wedding? Do you need a house that big?
It's ironic that people don't realize that NOT spending \$500 a month on their house and instead SAVING that money each month at a compounded 10 percent a year for the same 30 years (\$180,000) nets you back \$1.1M. Not only that, but NOT spending \$500 a month on your mortgage for the same 30 years, even at 5 percent, SAVES YOU \$172,000 in payments. That's how much you save with each \$100,000 you choose NOT to borrow for a mortgage!
Growing your wealth isn't just about making money in the markets. In 2015, Philstockworld will focus on wealth-building techniques which, combined with our winning investing strategies, can help put you on the path to a life of financial independence -- LEARN MORE HERE. | 1,135 | 4,259 | {"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-2019-04 | latest | en | 0.978209 |
https://scicomp.stackexchange.com/questions/36936/perturbation-problem-using-runge-kutta-4 | 1,653,184,651,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00060.warc.gz | 574,669,187 | 68,397 | # Perturbation problem using Runge-Kutta 4
I'm trying to evaluate the perturbations magnitude between 2 body orbiting a central one in three dimensions. In order to do this I need to have an estimate of the error, which I did using Richardson extrapolation as described here. I'm using Runge-Kutta 4 and also velocity Verlet. Things seems good in the unperturbed problem, which means I obtain an error $$O(10^{-7})$$ and $$O(10^{-12})$$ for Verlet and Runge-Kutta respectively by using a time step of $$0.001$$. When I consider the perturbations, only velocity Verlet still give me the expected error, while Runge-Kutta give me an error of the same order of Verlet, using the same previous time step. The same happens even if I reduce or increase the time step. Actually for the perturbed problem I already expected something bad related to the error, since that's what I obtained for the Richardson's fraction for Runge-Kutta
As it can be seen, the level of the figure should be about $$2^p = 16$$, but it's $$2$$ actually. That sort of spikes are present also in the same plot for velocity Verlet, but for the latter their magnitudes are much less than that in the plot above and the $$F_h$$ values are about $$4$$ anyway, as expected for Verlet. I checked all the Runge-Kutta code and the ode system code but I didn't find any error. Moreover I tested both methods with a simpler function obtaining correct results, so I can't understand the reason for such wrong error estimates for Runge-Kutta.
Edit 1
The function code is the following
vector<double> F(double t, vector<double> x, vector<CB> objs, vector<double> s) {
//objs store the masses of the objects
//s store the positions of the objects in the order [x1,y1,z1,x2,y2,z2,...];
//x store the positions and velocity of the target (perturbed) object in the order [x,Vx,y,Vy,z,Vz]
vector<double> p(x.size(),0);
double m, d;
vector<double> y(x.size()/2,0);
for (int i = 0; i < objs.size(); i++) {
//temporary store the mass of the i-th object
m = objs[i].CB::getmass();
//temporary store the position of the i-th object from the s vector
y = {s[i*x.size()/2], s[i*x.size()/2 + 1], s[i*x.size()/2 + 2]};
d = sqrt(pow((x[0] - y[0]),2) + pow((x[2] - y[1]),2) + pow((x[4] - y[2]),2));
//x system
p[0] = x[1];
p[1] += -G*m*(x[0] - y[0])/pow(d,3);
//y system
p[2] = x[3];
p[3] += -G*m*(x[2] - y[1])/pow(d,3);
//z system
p[4] = x[5];
p[5] += -G*m*(x[4] - y[2])/pow(d,3);
}
return p;
}
• What is your setup approximately? A starting configuration using real planetary data, or some artificial initial condition like one planet on a circular orbit of radius 1 and period 1, the other at radius 5, with a fixed proportion of masses, so that setting the mass factor turns the interaction on or off? Mar 2, 2021 at 10:59
• @LutzLehmann I use real planetary data taken from JPL Horizon system, cause I need to analyze different planetary configurations, of which the simpler one involve Sun, Jupiter and Saturn, with Jupiter perturbed by Saturn, which corrispondes to the perturbed case of my question. The motion of every planet involved in the simulation start from perihelion. The refernce frame is centered on the Sun.
– Zebx
Mar 2, 2021 at 12:44
• The errors oscillate like the periodic orbits, close to the zero-crossings you get the spikes. This can be prevented by taking the norm over all coordinate differences of the body. There must be something wrong with the method step that reduces the order to 1. How did you implement the perturbation? In the acceleration computation, or as something extra that is added to the method step? Mar 2, 2021 at 12:55
• @LutzLehmann The perturbations are implemented in the acceleration computation. They're "turned on" if more than one object is considered in the computations.
– Zebx
Mar 2, 2021 at 13:06
• Some hour ago I also compared my results to the one obtained using another Runge-Kutta 4 implementaion I found, but I got the same results for $F_h$.
– Zebx
Mar 2, 2021 at 13:32
To explain further what I wrote in the comments: Your function has the form F(t,x,co,xo) where x is the state, co the constants for the other objects and xo the dynamical data for the other objects.
Variant 1: You will get an order 1 approximation out of RK4 if you use the external dynamic in the RK4 loop as (in python notation)
dx1 = dt*F(t[k] , x[k] , co, xo[k])
dx2 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx1, co, xo[k])
dx3 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx2, co, xo[k])
dx4 = dt*F(t[k]+ dt, x[k]+ dx3, co, xo[k])
x[k+1]=x[k]+(dx1+2*dx2+2*dx3*dx4)/6
Variant 2: You will get an order 2 approximation if you use linear interpolation as in
dx1 = dt*F(t[k] , x[k] , co, xo[k])
dx2 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx1, co, 0.5*(xo[k]+xo[k+1]))
dx3 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx2, co, 0.5*(xo[k]+xo[k+1]))
dx4 = dt*F(t[k]+ dt, x[k]+ dx3, co, xo[k+1])
x[k+1]=x[k]+(dx1+2*dx2+2*dx3*dx4)/6
Variant 3: To get an error that goes down in 4th order with the step size, you need to integrate the other objects with double the rate and half the step size (or find an order 4 interpolation for the data) so that the loop then reads as
dx1 = dt*F(t[k] , x[k] , co, xo[2*k])
dx2 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx1, co, xo[2*k+1])
dx3 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx2, co, xo[2*k+1])
dx4 = dt*F(t[k]+ dt, x[k]+ dx3, co, xo[2*k+2])
x[k+1]=x[k]+(dx1+2*dx2+2*dx3*dx4)/6
Variant 4: Of course, having a piecewise polynomial interpolation xo_interp(t) of satisfying accuracy for the external data and using that same interpolation function for all step sizes in the RK4 experiment should also restore the 4th order
dx1 = dt*F(t[k] , x[k] , co, xo_interp(t[k]))
dx2 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx1, co, xo_interp(t[k]+0.5*dt))
dx3 = dt*F(t[k]+0.5*dt, x[k]+0.5*dx2, co, xo_interp(t[k]+0.5*dt))
dx4 = dt*F(t[k]+ dt, x[k]+ dx3, co, xo_interp(t[k+1]))
x[k+1]=x[k]+(dx1+2*dx2+2*dx3*dx4)/6
• Ok, so it seems I have at least two problems actually: the one which causes the order 1 problem when I consider a perturbing object and the one that causes the spikes. I modified the code in order to solve the first one as explained in your answer and now $F_h$ is better, meaning that at least the flat parts of the $F_h$ plot are around $16$, using variant 3. But why xo is "seen" as order $1$ if I evaluated it with RK4? In other words, why does variant $1$ give me an order $1$ x approsimation if xo was evaluated using RK4?
– Zebx
Mar 3, 2021 at 13:58
• Because the step function $x_o([t/dt]·dt)$ has error $O(dt)$ against $x_o(t)$, the difference is a saw-tooth function with height $\simeq x_o'(t)dt$. This gives an error $O(dt^2)$ in each step, which accumulates to an error of $O(t·dt)$ over the course of the integration. Similarly one knows that $\frac{x_o(t+dt)+x_o(t)}2-x_o(t+\frac12 dt)\simeq \frac12x_o''(t)dt^2$, so that variant 2 gives an error of $O(dt^2)$ in each step against a step with higher-order exact $x_o$ values. Mar 3, 2021 at 14:05
• Oh ok, I wasn't considering the error accumulation over time. Thank you very much!
– Zebx
Mar 3, 2021 at 20:57 | 2,224 | 7,103 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-21 | longest | en | 0.893947 |
https://www.longevitas.co.uk/site/informationmatrix/forecastingwithcohortsinaclosedportfolio.html | 1,542,294,492,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742779.14/warc/CC-MAIN-20181115141220-20181115163220-00082.warc.gz | 935,946,783 | 7,945 | ## Forecasting with cohorts for a mature closed portfolio
At a previous seminar I discussed forecasting with the age-period-cohort (APC) model:
$$\log \mu_{i,j} = \alpha_i + \kappa_j + \gamma_{j-i}$$
where $$\mu_{i,j}$$ is the force of mortality at age $$i$$ in year $$j$$; the parameters $$\alpha_i$$ , $$\kappa_j$$ and $$\gamma_{j-i}$$ are usually thought of as the age, period and cohort parameters respectively. These parameters are not identifiable from the formula above, so we need a set of constraints to fix unique estimates. Suppose we number the cohorts from 1 (oldest) to $$n_c$$ (youngest). One set of constraints is as follows:
$$\sum \kappa_j = 0,\; \sum \gamma_c = 0,\; \sum c \gamma_c = 0.$$
Forecasting the mortality table now depends on forecasting the estimates of the period and cohort parameters. During the seminar I pointed out some difficulties with forecasting the cohort parameters, specifically the high correlation between parameters which are projected as though they are independent. This prompted the following question from the audience:
"Suppose we have a mature annuity portfolio closed to new entrants. Do we need to forecast the cohort parameters as we have all the years of birth we need?"
We can illustrate how this might work with Office for National Statistics (ONS) data on males for years 1961–2010. Assume that all the pensioners in our portfolio are 65 or older, so we use the age range 65–95 and we seek a mortality forecast to the year 2040, i.e. up to thirty years for the members of the portfolio. Notice that we forecast the full 30 years for the age 65 members but, for example, only 5 years for the age 90 members.
Figure 1. Data and forecast regions: ONS male data, ages 65–95, years 1961–2010, years of birth 1866–1945. Closed portfolio with youngest members age 65. Forecast thirty years to 2040. Figure 1 shows the data and forecast regions.
Figure 1 tells us that we have estimates of all the cohort parameters that we require, while Figure 2 tells us why we want to avoid forecasting the cohort parameters: continuing the clear trend in the estimates of the cohort parameters quickly leads to absurd estimates of mortality; alternatively, forecasting the estimates with a zero-drift time series also seems highly implausible.
Figure 2. Estimates of the cohort parameters for ONS male data with ages 65–95, years 1961–2010
Now the way forward is clear! Fix the age and cohort parameters at their estimated values and forecast the period parameters.
Is there a catch? Suppose the above solution is proposed by actuary A, while, independently, actuary B has the following thoughts:
• B is worried that the estimates of the younger cohort parameters are based on very little data, eg, the cohort parameter for an annuitant age 65 born in 1945 is based on a single data point (and so in this cell the observed and fitted deaths are exactly equal). Actuary B solves this problem by using ages 55–95.
• B also sees little point in going back to 1961 where the mortality pattern is rather different from more recent times. Actuary B solves this problem by using data from 1971–2010.
Figure 3: Estimates of $$\alpha$$ by actuary A (red), by actuary B (blue)
Figure 4: Estimates of $$\kappa$$ by actuary A (red), by actuary B (blue)
Figure 5: Estimates of $$\gamma$$ by actuary A (red), by actuary B (blue)
At first sight, Figures 3, 4 and 5 look a little disconcerting. The parameter estimates vary enormously depending on which data set is being used. However, we can reconcile the two approaches quite easily. The difference in the estimates of
• $$\alpha$$ in Figure 3 is caused by actuary B excluding data with poorer mortality;
• $$\kappa$$ in Figure 4 is caused by the constraint $$\sum \kappa_j = 0$$ since actuary A centres around year 1985.5 while actuary B centres around year 1990.5;
• $$\gamma$$ in Figure 5 is caused by the constraints $$\sum \gamma_c = \sum c \gamma_c = 0$$ since actuary A centres around year of birth 1905.5 while actuary B centres around year of birth 1915.5
However, all is well since, for example, Actuary A estimates the log(mortality) of a male age 65 in 2010 as $$-4.34$$ while Actuary~B estimates it as $$-4.30$$. Indeed, the estimates of mortality made by A and B in 2010, the final year of observation, are very close for all ages. Thus both actuaries start their forecasts at the same values. From Fig 4 their forecasts of the period parameters are consistent and neither actuary needs to forecast age or cohort parameters. We conclude their forecasts of mortality will be consistent and the use of different data sets has not made a significant difference to the forecasts, as illustrated in Figure 6 (the plots for other ages show a similar consistency).
Figure 6: Forecast of log mortality by actuary A (red) and actuary B (blue) for the cohort aged 65 in 2010
Our conclusion is that forecasting with the age-period-cohort model in a closed portfolio does successfully avoid the need to forecast the cohort parameters. Forecasting the cohort parameters in Figure 5 is not something I would undertake with any degree of confidence!
Stuart McDonald
(Mar 28, 2014)
Thank you for this helpful blog post Iain. I agree with the rationale here - I see no reason why concerns about correlated parameters need force an actuary to rule out models with both cohort and period terms in the situation described.
In fact I believe it is possible to extend the arguement and apply it to open annuity portfolios when we are calculating measures that do not depend on assumptions about new business.
We might extend our ONS data so that we're including lives age 50 in 2010 (for consistency with your diagram). This would allow us to project with some confidence as far as the 1960 cohort.
Live born after 1960 (early 50s and younger) would not currently comprise material percentage of annuity liabilities - maybe 5% at most. So, perhaps with a small adjustment to reflect the additional uncertainty for the very young lives, an actuary may feel perfectly happy to use such a model to forecast mortality for the existing annuitants in an open portfolio.
Iain Currie
(Mar 31, 2014)
I always get very nervous when the subject of forecasting with a model with a cohort term is proposed so I welcome the idea that it is possible to forecast with such a model without the need to forecast future cohort parameters. Nevertheless, this approach requires the inclusion of younger ages and this of necessity alters the estimate of the period term; the forecast depends fundamentally on the estimated period term. In my blog, Actuary B extended the age range to a limited extent and we didn't seem to encounter a problem, in the sense that A and B obtained very similar forecasts. I don't think this is the same thing as saying one has carte blanche to extend the age range à volonté.
Stuart McDonald
(Apr 5, 2014)
Thanks Iain. This doesn't sound insurmountable then - in such a situation I might check the estimate of the period term for the extended range against that calculated using the ages I'm most interested in. If they aren't materially different (keeping in mind that even a small difference might compound up to something material over the projection period) then I can weigh up the slight change in the period estimate against my extra confidence in the cohort estimates (especially those with only a few observations) and decide how to proceed.
Matthew Smith, PLRe
(May 21, 2014)
Thanks Iain. If memory serves, I believe the quoted question was mine! The blog post (and subsequent one) is very clear and the conclusion is comforting despite the uncertainties and complications associated with projecting models including cohort effects.
My view would be that the advantages of working with models including cohort effects far outweigh the disadvantages associated with models that do not. This would be despite the additional complexity and care required with the former.
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Iain Currie is an Honorary Research Fellow in the School of Mathematical and Computer Sciences at Heriot-Watt University
##### APC in the Projections Toolkit
The Age-Period-Cohort (APC) model is one of a large number of stochastic projection models available in the Projections Toolkit. | 1,945 | 8,512 | {"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": 2, "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-2018-47 | longest | en | 0.871253 |
https://www.simscale.com/docs/content/simulation/model/OF_thermalModelsBoussinesq.html | 1,553,458,590,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203491.1/warc/CC-MAIN-20190324190033-20190324212033-00228.warc.gz | 897,820,347 | 4,860 | # OpenFOAM Thermophysical Models¶
When the Boussinesq approximation is ‘ON’ the fluid is considered to be incompressible. In that case the the density change is calculated based on the Boussinesq approximation. The fluid properties necessary for this approximation can be specified while selecting the material.
Thermophysical properties
## 1. Viscosity model¶
Viscosity model describes relation between viscosity and strain rate. Different viscosity models are available along with the most widely used Newtonian model.
## 2. Thermal expansion coefficient¶
The Thermal expansion coefficient is the ratio a material expands with respect to change in its temperature.
## 3. Reference temperature¶
The temperature at which the kinematic viscosity is calculated.
## 4. Laminar Prandtl number¶
Prandtl number is the dimensionless parameter defined as the ratio of momentum diffusivity to thermal diffusivity.
$Pr = \frac{\nu }{\alpha }= \frac{c_{p}\mu }{k }$
$$\nu$$ is momentum diffusivity
$$\alpha$$ is thermal diffusivity
$$c_{p}$$ is specific heat
$$\mu$$ is dynamic viscosity
$$k$$ is thermal conductivity
## 5. Turbulent Prandtl number¶
Turbulent Prantl number is the ratio of eddy momentum diffusivity to thermal eddy diffusivity.
$Pr_{t} =\frac{\varepsilon _{M}}{\varepsilon _{H}}$
Where $$\varepsilon _{M}$$ is the eddy momentum diffusivity and $$\varepsilon _{H}$$ is eddy thermal diffusivity.
## 6. Specific heat¶
Specific heat (Cp) is the amount of heat required to increase the temperature of a unit mass by one degree Celsius. | 388 | 1,553 | {"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} | 2.9375 | 3 | CC-MAIN-2019-13 | latest | en | 0.79924 |
https://frederic-wang.fr/thoughts-on-the-dihedral-hidden-subgroup-problem-part-2.html | 1,701,646,624,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00072.warc.gz | 310,328,106 | 3,826 | # Thoughts on the Dihedral Hidden Subgroup Problem (part 2)
In a previous post, I've proposed a way to solve the Dihedral Hidden Subgroup Problem using a superposition of oracle evaluations. However, it seems to require a way to inverse the functions $x↦f\left(x,.\right)$ which is a bit cheating since we do not have this feature for the classical problem. In this post, we will consider the case $M=\frac{N}{N\text{'}}=2$, $d=2d\text{'}+b$ and try to determine the parity $b$ of $d$. If $N$ is a power of two, this is enough to find $d$ recursively. As in the previous post, the main idea is to consider the superpositions
$\mid {\psi }_{0}⟩=\frac{1}{\sqrt{N\prime }}\sum _{i=0}^{N\prime -1}\mid f\left(2i,0\right)⟩$
and
$\mid {\varphi }_{0}⟩=\frac{1}{\sqrt{N\prime }}\sum _{i=0}^{N\prime -1}\mid f\left(2i,1\right)⟩$
Using the equality $f\left(g\left(d,1\right)\right)=f\left(g\right)$ we can rewrite the latter state
$\mid {\varphi }_{0}⟩=\frac{1}{\sqrt{N\prime }}\sum _{i=0}^{N\prime -1}\mid f\left(2i-b,0\right)⟩$
If $b=0$, the latter state is the same superposition as the former state, over $N\text{'}$ values of ${ℤ}_{N}$. Otherwise, $b=1$ and the latter state is the superposition over the $N\text{'}$ other values of ${ℤ}_{N}$. We let ${f}_{i}^{b}=f\left(2i-b,0\right)$ and measure the product state $\mid {\psi }_{0}⟩\mid {\varphi }_{0}⟩$ in the basis ${\mid x⟩\mid x⟩}_{x\in {ℤ}_{N}}$, ${\mid x⟩\mid y⟩+\mid x⟩\mid y⟩}_{x, ${\mid x⟩\mid y⟩-\mid x⟩\mid y⟩}_{x . It turns out that this allows to distinguish the two cases:
• If $b=0$, the product contains vectors with the same coordinates $\mid {f}_{i}^{0}⟩\mid {f}_{i}^{0}⟩$ for $i\in {ℤ}_{N}$. For the other vectors, we have a symmetry between the two coordinates. Hence we can group them by pairs $\mid {f}_{{i}_{1}}^{0}⟩\mid {f}_{{i}_{2}}^{0}⟩+\mid {f}_{{i}_{2}}^{0}⟩\mid {f}_{{i}_{1}}^{0}⟩$ for ${i}_{1}<{i}_{2}\in {ℤ}_{N}$. Finally, the product state belongs to the space spanned by ${\mid x⟩\mid x⟩}_{x\in {ℤ}_{N}}$ and ${\mid x⟩\mid y⟩+\mid x⟩\mid y⟩}_{x.
• If $b=1$, the product contains vectors $\mid {f}_{{i}_{1}}^{0}⟩\mid {f}_{{i}_{2}}^{1}⟩$ where the two coordinates are not equal. We never have two symmetric vectors i.e. which are the same after permutating the two coordinates. Moreover, each vector $\mid {f}_{{i}_{1}}^{0}⟩\mid {f}_{{i}_{2}}^{1}⟩$ is the expression of a sum/difference (according to the respective order of ${f}_{{i}_{1}}^{0}$ and ${f}_{{i}_{2}}^{1}$) of two vectors $\mid x⟩\mid y⟩+\mid x⟩\mid y⟩$ and $\mid x⟩\mid y⟩-\mid x⟩\mid y⟩$. So the product state belongs half to ${\mid x⟩\mid y⟩+\mid x⟩\mid y⟩}_{x and half to ${\mid x⟩\mid y⟩-\mid x⟩\mid y⟩}_{x.
Suppose that we have a procedure to create many states $\mid {\psi }_{0}⟩$ and $\mid {\varphi }_{0}⟩$. We measure a state in the space ${\mid x⟩\mid y⟩-\mid x⟩\mid y⟩}_{x with probability $\frac{b}{2}$. So repeating the procedure a constant number of times allow to determine $b$ with high probability.
The only gap in the previous algorithm is whether we can create the states $\mid {\psi }_{0}⟩$ and $\mid {\varphi }_{0}⟩$. As in the previous post, this is possible if the gates ${V}_{f,.}$ are available. Otherwise, we can create a superposition of states over ${ℤ}_{N\text{'}}$ and use the gate ${U}_{f}$ as well as a Quantum Fourier Transform to randomize the first coordinate. We get two random numbers ${j}_{1},{j}_{2}\in {ℤ}_{N\text{'}}$ and the states
$\mid {\psi }_{{j}_{1}}⟩=\frac{1}{\sqrt{N\prime }}\sum _{i=0}^{N\prime -1}{e}^{\frac{2i\pi {j}_{1}i}{N\prime }}\mid f\left(2i,0\right)⟩$
and$\mid {\varphi }_{{j}_{2}}⟩=\frac{{e}^{\frac{2i\pi {j}_{2}d\prime }{N\prime }}}{\sqrt{N\prime }}\sum _{i=0}^{N\prime -1}{e}^{\frac{2i\pi {j}_{2}i}{N\prime }}\mid f\left(2i-b,0\right)⟩$
Next, we can apply the same measurement as above. After a bit of calculation,we get the probabilities:
Space case $b=0$ case $b=1$
${\mid x⟩\mid x⟩}_{x\in {ℤ}_{N}}$ $\frac{1}{N\text{'}}$ 0
${\mid x⟩\mid y⟩+\mid x⟩\mid y⟩}_{x $\frac{2}{{N\prime }^{2}}\sum _{{i}_{2}=0}^{N\prime -1}\sum _{{i}_{1}=0}^{{i}_{2}-1}{\mathrm{cos}}^{2}\frac{\pi }{N\prime }\left({j}_{2}-{j}_{1}\right)\left({i}_{2}-{i}_{1}\right)$ $\frac{1}{2}$
${\mid x⟩\mid y⟩-\mid x⟩\mid y⟩}_{x $\frac{2}{{N\prime }^{2}}\sum _{{i}_{2}=0}^{N\prime -1}\sum _{{i}_{1}=0}^{{i}_{2}-1}{\mathrm{sin}}^{2}\frac{\pi }{N\prime }\left({j}_{2}-{j}_{1}\right)\left({i}_{2}-{i}_{1}\right)$ $\frac{1}{2}$
If we repeat the procedure several times, we need to take the means over the choice of ${j}_{1},{j}_{2}\in {ℤ}_{N\text{'}}$. Using some trigonometric identities, it is possible to write the sums of the first column $\frac{1}{2}+S+o\left(\frac{1}{N\text{'}}\right)$ where $S$ is a sum of cosinus. Unfortunately, evaluating $S$ with a computer suggests that $S=\Theta \left(\frac{1}{N\text{'}}\right)$ . Hence, we can not distinguish the two cases with only a polynomial (in $\mathrm{log}N$) number of samplings.
It is still open whether we can find a variation to make such an algorithm efficient... | 1,834 | 5,005 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 67, "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} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.761157 |
http://tptp.org/cgi-bin/SeeTPTP?Category=Problems&Domain=KLE&File=KLE038+1.p | 1,638,757,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00006.warc.gz | 84,716,426 | 1,676 | ## TPTP Problem File: KLE038+1.p
View Solutions - Solve Problem
```%------------------------------------------------------------------------------
% File : KLE038+1 : TPTP v7.5.0. Released v4.0.0.
% Domain : Kleene Algebra
% Problem : Star extensivity
% Version : [Hoe08] axioms.
% English : Star is extensive.
% Refs : [Koz94] Kozen (1994), A Completeness Theorem for Kleene Algebr
% : [Hoe08] Hoefner (2008), Email to G. Sutcliffe
% Source : [Hoe08]
% Names :
% Status : Theorem
% Rating : 0.31 v7.5.0, 0.38 v7.4.0, 0.17 v7.3.0, 0.31 v7.2.0, 0.34 v7.1.0, 0.26 v7.0.0, 0.33 v6.4.0, 0.35 v6.3.0, 0.33 v6.2.0, 0.44 v6.1.0, 0.43 v6.0.0, 0.39 v5.5.0, 0.52 v5.4.0, 0.54 v5.3.0, 0.59 v5.2.0, 0.50 v5.1.0, 0.52 v5.0.0, 0.58 v4.1.0, 0.52 v4.0.1, 0.48 v4.0.0
% Syntax : Number of formulae : 17 ( 14 unit)
% Number of atoms : 20 ( 12 equality)
% Maximal formula depth : 5 ( 3 average)
% Number of connectives : 3 ( 0 ~; 0 |; 0 &)
% ( 1 <=>; 2 =>; 0 <=; 0 <~>)
% ( 0 ~|; 0 ~&)
% Number of predicates : 2 ( 0 propositional; 2-2 arity)
% Number of functors : 5 ( 2 constant; 0-2 arity)
% Number of variables : 31 ( 0 sgn; 31 !; 0 ?)
% Maximal term depth : 4 ( 2 average)
% SPC : FOF_THM_RFO_SEQ | 611 | 1,467 | {"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} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.324015 |
http://temposchlucker.blogspot.com/2012/04/few-answers-to-bright-knight.html | 1,532,042,713,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00237.warc.gz | 345,538,531 | 23,349 | ## Friday, April 27, 2012
### A few answers to Bright Knight
BK said:
It is not at all clear that it is easier to improve by a fixed percentage on harder problems than it is on easier ones. Back in the days of the Bain Experiment, I found that reducing my solution times by a fixed percentage for my first pass through the first batch of problems gave a good fit to my solution times on the first passes through the later batches of problems. That appears to be roughly right for Woolum too.
I advocate to train every specific task that you have to perform in a specialized way. Optimized for that specific task. Preventing diminishing returns.
The first thing to do is to determine which tasks we are talking about. Every tasks has its own specialized sets of patterns associated with it, and if you don't find out which patterns to learn, chances are high that you don't learn these patterns at all.
The first task we have identified is that we have 20 - 30 elementary tactical themes, depending on the source. Let's round it off to 20 usefull themes. In my opinion these are trained best according to my speedtraining method:
Make per tactical element a separate problemset with rating <1200. Solve these problems as fast as possible, untill you reach a speed of <4 seconds per problem. 1000x per tactical element is probably enough to recognize it everywhere and fast. Let's say a problem takes 20 seconds at average. This leads to a total time of 20 elements x 1000 times x 20 seconds = 111 hours to master this task.
Doing the same with combinations I consider to be useless, since there are way too much combinations.
Doing problems higher rated than 1200 is useless for this task. Here the law of diminishing results sets in soon.
I identified duplo attacks as one of the three ways to gain wood. Traps and promotion being the other two. This means that you have to scan a position for all its duplo attacks. There are 5 different duplo attacks (double attack, discovered attack, pin, skewer, rontgen attack). The recognition itself of these 5 patterns is already taken care of when training for the first task. You only have to make it a habit to scan for all of these duplo attacks in stead of stopping short once you get distracted by one of them which looks promising.
I use high rated problems for this for a starter. Once I have identified all duplo attacks I store both the position and the duplo attacks as a flashcard in Anki and train them further with the aid of Anki, without going back to CT anymore.
Third task: making one of the duplo attacks work.
At average there are about 6 duplo attacks in a >2000 rated problem. Only one of them is winning. At this stage is is my task to find out which one and how I can make it work. For this I need all tactical elements that I have learned already in task one and that can act as some prelimanary move. Besides that, some conscious thinking is probably needed here in order to make it work.
Fourth task: identifying the focal squares.
Once I know the target, I must find the way back to the attacker, finding the focal points along the road. These focal points give me clues to the preliminary moves I ave to make before the combination will work. I must identify the defenders of the focal points and find a way to annihilate them. I must identify where my own pieces get in the way and clear them, etc..
I use high rated problems for a starter. Once I have identified the focal points I make a flashcard with the position and the focal points. From there I train it with Anki.
Fifth task: learning the mating patterns.
This has great similarity with task one.
If I'm not mistaken there are 29 different mating patterns that must be learned. Speedtraining with <1200 rated problems will do the job.
Sixth task: learning the patterns that are associated with promotion.
This has great similarity with task one.
Speedtraining with <1200 rated problems will do the job.
I just discovered this as a problem. It has its own associated patterns. First I must formulate a search algorithm. In order to find one I use high rated problems at CT which are mates. Usually> 4 movers.
Only once I have found a suitable algorithm, I can start to think about a method how to train them.
To give you an idea of the complexity of the matter: a search algorithm must guide you through this testposition in order to be usefull.
Under construction. See previous posts.
In summary:
Speed training with tagged low rated problems is suitable for most elementary patterns. High rated problems are used in two cases:
• As a starter to extract patterns that you will only find in complex problems. The actual learning is done with flashcards and Anki.
• During investigation. When you are investigating new tasks and defining new search algorithms. This is only temporarily.
It is true that you have an average of about 3 minutes per move in a standard time limit game, but how much time can you afford to spend on checking that your chosen move is proof against a tactical shot? I would have thought that 1 minute would be optimistic. Hoping for a sixty fold improvement in your solution time is too optimistic in my opinion.
That's all way too ridgid and too theoretical. We will find this out in an empyrical way.
What failure rate is acceptable when checking for tactical shots? 10%, 1% or less… There are typically about 40 moves in a game, and many or most of them could fall foul of tactical shots. If you accept a 10% failure rate, you are going to “blunder” once or more in most games. Even if you get your failure rate down to 1%, you will regularly lose to tactical shots.
Again way too theoretical. Not all positions call for an ability to solve tactics at 2300 level. But sometimes tactics just must work or you loose. If I look at the Polgar middlegame brick, I could flaw about 25% of the tactics that were presented there with the aid of Rybka. This means that both the grandmasters who played the game, the grandmasters that annotated the game in the Informator and Susan Polgar who checked the problems for the book overlooked the flaws. When both you and your opponent miss the same tactic, the tactic is irrelevant for the game.
1. The 80% correlation between blitz and standard time limit performance also suggests that both are equally difficult to improve beyond a certain point.
I meant the last two paragraphs to follow sequentially, rather than to be taken in isolation. Yes, difficult tactics that neither player spots are practically irrelevant. The point I was making is that what is most important is not to fall to simple tactical shots, and to capitalise on them when they are available to you. High success rate in 1 minute implies easy problems.
Diminishing returns are inevitable, but there is more to chess than tactical problems!
How did the top players get that good? In the past, there were no computers, so the answer was simply study books and play chess. There were very few problem books available in days gone by. How did Morphy train?
2. @BK,
The point I was making is that what is most important is not to fall to simple tactical shots, and to capitalise on them when they are available to you. High success rate in 1 minute implies easy problems.
To become a master, you need to be able to solve masterlevel problems with a succesrate of 50%.
The fact that you are able to solve low rated problems with a succesrate of >99% doesn't change that.
That what is typical for high rated problems can't be learned from low rated problems since the patterns aren't there.
How did Morphy train?
I don't know. But even if I did, I question that it is relevant. Since we are adults and he was young when he learned the game.
3. I am not saying that you should train only on the very easiest stuff, just that mastery of the easy stuff has more value than an incomplete understanding (or misunderstanding) of the complicated stuff. Learn to walk before you try to run! To become a master I have to learn lots of things. That takes time. Pick the low lying fruit first! I believe that the learning process is the same young or old. Morphy became a very strong tactical player with a few books written by weaker players and practice against weaker players. I am not saying that is optimal, but it worked.
4. I consider the readers of this blog to be ready with harvesting low hanging fruit. Otherwise I suggest to harvest first the low hanging fruit:)
5. Here is a Dan Heisman article that puts the case for easy problems well:
http://www.chesscafe.com/text/heisman109.pdf
Coakley emphasises the importance of training for accuracy rather than speed, but the two can be traded off to some extent. One way of getting more accurate is to remember things rather than having to work them out, which benefits both.
If it is obvious that tactics are in the air, you may be able to spend several minutes looking for them in a game. With “bolt from the blue “ tactics, you will be doing well if you can afford to look for a minute, particularly if you are on the receiving end. I believe that it is fine to routinely spend a few minutes on a problem. Solving a problem is more instructive than being told the solution. Much more than that? Occasionally, perhaps.
6. Both Morphy and Anderssen, huge tacticians back in their day, created tactical puzzles, wrote columns and went over games.
To create your own tactical problem from scratch is interesting because you have to have deep knowledge of the themes that you spell out in this post as well as "know" what makes a strong combination to challenge varying degrees of readers.
7. Given lately that my life takes on certain tangents that takes me away from the chess board, I have been deliberate in training in ways to minimize skill loss and still find a way to improve albeit sporadic intervals and levels of chess commitment. So, here I am coming back from an intense 5 month hiatus with absolutely no chess involvement. I am looking at my re-entry as an experiment in retention of skills from a cognitive perspective as much as possible as well as identify which are I lost the most and come up with a plan to “rust proof” that area. In reality it’s been about 8 months since I’ve really indulged heavily in the chess improvement circles and resurfaced briefly during Holiday breaks.
I am looking at the following areas of where my skills started to disappear: 1) tactics 2)Positional 3) Openings, and 4) endgames. I will save more of the details for a post on my blog.
Your post made me ponder the different layers of item 1, tactics, and where I may have lost the most.
Because of the past MDLM training and the unfocused 7 circles of hell , the best that came about from all that thrashing was the rudimentary tactics are still etched in my brain. After not looking at a chess board for months, anything beyond what you call “First Task” was challenging. I recognized from the start why this was the case. When I stormed the seven circles, I never took the time to create that mental dialog for the basic elements. “That one is a removal of the guard” “This one is a geometrical motif with a knight” etc. So as I “warm up” this time with the first task, I am making sure I do the deliberate mind mapping necessary to transfer the skill from a memory function to an actual skill in long term memory with a memory marker that I can recall through us of my own narrative.
I did this with positional motifs as a means to understanding my opening repertoire better and forcing myself to let the position dictate candidate moves than rely solely on rote memorization of move orders. It helped back in December when after not playing for several months, I was still able to play a decent game and not drop (much) in skill level.
So if I can apply this technique of creating memory maps through the right cognitive internal dialog in the four areas, then I stand a better chance of developing “chess skills” that are meant to last.
8. @BP,
I look forward to your post.
With hindsight, the path is pretty obvious. Strange that it takes so long to see it clear.
9. "How did Morphy train?"
He read over annotated games, sitting in his gazebo, with no board in sight.
I saw a short clip of Magnus Carlson at London. Hikaru moves, Magnus writes down the move and stares past his scoresheet, then makes a move on the board. Clearly he is playing the game blindfold and barely glanced at the board just to make the move.
10. Who really knows how Morphy trained. One thing is for sure Morphy probably would have killed to have the resources we have today to learn from. I'm a bit skeptical that all this defining of tactical problems will really do all that much for you. Just notice little things like open king position, pinned piece, weak square, that sort of thing and you'll spot tactics quickly. Whatever works for you though, but I'd much rather look at more Botvinnik games where there's plenty of tactics and awesome positional instruction...
11. Who really knows how Morphy trained. One thing is certain; he would have killed to have the resources we have today. | 2,822 | 13,132 | {"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-2018-30 | latest | en | 0.960634 |
http://www.singular.uni-kl.de/Manual/4-0-2/sing_1670.htm | 1,513,621,842,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948619804.88/warc/CC-MAIN-20171218180731-20171218202731-00306.warc.gz | 464,176,131 | 4,449 | # Singular
#### D.8.2.4 elimpartanyr
Procedure from library `presolve.lib` (see presolve_lib).
Usage:
elimpartanyr(i [,p,e] ); i=ideal, p=polynomial, e=integer
p: product of vars to be eliminated,
e =0: substitute from linear part of i (same as elimlinearpart)
e!=0: eliminate also by direct substitution
(default: p=product of all vars, e=1)
Return:
list of 6 objects:
``` [1]: (interreduced) ideal obtained by substituting from i those vars appearing in p, which occur in the linear part of i (or which can be expressed directly in the remaining variables, if e!=0) [2]: ideal, variables which have been substituted [3]: ideal, i-th element defines substitution of i-th var in [2] [4]: ideal of variables of basering, substituted ones are set to 0 [5]: ideal, describing the map from the basering, say k[x(1..m)], to itself onto k[..variables fom [4]..] and [1] is the image of i [6]: int, # of vars considered for substitution (= # of factors of p) ```
The ideal i is generated by [1] and [3] in k[x(1..m)], the map [5] maps [3] to 0, hence induces an isomorphism
``` k[x(1..m)]/i -> k[..variables fom [4]..]/[1] ```
Note:
the procedure uses `execute` to create a ring with ordering dp and vars placed correctly and then applies `elimpart`.
Example:
```LIB "presolve.lib"; ring s=0,(x,y,z),dp; ideal i = x3+y2+z,x2y2+z3,y+z+1; elimpartanyr(i,z); ==> [1]: ==> _[1]=x3+y2-y-1 ==> _[2]=x2y2-y3-3y2-3y-1 ==> [2]: ==> _[1]=z ==> [3]: ==> _[1]=y+z+1 ==> [4]: ==> _[1]=0 ==> _[2]=x ==> _[3]=y ==> [5]: ==> _[1]=-y-1 ==> _[2]=x ==> _[3]=y ==> [6]: ==> 1 ``` | 519 | 1,561 | {"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} | 2.890625 | 3 | CC-MAIN-2017-51 | latest | en | 0.827372 |
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Posted: Sunday 28th of Jul 18:53 I remember having often faced difficulties with factoring expressions, graphing and absolute values. A truly great piece of math program is Algebrator software. By simply typing in a problem from workbook a step by step solution would appear by a click on Solve. I have used it through many math classes – Algebra 2, Algebra 1 and Remedial Algebra. I greatly recommend the program.
daujk_vv7
Registered: 06.07.2001
From: I dunno, I've lost it.
Posted: Tuesday 30th of Jul 13:05 Okay. You can get it here: (softwareLinks). Further more they offer you definite satisfaction and a money-back guarantee. Bye Bye to your worries. | 961 | 3,917 | {"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} | 2.75 | 3 | CC-MAIN-2019-43 | longest | en | 0.915807 |
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# Ascorboc acid also known as vitamin C contains the elements
This answer was rated:
Ascorboc acid also known as vitamin C contains the elements carbon, hydrogen and oxygen it's know to contain 40.9% carbon and 4.58% hydrogen by mass
A. What is the empirical formula of vitamin C
B. A mass spectrometry study of vitamin C indicates that it's molar mass is between 170 and 190 Amy what is the molecular formula for vitmamim C?
Owen :
Hello
Owen :
The empiracal formula for Vitamin C is C3H4O3 and the Molecular Formular is C3(2)H4(2)O3(2) which is C5H8O6, and the molecular mass is 176.
Owen :
I hope this helps
Owen :
Please click on EXCELLENT Service (smile face)
Owen :
thank you
Owen :
I see I put C5H8O6 - -----change that to C6H8O6
Owen :
Total mass of C3H4O3 = (12 x 3) + (4 x 1) + (16x 3) = 88 and hence with the Molecular formula of C3(2)H4(2)O3(2) = C6H8O6*1 + 16*3 = 88
176 / 88 = 2
Therefore molecular formula =
Owen :
Then 88 x 2 = 176
Owen :
I hope this helps
Owen :
Do you have any more questions?
Owen :
Carbon = C = 12
Owen :
Oxygen = 16
Owen :
Hydrogen = 1
Owen :
Did you have any additional questions?
Customer : It's a two part question?
Owen :
You asked for the Empiral formula - that is C3H4O3
Owen :
that was part one
Owen :
Then part B - the question was what is the Molecular formula - that is C3(2)H4(2)O3(2) = C6H8O6
Owen :
Then to calculate the molar mass
Owen :
here is the work for that:
Customer : The numbers in the question are all subscripts
Owen :
Yes they are subscripts
Owen :
Sorry I can not put that here on this website like that
Owen :
subscripts do not work on the website - I wish they did
Owen :
Do you have any more questions?
Owen :
Hello - I'm still here
Owen :
Are you there?
Owen :
:)
Owen and 2 other Homework Specialists are ready to help you | 650 | 2,140 | {"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} | 3.046875 | 3 | CC-MAIN-2017-43 | latest | en | 0.916754 |
https://goobi.tib.eu/viewer/fulltext/856566209/157/ | 1,723,099,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00380.warc.gz | 219,518,858 | 36,369 | # Full text: The 3rd ISPRS Workshop on Dynamic and Multi-Dimensional GIS & the 10th Annual Conference of CPGIS on Geoinformatics
```144
ISPRS, Vol.34, Part 2W2, “Dynamic and Multi-Dimensional GIS”, Bangkok, May 23-25, 2001
Fig.3 Distribution Fig.4 Fiducials (a)
(b)
(c)
(d) Fig.5 Similar Object
Table 3 the candidates of each fiducial mark
Fiducial
X (pixel)
Y (pixel)
Correlation
coefficient
1
1
2935
205
0.47895
2
2935
169
0.42382
3
2886
205
0.41220
2
4
67
1598
0.55935
5
68
1655,
0.43236
6
67
1550
0.41243
3
7
2965
4453
1.00000
8
2965
4420
0.69096
9
2965
4508
0.69037
4
10
5817
1560
0.28147
11
5817
1609
0.22221
12
5918
1582
0.21994
Fig.4(a) is a good image of fiducial mark. Now one object similar
to the fiducial mark is added to generate the image in Fig.5. The
new image is used to test the algorithm. The candidates of
fiducial mark 3 are listed in Table 5. The point number of added
object is 8, whose correlation coefficient is also large. The
parameters of the algorithm are set the same as before. After
T<T ( „ the correspondence matrix is got as listed in Table 6. A
correct one-to-one correspondence is got.
Table 5 the candidates of the third fiducial mark
Fiducial
X (pixel)
Y (pixel)
Correlation
coefficient
3
7
2965
4453
0.99911
8
2913
4408
0.99513
9
2965
4408
0.70765
4 CONCLUSION
The interior orientation becomes much harder when fiducial
marks merge in image of objects. The self-adaptive algorithm
performs quite well without the usage of correlation coefficient.
ACKNOWLEDGE
This paper is supported by the National Natural Science
Foundation of China (Grand number: 40023004, 40001018)
Table 4 the correspondence matrix of real image experiment
m ai
1
2
3
4
5
6
7
8
9
10
11
12
Outlier
1
0.463
0.299
0.237
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
0.000
0.000
0.482
0.058
0.460
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
0.000
0.000
0.000
0.000
0.499
0.316
0.185
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.551
0.435
0.014
0.000
Outlier
0.537
0.071
0.763
0.518
0.942
0.540
0.501
0.684
0.815
0.499
0.565
0.986
Table 6 the correspondence matrix of the experiment with outlier
m a \
1
2
3
4
5
6
7
8
9
10
11
12
Outlier
1
0.376
0.298
0.326
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
0.000
0.000
0.408
0.184
0.407
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
0.000
0.000
0.000
0.000
0.393
0.242
0.365
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.502
0.372
0.126
0.000
Outlier
0.624
0.702
0.674
0.592
0.816
0.593
0.607
0.758
0.635
0.498
0.623
0.874
REFERENCES 1
(1) . Zhang Z., Zhang J.. 1996. Digital Photography. Press of
wtusm
(2) . Christian Heipke 1997 .Automation of interior , relative ,
and absolute orientation .ISPRS Journal of
Photogrammetry & Remote Sensing 52(1997)1-19
(3) . Kersten T and Haring T.1995 Automatic interior
orientation Internal report, ETH Zurich
(4) . Lue,Y, 1995. Fully operational automatic interior orientation
Proceedings Geolnformatics’95,HongKong,May
26-28,Vol-1,pp.26-35
(5) . Schcker.W. 1995. Ein operationeiles Verfahren zur
automatischen inneren Orientierung von
luftbildern .Z.Photogramm,fernerkundung,63(3):115-122
(6) . Strackbein,H.and enze.M.,1995.Automatische innere
Orientierung.Internal report .Landesvermessungsamt
Nordrhein Westfalen
(7) . S.Gold, A. Rangarajan, C. P. Lu, S Pappu, and E. Mjolsness.
New algorithms for 2-D and 3-D point matching : pose
estimation and correspondence . Pattern Recognition,
31 (8): 1019 1031, 1998
(8) . S.Gold, and A. Rangarajan. A graduated assignment
algorithm for graph matching. IEEE Trans. Patt. Anal.
Mach. Intell., 18(4):377-388,1996.
(9) . Haiti Chui and Anand Rangarajan 2000 A new algorithm for
non-rigid point matching , IEEE Conference on Computer
Vision and Pattern Recognition (CVPR),
(10) . Girosi, F.,Jones,M.,and Poggio,T.(1995). Regularization
theory and neural net work architectures. Neural
Computation, 7:219-269
(11) . Papadimitriou, C. and Rangarajan, K. (1982).
Combinatorial Optimization. Prentice-Hall,Inc.,Englewwood
Cliffd.NJ.
(12) . Gold,S.,Rangarajan,A.(1996). Softmax to softassign:
neural network algorithms for combinatorial optimizations
Journal of Artificial Neural Networks, in press
```
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3 Subtracting fractions with unlike denominators mazes
*** This product is included in the adding and subtracting fractions with unlike (multiple) denominators bundle ***
**This is also included in the adding and subtracting mazes bundle (27 included) **
This product includes 3 unique mazes with full solutions.
This is a more engaging way of practicing subtracting fractions with unlike denominators (where one denominator is a multiple of the other).
This activity is great because students can self-check the answers and most of the time they don’t realize they are answering multiple questions.
Students work their way through the maze, coloring the correct answers and following the path to the next question. This activity is color coded gold (medium).
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Number Series Questions With Solution For SBI Clerk.If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude. If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude.
Today, we are providing you with Quant Quiz on Number Series based on the latest pattern for your daily practice. This Quant subject quiz based number series is important for other banking exams such as SBI Clerk, SBI PO, IBPS PO, IBPS Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, and other competitive exams.
## Number Series Questions With Solution Set – 14
1. What should come in place of the question mark (?) in the following number series?
20, 80, ?, 1800, 9900, 59400
A 160
B 360
C 540
D 800
E 120
Show Correct Answers
B. 360
The series the following pattern:
20×4 = 80
80×4.5 = 360
360×5 = 1800
1800×5.5 = 9900
9900×6 = 59400
2. What should come in place of the question mark (?) in the following number series?
8, 18, 32, 50, ?, 98
A 70
B 76
C 80
D 72
E 60
Show Correct Answers
D. 72
The series follows the following pattern:
42 ÷ 2 = 8
62 ÷ 2 = 18
82 ÷ 2 = 32
102 ÷ 2 = 50
122 ÷ 2 = 72
142 ÷ 2 = 98
3. What should come in place of the question mark (?) in the following number series?
11, 19, ? , 41, 55, 71
A 29
B 31
C 27
D 25
E 23
Show Correct Answers
A. 29
The series follows the following pattern:
5×4 – 9 = 11
6×5 – 11 = 19
7×6 – 13 = 29
8×7 – 15 = 41
9×8 – 17 = 55
10×9 – 19 = 71
4. What should come in place of the question mark (?) in the following number series?
111, 89, 122, 100, 133, ?
A 159
B 111
C 123
D 135
E 99
Show Correct Answers
B. 111
The series follows the following pattern:
111 – 22 = 89
89 + 33 = 122
122 – 22 = 100
100 + 33 = 133
133 – 22 = 111
5. What should come in place of the question mark (?) in the following number series?
11, 36, 72, ?, 185, 266
A 111
B 101
C 121
D 100
E 98
Show Correct Answers
C. 121
The series follows the following pattern:
11 + 52 = 36
36 + 62 = 72
72 + 72 = 121
121 + 82 = 185
185 + 92 = 266
6. What will come in place of the question mark (?) in the following series?
9, 10, 23, 86, 425, ?
A 2542
B 2438
C 2546
D 2571
E None of these
Show Correct Answers
C. 2546
The series follows the pattern,
9 × 2 – 8 = 10, 10 × 3 – 7 = 23, 23 × 4 – 6 = 86, 86 × 5 – 5 = 425, 425 × 6 – 4 = 2546.
7. What will come in place of the question mark (?) in the following series?
18, 69, 22, 65, 31, 56, 47, 40, ?
A 59
B 61
C 53
D 63
E 72
Show Correct Answers
E. 72
The numbers at the odd places follow the pattern,
18 + 22 = 22, 22 + 32 = 31, 31 + 42 = 47, 47 + 52 = 72
The numbers at the even places follow the pattern
69 – 22 = 65, 65 – 32 = 56, 56 – 42 = 40
8. What will come in place of the question mark (?) in the following series?
23, 25, 28, 33, 40, 51, ?
A 78
B 10
C 103
D 98
E None of these
Show Correct Answers
E. None of these
The series follows the pattern:
23 + 2 = 25, 25 + 3 = 28, 28 + 5 = 33, 33 + 7 = 40, 40 + 11 = 51, 51 + 13 = 64
9. What will come in place of the question mark (?) in the following series?
2, 0, 6, 12, 42, ?
A 120
B 132
C 100
D 96
E None of these
Show Correct Answers
A. 120
The series follows the pattern:
(2 – 2) × 3 = 0, (0 + 2) × 3 = 6, (6 – 2) × 3 = 12, (12 + 2) × 3 = 42, (42 – 2) × 3 = 120
10. What will come in place of the question mark (?) in the following series?
69, 89, 111, 135, 161, ?
A 187
B 183
C 194
D 189
E None of these
Show Correct Answers
D. 189
The series follows the pattern:
69 + 20 = 89, 89 + 22 = 111, 111 + 24 = 135, 135 + 26 = 161, 161 + 28 = 189
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Computer Awareness | 1,486 | 3,908 | {"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} | 4.40625 | 4 | CC-MAIN-2023-14 | latest | en | 0.822706 |
https://www.diystompboxes.com/smfforum/index.php?topic=67213.msg535660 | 1,656,696,614,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00283.warc.gz | 772,106,066 | 8,745 | ### Author Topic: PT80 delay: DC voltages of opamp question. (Read 1994 times)
#### flo
##### PT80 delay: DC voltages of opamp question.
« on: April 21, 2008, 08:45:48 PM »
The following voltges were posted for the opamp in the PT80 delay:
http://www.diystompboxes.com/smfforum/index.php?topic=66619.0
TL072
1. 6V
2. 6V
3. 5.9V
4. 0
5. 3V
6. 6V
7. 6V
8. 12V
This what I'm measuring on my PT80 also and the thing works.
But what's bothering me is this:
If pin 5 (+input of opamp B of the TL072) is at 3V, why are pin 6 (-input) and 7 (output) then at 6V?
I expected pin 5 to be at 6V (=Vbias) and pin 6 also. I don't get it...
#### flo
##### Re: PT80 delay: DC voltages of opamp question.
« Reply #1 on: April 23, 2008, 09:36:47 AM »
How can the opamp work ok if its + and - inputs are 3V apart?
The output should have been at 0V then, so why isn't it?
The + input (pin5) should have been at 6V (Vbias) so why is it 3V (measured by me in my PT80 and by others in their PT80s)?
#### Pushtone
##### Re: PT80 delay: DC voltages of opamp question.
« Reply #2 on: April 23, 2008, 12:39:57 PM »
Perhaps the pot locations matter? LFO? I don't have the schem up in front of me.
Why not email the circuit designer?
Or even the manufacturer. The circuit is just a modified version of the application in the data sheet.
Dave
It's time to buy a gun. That's what I've been thinking.
Maybe I can afford one, if I do a little less drinking. - Fred Eaglesmith
#### slacker
##### Re: PT80 delay: DC voltages of opamp question.
« Reply #3 on: April 23, 2008, 01:08:20 PM »
Is pin 5 the + input that is the input to the effect?
If it is then the reason it is reading 3 volts is because your DMMs impedance is affecting the reading. A lot of DMMs have an impedance of about 1Meg ohm so when you try and measure the voltage at pin 5 the impedance of the DMM forms a voltage divider with the 1M resistor and pulls the voltage down.
Here's a dodgy drawing to make it clearer, because I'm bad at explaining things
Point R 6volts ------1M---------Pin 5
|
|
1M impedance of DMM
|
|
Ground
The voltage on the pin is probably really 6 volts or as you've said the circuit wouldn't work. If you need to measure it accurately you can make a buffer out of a Jfet input opamp connect the probe to the input of that and take the output to your DMM. The buffer will have an impedance of 100s of Megs so it won't load the voltage down.
« Last Edit: April 23, 2008, 01:13:15 PM by slacker »
#### flo
##### Re: PT80 delay: DC voltages of opamp question.
« Reply #4 on: April 23, 2008, 05:23:22 PM »
Is pin 5 the + input that is the input to the effect?
Yes it is.
If it is then the reason it is reading 3 volts is because your DMMs impedance is affecting the reading.
That must be it: The impedance of the DMM.
My DMM and my oscilloscope probe both measure 3V so they probably both have a resistance around 1M.
So if I lower the 1M bias resistance that goes from Vbias to pin5 temporarily to 100k, I should measure 6V again...
And indeed! I just put a 100k resitance parallel over the 1M bias resistor and measured the voltage on pin5: 5.56V.
@slacker: Thanks a lot! The world of electronics makes sense again... | 973 | 3,197 | {"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-2022-27 | latest | en | 0.891454 |
https://artofproblemsolving.com/wiki/index.php/AoPS_Wiki_talk:Problem_of_the_Day/June_6,_2011 | 1,713,527,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00187.warc.gz | 102,071,980 | 9,443 | # AoPS Wiki talk:Problem of the Day/June 6, 2011
## Solution
Just rewrite the numbers over and over and get rid of every 9th:
1 2 3 4 5 6 7 8 9 Remove the 9
10 11 12 1 2 3 4 5 6 Remove the 6
7 8 10 11 12 1 2 3 4 Remove the 4
5 7 8 10 11 12 1 2 3 Remove the 3
5 7 8 10 11 12 1 2 5 Remove the 5
7 8 10 11 12 1 2 7 8 Remove the 8
10 11 12 1 2 7 10 11 12 Remove the 12
10 11 1 2 7 10 11 1 2 Remove the 2
7 10 11 1 7 10 11 1 7 Remove the 7
10 11 1 10 11 1 10 11 1 Remove the 1
10 11 10 11 10 11 10 11 10 Remove the 10
The last number to be destroyed is $\boxed{11}$. | 294 | 575 | {"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": 1, "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-2024-18 | latest | en | 0.610454 |
https://music.stackexchange.com/questions/11213/is-it-correct-to-say-that-all-common-chords-are-built-using-the-major-scales | 1,701,711,755,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00137.warc.gz | 453,075,306 | 46,290 | # Is it correct to say that all common chords are built using the major scales?
I'm creating an abbreviated music theory course for a guitarist buddy of mine. At this point, we're covering chord formulas, e.g., 1-3-5 in the C Major scale = C-E-G.
I found myself about to write a sentence (marked in bold below) in the lesson I'm working on:
"To determine the notes within a chord, we need two pieces of information: 1) the chord formula (e.g., 1-b3-5 for the minor chord), and 2) the key of the scale from which the chord is drawn (e.g., the C Major scale). All chords, by the way, are drawn from the major scales, versus from the minor, pentatonic, blues, or diminished scales.
Thus, the notes for the minor chord in the C Major scale are........"
• Even your example, of a minor tonic, requires a note not in the major scale. I recommend you grab a book on introductory music theory rather than trying to reinvent the wheel. Jul 5, 2013 at 16:51
• I'm not sure I understand what you mean by a chord being drawn from a scale. Have you considered such relatively common chords as 1-3b-5-7 (minor with major 7), 1-3b-5b-7bb (diminished) or 1-3-5# (augmented)? Jul 6, 2013 at 9:15
• I've cleaned up the comments a little here.
– user28
Jul 8, 2013 at 14:27
• It would be more appropriate to say "Most commonly used chords" and then add reference to more detail. Jul 10, 2013 at 15:12
Good question, but your assertion is entirely incorrect. Chords are derived from intervals, not particular scales. Your presented definition of chords would not hold up under 12-tone theory, chromaticism, clusters, pandiatonicism, non-functional harmony, or a host of other theoretical contexts.
In addition, different chords are considered "common" in music theory during different time periods. Major, minor, and diminished chords have been common for the past few hundred years (about 600 or so,) but many other chords are now considered common. For this reason, it is also good to specify which time period your talking about, or at least clarify which common chords you're going to discuss.
Therefore, it is for the above reasons that it would be incorrect to say that "all common chords are built using the major scales" as it is non-specific and would only hold up under the umbrella of functional harmony.
Hope that helps!
• But if you gentlemen were confused by the statement, there’s a good chance my guitar buddy will be confused, also. Let’s see. How can I clarify what I’m trying to say in the lesson (and thus, ask in this post)? How about the following? All chords, by the way, are built upon the root of one of the twelve major scales, versus the root of any other scales, such as the minor, pentatonic, blues, or diminished scales. AND, I’m still awaiting confirmation on the point, because the minute I say something like, “all chords,” I’m going to find out later that I’m wrong. Thank you, --Otis Jul 6, 2013 at 1:14
• @OtisGilchrist - I was not confused by your statement - it is simply incorrect. You will be hard-pressed to find a respectable confirmation when your own point is in itself incorrect. I agree with Carl's comment above - refer your friend to an actual theory textbook and interpret / clarify when you need to. Jul 6, 2013 at 13:06
• Well, I've taken the advice of revamping my abbreviated course and basing it on Peter Vogl's The Guitarist's Music Theory Book. I now feel more confident that I'm guiding my buddy based on a higher understanding of theory. Jul 13, 2013 at 15:37
• @OtisGilchrist - good for you! My comments / answer were / was nothing personal, it's just, why work harder than you need to? Jul 13, 2013 at 17:47
I would reword your statement to say "Since all chords are organized collections of tones, all chords can be associated with a scale within the modern diatonic & chromatic systems of harmony. Major triads, for example, can be associated with the major scale since both collections contain a root, major 3rd and fifth. Although there are many scales which align themselves with a major triad, the major scale is a more well-known choice and therefore a more digestible association for beginning instrumentalists."
Although there are many theories in music and a whole host of nomenclature and syntax that exist, it all boils down to the same set of 12 tones (in modern Western diatonic & chromatic harmony, that is.) I have found that the easiest way to chords, arpeggios and improvisation to students is to associate the chords with common scales. For example:
CHORD => SHARED TONES => ASSOCIATED SCALE(S)
C Major Triad => 1, 3, 5 => Major, Lydian
C Minor Triad => 1, b3, 5 => Natural Minor (Aeolian), Dorian
C Diminished Triad => 1, b3, b5 => Whole/Half Diminished, Half/Whole Diminished
C Augmented Triad => 1, 3, #5 => Augmented, Whole Tone, 3rd Mode Melodic Minor
C Dominant 7 => 1, 3, 5, b7 => Mixolydian, 5th Mode Harmonic Minor, Super Locrian
These are just a few examples. Every scale contains hosts of possible chord combinations, just as every 3 or 4-note chord is contained within several scales. These simply serve as mechanisms to associate chords with a set tonality and provide a degree of functionality for beginning & intermediate students of music. I hope this helps!
• Nate I'm going to have to disagree with you in part here. I think your definition works pretty well for functional harmony, but it is not an all-inclusive definition. For example, it would not hold up under quartal or quintal chord theory or even microtonal chord theory. As I mentioned in my own answer, your definition also leaves out cluster chords and sound-mass textures as well. The definition is okay, but it is very limited. Jul 6, 2013 at 13:11
• @jjmusicnotes - I agree with you completely. That's why I explicitly stated that this is for modern diatonic & chromatic systems. I went on a limb and assumed that Otis was not looking for a definition that covered the entire landscape of music theory, but something that served as a practical & usable starting point for his abbreviated music course. Jul 6, 2013 at 17:37
• @NateKimball - agreed, I assumed as much as well. I think Otis would be fine if he just specified the limitations of his scope. Jul 6, 2013 at 19:00
• I’ve taken the advice and have revamped my abbreviated course to base it on Peter Vogl’s The Guitarist’s Music Theory Book. I feel more confident now that I’m guiding my guitar buddy on the basis of a higher understanding of theory. Nate, I’ll be happy to upgrade your answer. How do I do that? I appreciate all your comments/answers, gentlemen. --Otis Jul 13, 2013 at 15:42
Nate gave a great answer, though if we're thinking of only chords that can be built from the major scale, the augmented triad (C, E, G#) doesn't fit. It can be constructed from the minor scale, for instance, in C-minor, the notes, Eb-G-Bnatural are available. However, it is a rare chord so maybe should not count. (It gets more acknowledgment mainly because it forms a natural parallel with the diminished triad, which IS in the major scale).
The (fully-)diminished seventh chord (C, Eb, Gb, A-natural) is however a pretty common chord that can only be formed with minor mode scales, not major. For instance, in C-minor, B, D, F, Ab.
Then there are chords that are less common than the diminished-seventh, but still more common than the augmented triad, that cannot be formed from either the major or minor scale. For instance, the ("Italian") augmented sixth chord, Ab, C, F# isn't available in either the notes of the major or minor scale. Nor are the "French" (add D) or "German" (add Eb) or "Swiss/Alsatian/etc." (add D#) versions of the same chord.
I suspect that you are talking more about the way were refer to the notes within a chord or scale, such as 1, 3 and 5 make up a major chord and 1, b3, and 5 make up a minor chord or 1, b2, b3..... make a Phrygian scale. If this is the case I would word it so that you're saying the nomenclature is derived from its relationship to a natural major scale (Ionian). Using accidentals (#/b) to describe a note shows how it is different from Ionian or a major triad.
When intervals are referred to as major or minor they draw their name in relation to a major scale as well. 2, 3, 6 and 7 can be referred to as major or minor but you should be careful to notice that not all the minor intervals apply to the minor scale (Aeolion), because it has a major 2. This is for the same reason as with the earlier example, it is based on major and declaring it minor designates that it is a half-step lower than the interval would be in the major scale. 4 and 5 have been deemed perfect for their consonant qualities and are deemed augmented (#) or diminished (b). All naturally occurring modes but Lydian and Locrian have perfect 4 and 5 and they both have one or the other.
I'm glad to see that you want to word things properly for your students!
There's a shred of truth in there. The chord 'C' means C major, the 1st, 3rd and 5th notes of a C major scale. If we want something different - Cm, Cdim, Caug (and that's just the triads) - we have to say so. And when we name intervals, our basic framework is the major scale starting on the lower note. Whatever key or mode a piece is in, D to A is a Perfect 5th, because A is the 5th note of D major scale. It's also the 5th note of D minor scale, but that doesn't change its name.
Guitarists often seem interested in what scale 'goes with' a chord. That's looking at chord theory from another angle. | 2,378 | 9,513 | {"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} | 2.671875 | 3 | CC-MAIN-2023-50 | longest | en | 0.942382 |
https://quadernity.blog/aside-four-phases-from-metaphysical-potentiality-to-physical-actuality/ | 1,606,252,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00156.warc.gz | 467,429,535 | 50,317 | # Four Phases from Metaphysical Potentiality to Physical Actuality
A coin is a symbol of duality. A duality is a unity of two/duo.
A coin’s two faces are mutually exclusive, being opposite one another; and they are mutually dependent, as they cannot be separated from one another without destroying the coin that comprises them.
A coin sitting on a table or stuffed in one’s pocket is at Rest. Within this Ground state (the state of minimum energy), there is potential for the coin to be activated. When someone calls, “Heads you win, tails you lose!” the coin is employed in a game that punctuates the Ground state that exists before and after the activity.
While a coin is actively spinning through the air, heads or tails are BOTH possibilities; however, we have NEITHER heads nor tails decisively.
Simply grasping a coin mid-air reduces the indeterminate potentiality of BOTH/NEITHER to EITHER heads OR tails, though we know not which until we sneak a peek.
The pre-determinate EITHER/OR potentiality is further reduced when we finally observe the result of the coin toss; at the Final Outcome a determinate actuality takes its place.
Thus, the four states of the coin shall be called:
Ground >> BOTH/NEITHER >> EITHER/OR >> Final Outcome
Below, the three activated states are represented hierarchically. From top down on the inverted triangle we follow the trajectory from mingled-in potentialities to a singled-out actuality. From a maximum potential in the top tier, the descent undergoes a vanishing potential, eventually reaching a limit/point where the potential becomes nil and the Final Outcome is realized.
##### GROUND
Whereas the Ground state that transcends the triangle (and is labeled above and outside it) has no significance to the game, the Final Outcome (at the triangle’s lowest point, but labeled beneath it for legibility) holds the game’s ultimate meaning.
Players register a reaction to the outcome. If one wins the coin toss she is happy; her energy/frequency goes up and receptors open. If one loses he is unhappy; his energy/frequency goes down and receptors recoil.
The final state is the point of the game (pun intended)!
#### Descent Through States of Consciousness
As human Observers we may reflect upon our descent through states of consciousness beginning with a pre-cognitive (or instinctual, deep-sleep) state, transitioning through the subconscious (impulsive, ideational, dreaming) state and arriving at the fully cognizant (intentional, intellectual, wakeful) state.
1. Seeing is the pre-cognitive interaction between the INformable (LLQ) and the INformative (LRQ) while they are mingled-in together in the BOTH/NEITHER state.
2. Looking For reduces potentiality, narrowing the field down to the EITHER/OR state. Attention is on alert for content that is still mingled in with its companion context in a EITHER/OR state. At some point while Looking For, the INformable state gets satisfied. Something is found. Content is counted and context is discounted. Attention is paid to the content only. Paying Attention to one thing involves the ignoring of other possibilities. The EITHER/OR state collapses; the content is singled-out while its companion, the physio-spatial context, is overridden by the Observer, who re-presents the content and re-contextualizes it in accordance with the context of His own psycho-spatial backdrop. The subconscious but causal act of Paying Attention reduces the EITHER/OR state to its Final Outcome.
3. Looking At the Final Outcome is the reflexive act of the Observer Creator (LLQ). The Final Outcome is the Observed Created (LRQ) that was singled out as a Particularity by the Observer/Creator. (Quadernity calls a particular-entity a Particularity.) In reviewing the perceived content within the context of an assumed reality (the Observer‘s unquestioned mindset) the Observed seems to be outside of and other than the Observer. This is a projection, of course, a trick of the Observer’s mind, for without apparent differentiation, assessment and evaluation of the selected content is impossible.
Because Seeing and Looking For both occur in the precognitive intelligence of the INformable, but not in His conscious intellect, He is unaware of His role in the reduction of potentiality. He recognizes the Patterned Matter after the fact, but He cannot recall His pre-conscious participation in its OUTformation.
The following quote is from the Upanishads, an ancient Indian text written thousands of years ago:
Whether we know it or not, all things take on their existence from that which perceives them.
What did our ancient ancestors know that we are just now remembering or re-learning?
The Informable Observer has little idea that what Mattered (psychologically) to Him Mattered (physically) for Him. Regardless, He reacts to what He singled out. This reaction produces altered wave expressions from the Observer. These waves are reflected (not generated) by the Particularity, and the consequent interference Patterns are recorded in the outside Informative space of the Particularity.
The causal Creator seeds/Structures the raw Substance. The INformative Patterns in space contribute to the probability of whether Matter will recursively emerge according to that same Pattern, or a somewhat similar one, or a radically different one.
#### Coin Toss Game and the Double-Slit Experiments
Like the coin toss game, quantum physics involves a Ground state that is punctuated by:
• a BOTH/NEITHER state (wave-particle is indeterminate; the maximum-potentiality, minimum-actuality state; Seeing);
• an EITHER/OR state (wave-particle is a pre-cognized duality; midway between potentiality and actuality; Looking For);
• and a singled-out Final Outcome (Particularity is the re-cognized minimum-potentiality, maximum-actuality state; Looking At).
Between the mingled-in BOTH/NEITHER state and the singled-out Final Outcome, there is a transitional zone wherein pairs of complements (EITHER/OR) coexist. The complements can never be simultaneously ascertained, or singled-out. The more you know of one, the less you can know of the other.
In physics, the more you know the position of a particle the less you can know about its momentum, and vice versa. The same holds true for content and context (introduced in the previous section and exemplified in the final paragraph of this section). The inverse relationship of pairs of characteristics is known as Complementarity.
Catching a coin mid-air reduces its indeterminate state of BOTH/NEITHER to a predetermined state of EITHER/OR. For any and all coin tosses, only one side at a time pertains at the Final Outcome.
The famous double-slit experiment can be set up to make evident either the wave aspect or the particle aspect of quanta, such as electrons and photons. Before setting up the double-slit experiment, a decision is made as to which aspect of the quanta we are Looking For; therefore, the quanta emitted in a double-slit experiment are already reduced to their EITHER/OR duality, which enjoys superposition (the combination of the wave aspect (think context) and particle aspect (think content)). Again, only one of the dual aspects is evident in the Final Outcome for each run-through of the experiment.
What we actually Look At in the Final Outcome is merely a remnant of a four-phase reduction of potentiality.
#### Double-Slit Experiments
Before we get into the mechanics of the double-slit experiment, it is important to know that all experiments in quantum physics assume the system being scrutinized is closed, meaning it is in isolation from outside influences.
The assumption that a quantum system is secured from all influences leads to inexplicable experimental outcomes. In order to explain the bizarre implications of the double-slit experiments, it has been suggested that nonphysical human consciousness (somehow?) intrudes upon the physical system, or that particles themselves are psychic tricksters, or that our choices cause the universe to split off into countless parallel universes with their own realities, or divine intervention is perhaps at play. A reasonable assumption is that hidden-variables are producing the provocative results.
Quadernity shows us that closed systems are impossibilities, as there is nowhere that space is not. Space is not empty; it is full of energetic fields. All physical systems endure (persist through some amount of time) within a spatial environment. Even in a vacuum, a physical system being tested would be under the influences of quantum fluctuations.
#### The Setups
As photons or electrons are emitted one-at-a-time in the double-slit experiment; they are already reduced from their indeterminate BOTH/NEITHER state to their EITHER/OR state as the experiment commences.
The EITHER/OR state includes the particle and wave as a duality of unknowns. The question we have is about how a quantum is traveling when it hits the back screen and comes to its demise. Was it traveling all spread out, with its wave extended/non-localized, or did it travel more like a bullet, with its wave contracted/localized?
When only one slit is open:
Final Outcome reveals that the absorption of quanta by the screen happens directly behind the single open slit.
This distribution tells us nothing about whether the quanta traveled like waves or more like bullets. Regardless of which way the quanta travel, there is no possibility of interference with only one slit open.
With two slits open, there is a possibility for interference, but it is not a sure thing.
A) When interference/observation occurs before the quantum reaches the back screen:
The Final Outcome reveals that nearly all absorption of quanta occurs directly behind the two slits, same as happened behind the one slit experiment.
This tells us that the EITHER/OR state was reduced to a particle/bullet before the quantum reached the back screen.
In Wheeler’s Delayed Choice experiment, the choice as to whether or not to attempt detection is delayed until after a quantum passes through the slit screen. If detection is employed when the quantum is between the slit screen and the back screen, it still gets reduced from its EITHER/OR state to its Final Outcome Particularity before hitting the final screen. As long as the collapse happens before absorption, the distribution will appear as if the quanta had traveled all along like a bullet.
B) When interference/observation DOES NOT occur before the quantum reaches the back screen:
The Final Outcome reveals multiple, vertical bands of absorption all across the back screen.
This distribution into multiple bands indicates that the EITHER/OR state persisted until the quanta hit the back screen. Particles rode their waves until they happened upon the screen. Upon impact, the waves collapsed, leaving a mark only where the particle was absorbed by the back screen.
#### Speculative Results from Double-Slit Experiments
Below are two wildly different videos that tell of double-slit experiment results. The first suggests interpretations that the second will dispel.
Double Slit Explanation, Video #1:
This video leads its viewers to conclude that observation/detection alters the behavior of quanta. This idea is not false; it is true that if quanta are observed before reaching the back screen the distribution array will suggest that the electrons traveled through the slits as particles or like little bullets. Otherwise, the array looks as if the quanta traveled through the double-slits as waves. Emerging from each slit are secondary waves that produce an interference pattern as they overlap while out-of-phase.
I disagree, however, with Dr. Al-Khalili’s suggestion that quanta, electrons in this case, somehow “know” whether or not they are being watched, and then mysteriously alter their behavior accordingly.
The difference between finding particle clumping in a single band per slit, or in multiple bands spread out across the screen, is determined by whether or not there was influence/interference or trajectory interruption at any point prior to the quantum hitting the back screen. This hasn’t anything to do with who (or what) detects which opening the quanta passed through; and it doesn’t have anything to do with the quanta’s foreknowledge about the experiment’s setup.
Al-Khalili’s suggestion is also disputed in Video #2, wherein a couple other valid points are made:
In the second video, the comical presenter makes three points.
• He assures us that the universe is not dependent on human-consciousness.
• He disabuses us of the concept that quanta know or care if humans are watching them. They do not choose circumstantially which laws of physics they will obey.
• He differentiates between quanta and particular bits of matter. Quanta, he says, are always only waves that shrink/collapse when interacted with; they are never particles.
The last point deserves some nit-picking, but before diving deep into that we should include one more opinion — this one from the late, highly esteemed John Archibald Wheeler:
The thing that causes people to argue about when and how the photon learns that the experimental apparatus is in a certain configuration and then changes from wave to particle to fit the demands of the experiment’s configuration is the assumption that a photon had some physical form before the astronomers observed it. Either it was a wave or a particle; either it went both ways around the galaxy or only one way. Actually, quantum phenomena are neither waves nor particles but are intrinsically undefined until the moment they are measured.
Three people, all of whom make a living studying and teaching in the field of quantum physics, have offered irreconcilable opinions.
Essentially, the first video suggests that quanta are BOTH waves and particles. On the contrary, Dr. Wheeler says quantum phenomena are NEITHER waves nor particles. Quadernity offers a BOTH/NEITHER option which reconciles these two opposing views.
The second video says that quanta are always waves and never particular objects. The use of the word quanta makes all of this tricky; anything quantized is what Quadernity would call a Particularity, something with its particular nature predominant. However ironic the semantics, the presenter means that the wave aspect, being non-localized, exists in the realm of probabilities (meta-physical, ULQ), while the particle nature, being very-nearly localized (though not perfectly so, due to Heisenberg’s Uncertainty), exists as an object (a Particularity, LRQ). Quadernity maps out the differences of these two phases by showing them in different quadrants. To move from one quadrant to another involves a transition, which is overlooked by quantum physicists who are not yet looking through the lens Quadernity.
The intermediate state (the EITHER/OR wave-particle duality) transitions between what Wheeler calls the “intrinsically undefined” (what I call the indeterminate, BOTH/NEITHER state) and the particle distribution eventually Looked At (Final Outcome).
The EITHER/OR state holds the potentiality for the particle aspect to be revealed at the Final Outcome. The particle aspect bounces around randomly throughout the electron’s wave field. They are together in the EITHER/OR state, a.k.a. superposition.
The only way a quantum can be seen/measured/observed is for something to interact with it. Interaction = Observation. Interaction does not require self-reflective consciousness; no humans are necessary. Whether captured by ions in a gas or forced to pass through a filter, the disturbance/observation of quanta destroys the predetermined, non-localized EITHER/OR state, and sometimes even absorbs the particle. If there is interference before the quantum hits the screen, the quantum is not immediately absorbed; however, if the wave-particle hits the back screen of the double-slit experiment, the wave collapses and simultaneously the particle is absorbed.
If the particle is not absorbed upon collapse of the EITHER/OR state, it remains as merely a carcass of the lost potentiality. With potentiality reduced to the minimum, the actuality of the particle is raised to the maximum. Thus, the particle appears as an emergent actuality.
We humans then consider the localized particle of inert matter from the outside-in. Such matter particles are given ultimate importance by physicists who subscribe (officially or unofficially) to philosophies that categorize them as realists, positivists, materialists or physicalists.
#### What Counts as Observation?
Physicists do not use Quadernity’s coined phrases mingled-in and singled-out; instead, they use the more sophisticated terms superposition and decoherence to describe these states. A pattern is decohered, or singled out, by an Observer via disturbance/interruption of the mingled-in state. In terms of Quadernity, this process of INformation leads to OUTformation of some Particularity with measurable parameters.
To make the quote below more comprehensible, you may want to translate “superposition” to mingled-in and “path information” to singled-out.
Anton Zeilinger discusses superposition and interference patterns:
[T]he superposition of amplitudes … is only valid if there is no way to know, even in principle, which path the particle took. It is important to realize that this does not imply that an observer actually takes note of what happens. It is sufficient to destroy the interference pattern, if the path information is accessible in principle from the experiment or even if it is dispersed in the environment and beyond any technical possibility to be recovered, but in principle still ‘‘out there.’’ The absence of any such information is the essential criterion for quantum interference to appear.
Dr. Zeilinger states that having no path information about a particle is the “essential criterion for quantum interference to appear“. His statement makes it sound like the interference pattern appears as a result of the lack of knowledge about the path. Even though it is true that in the absence of path knowledge superposition remains, it is not a consequence of having no path knowledge. Instead, the interference pattern that shows up on the back screen of the double-slit experiment is merely indicating that the mingled-in state (superposition), which exists by default prior to any Observer attempting to derive path knowledge from the system, was preserved up until absorption of the quantum by the screen.
Observing is the act of singling-out something (a behavior, property, trajectory, etc.) from a mish-mash of alternatives, and thereby, condensing the selected content (particle-aspect) from its natural context (wave aspect). Upon disturbance by means of observation (deflection or absorption), the quantum is disturbed and its wave contracts. It is the disturbance that destroys the coherence (mingled-in state) of the particle’s wave with its environment, not the lack of path knowledge.
Once the deBroglie wave aspect is collapsed/condensed into a particle, whether before or upon absorption, scientists see only evidence on the back screen that there was existence of a particleSingled-out particles are all we ever see recorded on the back screen, regardless of the pattern of cumulative distribution, in clumps or in bands. The wave itself is never captured by, or recorded on, the back screen! Because particles travel like bullets, scientists expect the particles to have arrived by traversing a straight path between the points of emission and absorption.
The wave aspect offers a multitude of potential positions for the particle. Before disturbance, the particle and wave are conjoined as a quantum system that is coherent with its environment. If particles, one by one, are absorbed by the back screen without prior disturbance, the interference pattern that gradually forms on the back screen, particle by particle, indicates that each arrived as if carried by a wave (Bohm’s pilot wave), or as if traveling itself as a wave (as its wave-aspect would naturally travel if undisturbed). While traveling in/as a wave, direct path knowledge about any given particle is meaningless. Countless possible positions endure until absorption or disruption.
Interruptions/disturbances happen all the time without any person knowing about it. It is not that human Observers have path knowledge (a.k.a. which-way information), but the act of disturbance itself, that produces measurability. Measurability implies only that something (not necessarily a person) can become aware of something else.
Since all things/Particularities emerge within an environment, it is reasonable to assume that the potentiality for that Particularity to emerge singled-out is present within the environment into which it presents itself. The wave aspect is coherent with the environment. Environments contain all sorts of particles and energetic fields that can disturb potentials or provide gradients; even the vacuum has an energy flux which produces virtual particle eruptions. If something in the environment does disturb the coherence, or mingled-in potentiality, a Particularity gets singled-out. More often than otherwise, the environment (a non-empty space) is the Observer that Observes the Observed.
Here linked is the abstract of a 2007 article* from ScienceMag.org. It explains how a single electron can constitute an environment/Observer for a minimum particle system, and also explains how the conversion from EITHER/OR to Final Outcome can be a result of quanta coupling to their environment.
*To view the full article as a downloadable pdf, which I recommend, you must sign up with AAAS. I did this without cost, complication or repercussions; just follow the link provided near the abstract.
[O]ne particle can be considered an observer that carries partial information about the other particle and its path through the double slit. The amount of which-way information exchanged between the particles is limited by the observer particle’s deBroglie wavelength.
The smaller the particle, such as an electron, the bigger is its deBroglie wavelength. Conversely, the deBroglie wavelengths shrink/recoil as the particle‘s mass/energy/size/complexity increases, which is why we do not encounter the matter-waves of macroscopic entities.
Complementarity tells us that the greater the momentum of the measured wave field, the more evasive is the position of the particle aspect. Conversely, zeroing out the momentum of the wave-field when it hits the back screen allows us to know the position of the particle.
The back screen of the double-slit experiment never captures the actual wave aspect of the electron! Quanta are only ever observed as particles. As the quanta is absorbed upon contact with the screen, the screen provides a record of the particle‘s position at the moment of impact. The screen always only shows a distribution pattern of where individual particle positions were recorded by the photo-sensitive film. We can never know anything decisive about a quantum prior to its detection/interaction/observation.
In the video below, electrons interact with photons. Between them, they exchange packets of energy, either speeding up or slowing down accordingly. The particle that ‘Paid’ Attention, or gave a packet of energy, has its frequency reduced (longer wavelengths), and the particle that receives Attention/energy has its frequency increased (shorter wavelengths).
As soon as the quantum system receives attention/energy, its frequency increases. Video #2 from further above nicely described how an photon can be observed by a electron. One of its animations showed how the interaction of the photon with an electron compresses the wavelength of the electron, making it more localized. Shorter wavelengths have greater frequency and more energy.
Before the influence, the wave aspect was coherent with, or spread out through its environment (non-localized, low frequency). After the influence, the wave aspect tightens (localized, high frequency). The good folks at EPFL found a way to capture an image DURING the moment of influence, before the particle aspect is singled out from the wave.
A likely scenario: If the EITHER/OR wave-particle is influenced/interrupted/observed by anything (including other quanta) before it is absorbed by the back screen, its combined mass/energy would likely spike. This jolt of energy would shrink the deBroglie wavelength, giving the particle aspect predominance, making the quanta travel as a bullet.
Further speculation: very tight waves closed around a particle could act as a protective sheath, like the atmosphere around earth, also like the so-called “personal space” we have around ourselves, which monitors influences from the environment at large.
Check out Zenneck Surface Waves from which a company in Bahrain is working to achieve transmission of wireless power, similar to what Nicola Tesla attempted to do in the early 1900s.
The video below demonstrates quantum behavior in non-quantum silicon spheres. Listen for the statement about the separation of the droplets from the substrate, and how the waves in the substrate carries a memory of the paths taken by the droplets. Each particle has its own guiding wave.
As planets move follow the sun they are surrounded by layers of protection. The magnetosphere is shown below.
In addition to the magnetosphere, Earth has atmospheric layers of protection.Check out how spiders can fly, by using the voltage in our atmosphere. Article from the Atlantic Magazine.
And, what do ya know! An “atmosphere” surrounding quanta has recently been detected. The atmosphere, according to Nobel prize winner, Frank Wilczek, is a “thin layer of influence”.
The Rig Veda proclaims the three spaces are the Sky, the Atmosphere and the Earth.
And according to the Tao te Ching:
The space between heaven and earth is like a bellows.
Okay, then! Let’s say the wave aspect expands and contracts to allow or prevent influences. It acts like a cell wall, or like the magnetosphere surrounding our solar system. It is like a boundary between the inside space of the particle and the outside space of the environment. The wave, as a boundary, both connects and divides the particle and the environment.
The EPFL article’s abstract is quoted below, augmented with my bolded, underlined links to give convenient access to definitions for words not well known outside the field of quantum physics:
[P]articles couple to their environment (for example, by gravity, Couloumb interaction, or thermal radiation). These couplings shift the wave phases, often in an uncontrolled way, and the resulting decoherence, or loss of phase integrity, is thought to be a main cause of the transition from quantum to classical behavior. How much interaction is needed to induce this transition? Here we show that a photoelectron and two protons form a minimum particle/slit system and that a single additional electron constitutes a minimum environment.
Restatement translates to jargon of Quadernity: “[D]ecoherence, or loss of phase integrity, is thought to be a main cause of the transition from quantum [EITHER/OR wave-particle] to classical behavior [Final Outcome Particularity]”.
Prior to disclosing the Final Outcome, the EITHER/OR state holds. In the EITHER/OR state the content/particle remains with the context/wave until it is adopted into a new context. The new context (coupling to the environment, or absorption at the back screen) replaces the previous context (carrier-wave) as the content (quanta’s particle aspect) is singled out in the Final Outcome.
The quote above refers to the quantum state; however, there is no discrimination between Wheeler’s idea that the quantum state is indeterminate (the BOTH/NEITHER state) and the EITHER/OR wave-particle duality.
Overlooking this subtle, but significant, transition is like overlooking the fleeting transition between inhalation and exhalation. Inhalation draws oxygen in and down into lungs. Exhalation pushes CO2 up and out of lungs. Within the all-important transition is a change of direction and an exchange of gases! Now that we know of it we pay close attention to our breaths; even so, this hardly helps us notice what is obviously there.
#### Connecting Consciousness and Corporeality
In the graphic below, the green-colored levels of the wave form represent Corporeality. The blue-colored levels represent ConsciousnessJust as an inhalation (Pull) and exhalation (Push) are joined by a Transition (explored thoroughly in Cylindrical Lattice Model), there is a Transition between Seeing the Possibilities and Looking At the Final Outcome. In the center of the graphic above, where the green and blue levels meet, attention is ‘paid’ (as a toll or re-source) by the causal Subject, the Observer, to its effective Object, the Observed.
When detected, interfered with, or ‘paid’ attention, quanta are reduced from their non-local wave to the inevitability that a localized particularity will be present at Final Outcome. The Final Outcome is only re-cognized (actually Looked At) after the experiment is concluded.
The graphic is reprinted below. In the green troughs, locate the BOTH/NEITHER Possibilities (left side) and the singled-out Final Outcome (right side). In the blue crests, locate Seeing the Possibilities (left side) and the Looking at the Final Outcome (right side).Now find the following alignments in the graphic above.
• Labels for BOTH/NEITHER Possibilities and pre-cognitive Seeing are vertically aligned.
• The labels for Receiving Attention and Paying Attention are vertically aligned in the center of the X-shape.
• Labels for Final Outcome and Looking At are also vertically aligned.
In the chalkboard model of Quadernity below follow the sequence of descent, the process of OUTformation: The Possibilities are Creatable (ULQ). She Pulls a seed from the INformed Collective Consciousness (URQ) from which She Structures Her Substance/Matter (Transition of Creating/OUTforming, diagonally downward). After Transition, She acquires the label, INformer (ULQ) of Patterned Matter, Her Created Object (LRQ), and the seed of the INformed (URQ) acquires the label, Creative (URQ).
In the chalkboard model of Quadernity above follow the sequence of ascent, the process of INformation:
The Seeing INformable (LLQ) is unconsciously receptive (Pull) to the (Pushed) INformative (LRQ) essence of the Substance being Structured
Thus re-minded, the INformable semi-consciously Looks For His Pattern to be revealed in some forthcoming Matter. With attention piqued, the INformable narrows His receptors, thus specifying/determining the inevitability that something re-cognizable will appear at the Final Outcome (Created LRQ). The INformable’s (LLQ) energetic reaction (the ‘paying’ of attention) INforms (Transitions diagonally upward) the Collective Consciousness, represented by the term INformed (URQ).
The outline of the process continues below the chart reprinted for easy access.
Because the INformable had singled out what will be Created (LRQ) and eventually (Looked At), He shall be known as the Object’s Subjective* Creator (LLQ). Having given His attention to something (an expenditure of energy), the Creator is finally alerted/awakened (after the fact) into His conscious state, from which He Looks At the Patterned Matter (LRQ) to which He had previously given His attention and unknowingly designated/selected/singled-out to make an Objective** appearance.
*Subjective, as in At-the-Cause of…
**Objective, as in At-the-Effect of…
With the INformed (URQ) updated, a Creative (URQ) seed re-members the INformer (ULQ), who may then receive it (Pull what was Pushed) to instigate another round.
#### Feedback between Mother/Matter and Father/Pattern
An Observer is any functional agent (not necessarily a person) that interactively reduces the indeterminate BOTH/NEITHER potential to a pre-determined EITHER/OR potential, and then further, to its Final Outcome.
Hedda Hassel Mørch is a Norwegian philosopher and postdoctoral researcher hosted by the Center for Mind, Brain, and Consciousness at NYU. Quoted from her article, Is Matter Conscious? published in Nautilus magazine.
Conscious experiences are just the kind of things that physical structure could be the structure of.
Observation is the process that temporarily singles out some specifically designated Patterned Matter (a Particularity, something made particular). Observation involves the Structuring of Substance into Patterned Matter. Observation is a process that includes more than one conscious experience: Seeing, Looking For and Looking At.
All Matter is PatternedFemale and Male are inseparable.
Matter – Mater – Mother
Pattern – Pater – Father
Patterns make Matter meaningful to the INformableMatter makes Patterns re-cognizable to the Creator.
For anything to be recognized as real it must endure the ravages of spacetime. Without the constant sustenance of feedback nothing (not even waves) can exist perpetually.
Mother and Father are inter-actors; without feedback between these cooperative agents there could be neither Corporeal Matter persisting in space nor Patterns of Consciousness recurring in time. Active inter-Subjectivity (alternating of give and take between Subjects, the INformable/Creator and the Creatable/INformer) perpetuates the greater system in which they are enfolded.
Neither the Created/Observed nor the Creator/Observer is ever fixed; each is constantly becoming. They are mutually dependent. Each partner relinquishes some of His/Her potentiality for the sake of the other’s continuity.
Universal Mother says, “If you will re-member me, I will re-mind you.” Re-membering, in this case, means the Father/Creator/Observer would return to the Mother a vibrational energy, reflecting that which has influenced Him from the physical realm. In return, Mother re-minds Father by once again Substantiating the Pattern with Matter that vibrates in such a way as to resonate with His receptivity. The reciprocity, if it holds, perpetuates the dynamic system of any Observed entity and its Observer. This, of course, does not mandate a one-to-one relationship, but happens also at the collective level, accounting for consensual “reality”. As will be demonstrated by the Manipulable Model, the Creator represents the omniscient Observer, the collective Consciousness, while the INformable is a local, limited Observer.
#### Resonance and Mathematics
What is finally realized, or Looked At, has been organized to attune with the available, receptive INformable.
Says Max Planck in 1944:
As a man who has devoted his whole life to the most clear headed science, to the study of matter, I can tell you as a result of my research about atoms this much: There is no matter as such. All matter originates and exists only by virtue of a force which brings the particle of an atom to vibration and holds this most minute solar system of the atom together. We must assume behind this force the existence of a conscious and intelligent mind. This mind is the matrix of all matter.
As everything vibrates, resonance is a means of communication between Consciousness and Corporeality, and between potentiality and actuality, as well.
Lynn McTaggart refers to a resonance between brains and perceived objects.
When we observe something in the world, certain portions of the brain resonate at certain specific frequencies. At any point of attention, our brain only presses certain notes, which trigger strings of a certain length and frequency. We perceive an object by “resonating’ with it, getting in ‘synch’ with it. To know the world is literally to be on its wavelength.
Patterned Matter has a linear history between its origin and demise, the end points of its standing wave. The active wave exists entirely within the stillness of the Ground state.
1. The BOTH/NEITHER state consists of oscillating peaks and troughs (like the coin flipping in the air).
2. The graphic above, and reprinted again below, includes two interacting waves, one above the other. Either of the two waves can be thought of in terms of the standing wave currently in discussion. Looking at only one of the two waves, notice where the green trough half of the wave cycle meets the blue crest half. These nodes are situated upon the axis of the dashed lines. EITHER the green half of the wave cycle Pulls and the blue portion of the wave Pushes, OR vice versa, repeating in alternation. In the complex of both waves, exchanges of energy/attention occur between peaks and troughs where the two waves touch. EITHER the trough of one wave cycle Pulls and crest of the other Pushes, OR vice versa.
3. In either case, the wave collapses as it is absorbed/resolved back into the Ground state. Only at this Final Outcome is a trace of its particular existence ever found.
Accounting for linearity involves adding one number to another. We are taught to add on a one dimensional line with a move to the right.
0 + 2 = 2 = 2¹ (one dimensional: line)
All Patterned Matter is made of smaller Parts and Participates in the constitution of greater wholes.
For more on the hierarchy of Patterned Matter, see The Holarchy.
Accounting for the hierarchical relationships of Patterned Matter involves exponential dimensionality. Beyond the one-dimensional line we achieve two-dimensional area by multiplying two numbers together and three-dimensional volume by multiplying three numbers together.
2 x 2 = 4 = 2² (two dimensional: area)
2 x 2 x 2 = 8 = 2³ (three dimensional: volume)
Rhythm involves recurring waves. Counting peaks per second, or frequency (in time), of a standing wave involves arithmetic.
A wavelength is measured from peak to peak. Measuring distance between peaks (in space) involves geometry.
The necessity of integrating Patterns and Matter involves two forms of mathematics!
For how these two aspects of Patterned Matter synchronize, see Quadernity’s Nature-Based Number Line.
#### Quadernity Acknowledges Four Phases Applicable to All Scales
The Ground state is a minimum energy state of Rest. The Female‘s Ground state exists before She Pulls and after She has Pushed. Both Female and Male Pull * Transition * Push * Rest in a 180 degree offset, which aligns Pulls with Pushes and Transitions with Rests.
#### Four Cyclic Phases from Instigation to Iteration
1. Female Pulls. INformer acquires Structure; movement/disturbance is the initial act (Male Push), that seeds the potential of the Ground state, thereby activating its indeterminate BOTH/NEITHER state (mingled-in). Only the initial insemination/influence comes from beyond/above the nascent system. Required is the taking in of “spirit” (inspiration); spirit translates in most languages to breath. In other scriptures it is known as Vach, the Word, or mystic sound OM. In any case, some disturbance of the otherwise still ‘waters’ (the raw Substance) provides the Female/INformer with the impetus to organize Her Matter.
2. Female Transitions and Male Rests (pre-cognitive). Matter/Substance is INformed/Structured by the seed‘s Creative Pattern. How the INformer organizes Her output is esoteric. Raw Substance is the super-saturated fluid, the plenum, or the full potentiality of energy/matter. The aforementioned movement/disturbance/de-still-ation induces the solution to expel solutes (as crystals form within a seeded super-saturated solution*). Is it no wonder that the Virgin Birth is said to deliver a Christ (crystallized light)! The Mother had not yet been influenced by the reaction(s) of man (the local INformable(s)).
3. Female Pushes and Male Pulls. Patterned Matter is OUTformed and singled-out by Creator, whose reception occurs subconsciously to the local INformable, the adoptive Father (think Joseph in the manger). The BOTH/NEITHER state is reduced to the EITHER/OR state.
4. Female Rests and Male Transitions. The INformable‘s waking consciousness finally re-cognizes the singled-out Patterned Matter. EITHER/OR is reduced to its Final Outcome by the INformable Father (e.g., Joseph) who locally attends the birth/delivery/appearance and divides the light (crystal/Christ/son-sun-sol-soul) from the darkness (waters/Mary/Mother), by cutting the umbilical cord. The Mother can now return to Her Ground state and Rest. The INformable Father-Observer assigns exoteric meaning to the OUTformed He witnesses.
The meaning made causes a reaction that ripples upon the now Resting Female, who is thereby re-seeded. The cycle begins again with Step One wherein the Female Pulls the seed of the Male Creator. As opposed to Pulling the seed of a transcendent Spirit/Word, as was the case in the initial round, in all iterative cycles the Female Pulls a seed from Her Male counterpart and co-creator of all OUTformed impurities, each retaining at least a trace of the original Spirit/Word.
*Below is a video showing how crystals form in super-saturated fluid.
#### Macroscopic Examples of the Four-Phase Descent into Empirical Existence
Human Reproduction:
1. Inception: egg receives/Pulls sperm, BOTH female and male gametes are mingled-in; fetus NEITHER a female nor a male.
2. Gestation: EITHER male OR female fetus is developing/Transitioning, still mingled-in with mother.
3. Birth/Push: usually one sex or the other is evident as baby is born (Final Outcome); singled-out from mother.
4. Public is informed of baby’s sex, name, weight, height, time of birth, etc, while the mother Rests, returns gradually to Ground state.
Crystal Formation:
1. Substance/super-saturated solution (BOTH/NEITHER solutes and solvent are indifferentiable, mingled-in) is Structured by contaminant/disturbance/influence.
2. Solutes fall from solution and agglomerate within it (differentiable EITHER/OR remains mingled-in).
3. Crystal is retrieved from solution (singled-out, Final Outcome).
4. Observers admire the crystal and overlook the solution from which it came. Solution returns to Rest in its Ground state.
#### Scriptures also Align
Tao te Ching reveals knowledge of these four phases in the two following quotes.
The ten thousand things come from being. Being comes from not being.
According to Quadernity, the ten thousand things are the INformative Createds (LRQ) observed by the INformable/Creator (LLQ), who, in counting them, discounts the coupled potentials, the Creative/INformed/Structure (URQ) and Creatable/INformer/Substance (ULQ), who together are the ‘being’ from which every one of the ten thousand things emerges.
The Tao begets one; one begets two; two begets three; and three begets the ten thousand things.
The Tao is the Female‘s Ground state. One is the indifferentiable BOTH/NEITHER state (She is infused/inspired by a spirit/breath). Two is the EITHER/OR state. Three is the revealed final state (counted content), the hidden alternative (discounted context), and the necessary Observer.
#### Seeing vs. Looking At as Applied to Our Human Experience
I have the habit of persistently and passionately asking questions (carefully setting my receptors to what I am looking for) especially before bedtime. Responses are offered up to me from the transcendent realm, of which we are largely unconscious. These responses come in the form of symbolic imagery that dances through my surreal dream-state. Each morning I must keep my judgmental mind at bay — absolutely still and quiet — so I can linger longer at the portal between my focused attention and non-focused awareness and spy on the insights arising in my liquidly lucid dream-state. Shuttling the impressions I “see” over into my long-term memory bank requires that I resist all temptations to interpret, reorganize, evaluate or associate the images with previous experiences. It is only after these impressions are fixed into my active memory bank, that I am able to look at them, to evaluate them and contemplate their meaning.
When one falls asleep the individual’s point-of-view slips away without reluctance. When one dies, the point-of-view is relinquished, perhaps with more reluctance. When one meditates, his/her point-of-view must persistently be surrendered at the threshold of the realm of unconditional seeing.
My experiences of knowing without thinking or learning occur as I practice voyeurism at the cusp between dreaming and wakefulness. As I just begin to awaken, I am of two minds: one is in a de-focused dream-like state, and the other is actively aware that I am dreaming. If my self-conscious mind is allowed to judge, reason, or anticipate, my portal for seeing closes. Once the ego arrives, it gains a momentum that is hard to harness. The dream-state cannot abide the waking state, and the waking state cannot function within the dream-state.
Between the precognitive Seeing of sensory INput (as mingled-in potentiality) and the cognitive Looking At the Final OUTput (the singled-out actuality), there exists a subconscious receptivity/Pull, a Looking For.
From our points-of-view, we often Look For outcomes that most readily perpetuate our familiar patterns and routines. This can be unfortunate because our freedom is in choosing how to Look For something; no freedom remains once we Look At something!
Perhaps you are wondering if there can be an even greater freedom in Seeing than there is in choosing what we Look For.
Seeing, or knowing, may be likened to the the active intuitive abilities of shaman, indigenous peoples, or genius savants like Daniel Tammett, a young man from the UK who sees and knows things without having to think. Here is the link for Daniel Tammett’s TED talk. (18 minutes long) I’m sure you will find it intriguing and worthy of your attention.
Opening up the mind to the level where a choice can be made about seeing can challenge the usual ego-centric point-of-view. De-focusing our sensory acuity could feel awkward, or even threatening. While some might quickly abandon the endeavor and return to their more comfortable and automatic routine, others may seek out tools that aid the practice de-focusing superficial sensory input. Floatation tanks are an increasingly popular means of reducing focus on sensory input. For those who persistently practice the art of de-focusing, spiritual freedom is surely increased.
Breaking old habits is not easy; in the brain, overcoming undesirable habits is accomplished by consciously installing a better habit in its place. Some habits are not even recognized as habits; for instance, we are so accustomed to having our eyes keenly focused that it takes some effort to de-focus them. This may apply to listening and reading, as well. Hearing words, or reading words on a page, and knowing the words’ common definitions, we instantly deduce meaning from the communication, hardly questioning the speaker’s or writer’s intention. An alternate or less popular definition of a word could add a new twist to our understanding. And it is often body language, intonation or what is intimated between the lines that offers us a fuller comprehension. One who simply accepts something at face value may be discounting a broader range of values.
Expanding our minds is akin to de-focusing our eyes. To practice de-focusing your eyesight, here is an exercise. Notice the two pink squares above the jelly beans. Simply cross your eyes slightly as you look at the two pink squares until it looks as though there are three squares rather than two. Then gently direct your gaze down at the jelly beans. By de-focusing your eyes, did you discover a deeper dimensionality and an overall richer impression of the jelly beans?
With our eyes focused narrowly as usual, we see jelly beans arrayed two-dimensionally before us. By intentionally de-focusing our eyes, we see the jelly beans hovering before us as if suspended in three-dimensional space. The narrow focus gave us less information than the de-focused view.
This jellybean image gives us an opportunity to discover that by habitually locking down our point-of-view we forfeit, by default, the very choice-making process that affects our realizations (what we bring into existence and/or what we make real unto ourselves).
It is empowering to learn that what becomes our reality is determined by how we consider the options before us. The trick is to open our minds and heighten our sensitivity so the intervals in which choices can be made do not flit by unnoticed. | 10,194 | 48,727 | {"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-2020-50 | longest | en | 0.919082 |
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Magma
IsAlternative
test whether a magma is alternative
Calling Sequence IsAlternative( m )
Parameters
m - Array representing the Cayley table of a finite magma
Description
• The IsAlternative command returns true if the given magma satisfies the laws (X*X)*Y = X*(X*Y) and X*(Y*Y) = (X*Y)*Y. It returns false otherwise.
Examples
> $\mathrm{with}\left(\mathrm{Magma}\right):$
> $m≔⟨⟨⟨1|2|3⟩,⟨2|3|1⟩,⟨3|1|2⟩⟩⟩$
${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {1}\\ {3}& {1}& {2}\end{array}\right]$ (1)
> $\mathrm{IsAlternative}\left(m\right)$
${\mathrm{true}}$ (2)
> $m≔⟨⟨⟨1|2|3⟩,⟨2|3|3⟩,⟨3|1|2⟩⟩⟩$
${m}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {2}& {3}& {3}\\ {3}& {1}& {2}\end{array}\right]$ (3)
> $\mathrm{IsAlternative}\left(m\right)$
${\mathrm{false}}$ (4)
Compatibility
• The Magma[IsAlternative] command was introduced in Maple 15. | 369 | 978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "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-2021-49 | latest | en | 0.321768 |
http://oeis.org/A019668 | 1,500,921,894,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424909.50/warc/CC-MAIN-20170724182233-20170724202233-00265.warc.gz | 230,370,979 | 3,342 | This site is supported by donations to The OEIS Foundation.
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A019668 Decimal expansion of sqrt(Pi*e)/24. 1
1, 2, 1, 7, 6, 1, 7, 6, 5, 2, 2, 1, 7, 6, 1, 5, 7, 7, 6, 8, 9, 2, 3, 4, 2, 9, 4, 8, 3, 7, 8, 2, 0, 0, 9, 6, 4, 7, 4, 9, 1, 2, 8, 1, 4, 6, 9, 3, 6, 0, 6, 9, 5, 2, 4, 5, 9, 0, 7, 2, 5, 0, 0, 4, 7, 7, 9, 9, 8, 1, 7, 3, 8, 4, 8, 4, 5, 6, 5, 3, 6, 8, 5, 1, 8, 3, 1, 5, 2, 3, 6, 5, 6, 9, 6, 0, 7, 8, 8 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Ivan Panchenko, Table of n, a(n) for n = 0..1000 MATHEMATICA RealDigits[Sqrt[Pi*E]/24, 10, 120][[1]] (* Harvey P. Dale, Jan 24 2016 *) CROSSREFS Sequence in context: A082791 A077230 A244238 * A091700 A157743 A135895 Adjacent sequences: A019665 A019666 A019667 * A019669 A019670 A019671 KEYWORD nonn,cons AUTHOR STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 546 | 1,148 | {"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-2017-30 | latest | en | 0.571173 |
http://www.docstoc.com/docs/96830721/Goodness-of-fit-methods-for-regression-models | 1,395,061,740,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678705611/warc/CC-MAIN-20140313024505-00038-ip-10-183-142-35.ec2.internal.warc.gz | 272,245,682 | 17,140 | # Goodness-of fit methods for regression models
Document Sample
``` Goodness-of fit methods for regression models
Jelle Goeman
Department of Medical Statistics and Bioinformatics
Leiden University Medical Center
Leiden, the Netherlands
Goodness-of fit methods for regression models – p. 1/3
Regression models
n subjects with data (xi , Yi )
Outcome Yi is related to covariates via xi β
with
xi vector of covariates
β vector of regression parameters
Examples:
• Linear regression
• Poisson regression
• Logistic regression
• Multinomial logistic regression
• Cox regression
Goodness-of fit methods for regression models – p. 2/3
How to determine the fit of such a model?
Binary outcome Y with values 1 and 0
Model for pi = P (Yi = 1) :
exp(xi β)
pi =
1 + exp(xi β)
logit(pi ) = xi β
Goodness-of fit methods for regression models – p. 3/3
Goodness-of-fit for logistic regression
Many methods exist:
Hosmer-Lemeshow test
A chi-squared type test
10 groups are made, based on predicted probabilities
Implemented in standard software (SPSS, SAS)
Tests based on extending the model
Reweighting cases and controls Nagelkerke et al (Stat in Med
2004)
Goodness-of fit methods for regression models – p. 4/3
More goodness of fit methods
smoothing methods
Compare model to smooth alternatives
Consider smoothed residuals
Goodness-of fit methods for regression models – p. 5/3
Our approach uses smoothing residuals
Residuals
ei = Yi − pi
with pi the predicted probability that Yi = 1.
Smoothed residual:
˜
ei = wij ej
j
with wij smoothing coefficients.
A weighted sum of residuals in the neighbourhood of i
Goodness-of fit methods for regression models – p. 6/3
Example: data from early arthritis clinic
570 patients with undifferentiated arthritis
Outcome Reumatoid arthritis (RA)after 1 year yes/no
Logistic model with CRP ( C-reactive protein, blood test which
indicates inflammation) as covariate
yields: logit (pi )= -1.063 + .012 CRP
Does the model fit?
Goodness-of fit methods for regression models – p. 7/3
Plot of residuals, with kernel smooth
Goodness-of fit methods for regression models – p. 8/3
Test based on smoothed residuals
T = e2
˜i
i
In matrix form
T = (Y − p) W W (Y − p)
with W matrix of smoothing coefficients wij .
Goodness-of fit methods for regression models – p. 9/3
Distribution of T
T is quadratic form. Under H0 , E(T) and var(T) can be
calculated exactly assuming that β is known
Correction for estimation of parameters. Shrinkage of E(T)
and var(T).
Distribution can be approximated by scaled chi-square, cχ2 ,
(ν)
with c = var (T )/(2E (T )) and ν = 2(E (T ))2 /var (T ).
Goodness-of fit methods for regression models – p. 10/3
How to smooth?
Choice of smoothing method – We use Kernel smoother
(uniform kernel)
Choice of distance measure
We standardize numeric covariates
Mixture of Euclidean distance (numeric covariates) and
number of equal values for discrete covariates
Goodness-of fit methods for regression models – p. 11/3
Choice of Bandwidth
Choice of bandwidth is important
Small bandwidth: no power
Large bandwidth: deviations are smoothed away
Suggestion
1. Use 0.5* average distance between observations
2. Compute test statistic for different bandwidths and look at
distribution of maximum. Test statistics are highly
correlated, so this approach is quite powerful.
Goodness-of fit methods for regression models – p. 12/3
Remarks
Easy to calculate (Software is available in SAS and R)
Sample size can be large: 1600 observations is no problem
Handles continuous and categorical covariates
Could be used as global gof test (smooth in all directions)
Could be used in a specific direction (smooth for one
covariate)
if W is identity matrix, T equals Copas unweighted sum of
squares statistic (which performs quite well in simulations of
Hosmer et al (Stat in Med 1997, 2002))
Goodness-of fit methods for regression models – p. 13/3
The RA example: model with only CRP as covariate
P-value of T for different bandwidths
0.001
pvalue
0.5*average distance
0.01
0.05
0.10
0.15
0.0 0.5 1.0 1.5
Bandwidth
Goodness-of fit methods for regression models – p. 14/3
Several models in the RA example
Covariates p-value test Deviance
CRP 0.02 689.80
CRP CRP2 0.01 678.28
log(CRP) 0.11 677.36
CRP in 3 categories 0.79 672.63
Goodness-of fit methods for regression models – p. 15/3
Extension to multinomial regression models
Outcome Y has g categories
Model for
exβs
P (Y = s) = g xβt
, s = 1, · · · , g
t=1 e
Here
g n n
2
T = wij ([Yj = s] − pjs )
s=1 i=1 j=1
Derivations more complex
Submitted for publication (Goeman and le Cessie(2005))
Goodness-of fit methods for regression models – p. 16/3
RA example: multinomial regression
Outcome at 1 year with 4 categories: RA (177), Probable RA(93),
Other disease(94), Undifferentiated arthritis(205)
Results goodness of fit test:
Covariates p-value test Deviance
CRP 0.0003 1484.8
log(CRP) 0.11 1471.5
CRP in 3 categories 0.79 1465.3
Goodness-of fit methods for regression models – p. 17/3
Residual plots, model with CRP as covariate
outcome = RA outcome = probable RA
1.0
1.0
0.5
0.5
residual
residual
0.0
0.0
-0.5
-0.5
-1.0
-1.0
0 50 100 150 200 0 50 100 150 200
CRP CRP
outcome = Other disease outcome = Undifferentiated arthrites
1.0
1.0
0.5
0.5
residual
residual
0.0
0.0
-0.5
-0.5
-1.0
-1.0
0 50 100 150 200 0 50 100 150 200
CRP CRP
Goodness-of fit methods for regression models – p. 18/3
Relation with random effect models
If model does not fit
0.05
0.04
0.03
FITTED MODEL FITTED MODEL
Y
IS TOO LOW IS TOO LOW
0.02
0.01
FITTED MODEL
IS TOO HIGH
0.0
0.0 0.2 0.4 0.6 0.8 1.0
X
Neighbouring points deviate in the same direction
Positive correlation between neighbouring residuals fit methods for regression models – p. 19/3
Goodness-of
Use random effect models
Is model logit(pi ) = xi β adequate ?
Embed model in family of random effect models
logit(pi ) = xi β + ri
with ri random effects, E(ri ) = 0; cov (ri ) = τ 2 R
Goodness of fit test: score test for H0 : τ 2 = 0
Let R depend on the distance between observations in
covariate space.
Close observations have large correlations
Goodness-of fit methods for regression models – p. 20/3
Score test statistic has the form:
T = (Y − pi ) R(Y − pi )
This is exactly the same as our smoothed residual test
with R = W W
For multinomial regression the same relation exists (more
complex: multivariate random effects, see our paper)
Goodness-of fit methods for regression models – p. 21/3
It is clear to what alternatives our goodness of fit test has
optimal power
Can be extended to other areas (GLM, Survival, Genetic
family trees, micro array data)
Goodness-of fit methods for regression models – p. 22/3
Why not fit the random effect model?
Can be done with standard software
No problem with choice bandwidth (use the one
corresponding to largest likelihood)
Use likelihood ratio test as goodness of fit test (distribution
0.5(χ2 + χ2 ))
(1) (2)
Posterior means of random effects as diagnostic tool
Goodness-of fit methods for regression models – p. 23/3
Linear regression
Data of 702 infants from Malawi (birth weight vs gestational age)
Goodness-of fit methods for regression models – p. 24/3
Does linear model fit well ?
Use proc mixed in SAS to fit model with random terms
Different Rs:
Exponential correlation function
Rij = exp(−(||xi − xj ||/h))
Gaussian correlation
Rij = exp(−((xi − xj )2 /h))
proc mixed data=malawi cl method=ml ;
class babycode;
model gew= age/solution outp=posterior
outpm=model residual;
random babycode/ type=sp(exp)(age)solution ;
run;
Goodness-of fit methods for regression models – p. 25/3
Results
Sometimes convergence to boundary → try different starting
values
Deviance model without random effect 3997.2
Deviance model with exponential error structure 3991.0
Deviance model with Gaussian error structure 3990.5
Estimate of τ 2 was 0.6041 (95%CI 0.18 ; 12.6)
Goodness-of-fit test:6.2, p=0.02
A squared term in gestational age is significant (p=0.0047)
Goodness-of fit methods for regression models – p. 26/3
Plot of posterior means xi β + E[ri |Y ]
et al fit methods for
Relation with non parametric smoothing (Verbyla Goodness-of 1999)regression models – p. 27/3
Logistic regression
Fit alternative model with correlated random effect
Use SAS PROC GLIMMIX
Model almost always converges to boundary (either 0 or
infinite estimates)
Did not work out
Goodness-of fit methods for regression models – p. 28/3
Back to the RA example
In total 21 possible predictors, with 28 parameters
After variable selection 11 predictors, 13 parameters
Goodness-of fit methods for regression models – p. 29/3
Goodness of fit test is not significant
P-value of T for different bandwidths
0.001
pvalue
0.01
0.05
0.5*average distance
0.10
0.20
0.30
0.40
0 1 2 3 4 5 6
Bandwidth
Goodness-of fit methods for regression models – p. 30/3
Is it a clinical useful model?
Histogram of predicted probabilities
Goodness-of fit methods for regression models – p. 31/3
Is it a clinical useful model?
Look at predicted probabilities of RA
53 % has pred < 0.2 ⇒ probably no RA ⇒ No treatment
10 % has pred > 0.8 ⇒ probably RA within year ⇒ Start
treatment
36 % has pred between 0.2 and 0.8. We do not know what will
happen: unclassifiable
Hence we could make a "Tailor made" prediction for 63% of the
patients.
36 percent can not be classified
Difference between fit and predictive value
Goodness-of fit methods for regression models – p. 32/3
Conclusions
A simple goodness-of fit test for regression models
Goodness of fit test using smoothed residuals is equivalent to
score tests for random effects
Yields residuals for further diagnostics
It would be nice to fit the alternative model
Then posterior means of random effects can be used as
diagnostic tool
This is a kind of non parametric smoothing
In practice not easy to do
Software on our website
Goodness-of fit methods for regression models – p. 33/3
```
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views: 16 posted: 9/27/2011 language: English pages: 33 | 3,060 | 10,696 | {"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-2014-10 | longest | en | 0.774039 |
https://community.esri.com/t5/geoprocessing-questions/increase-raster-resolution/m-p/323372 | 1,652,907,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00544.warc.gz | 243,071,489 | 42,766 | # Increase raster resolution
2085
4
04-05-2017 02:55 PM
New Contributor II
I have a raster layer where each pixel is 10 meters. I would like to reduce the pixel size to 5 meters. I have not had any success finding the right tool (if there is one). I have converted the 10 meter raster to a point feature and tried to re-convert it at a 5 meter cell size but that just left voided areas. Maybe I'm missing a step? I have spatial and 3D analyst tools. Can someone point me in the right direction?
Tags (1)
1 Solution
Accepted Solutions
MVP Esteemed Contributor
4 Replies
MVP Esteemed Contributor
MVP Esteemed Contributor
If you are upsampling, for example, from 5 meter to 10 meter, the tool of choice is Aggregate. Unless of course you want to do it in numpy, Dan!
MVP Esteemed Contributor
Avoid the aggrevation of Aggregation using Numpy .... from this blog post
and to avoid repeating myself... and just halve the cell size when bringing back into arcmap
``````a
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
np.repeat(np.repeat(a, 2, axis=0), 2, axis=1)
array([[0, 0, 1, 1, 2, 2],
[0, 0, 1, 1, 2, 2],
[3, 3, 4, 4, 5, 5],
[3, 3, 4, 4, 5, 5],
[6, 6, 7, 7, 8, 8],
[6, 6, 7, 7, 8, 8]])
``````
New Contributor II
Thanks. This is it | 456 | 1,273 | {"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} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.797287 |
https://www.jiskha.com/search?query=there+are+4+packets+of+muesli+sold+for+every+3+packets+of+bran.The+owner+sold+84+packets+of+muesli+and+bran+today.how+many+packets+of+muesli+did+the+store+sell+today | 1,534,638,838,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213903.82/warc/CC-MAIN-20180818232623-20180819012623-00338.warc.gz | 929,096,108 | 14,231 | # there are 4 packets of muesli sold for every 3 packets of bran.The owner sold 84 packets of muesli and bran today.how many packets of muesli did the store sell today
13,967 results
1. ## math
there are 4 packets of muesli sold for every 3 packets of bran.The owner sold 84 packets of muesli and bran today.how many packets of muesli did the store sell today
2. ## Math
A truck was packed with cartons each containing packets of 500g and 250g. The total mass of the packets was 900kg. If the number of 500 packets was 1000, what was the number if 250 packets?
3. ## math
A shopkeeper bought cooking fat in 30 one kilogram packets and 20 five hundred packets. She repacked the fat in packets of 125 grams each. How many packets did she get?
4. ## Statistics
Machines are used to pack sugar into packets supposedly containing 1.20 kg each. On testing a large number of packets over a long period of time, it was found that the mean weight of the packets was 1.24 kg and the standard deviation was 0.04 Kg. A
5. ## statistics
Machines are used to pack sugar into packets supposedly containing 1.20 kg each. On testing a large number of packets over a long period of time, it was found that the mean weight of the packets was 1.24 kg and the standard deviation was 0.04 Kg. A
6. ## SMU
Ans:- Machines are used to pack sugar into packets supposedly containing 1.20 kg each. On testing a large number of packets over a long period of time, it was found that the mean weight of the packets was 1.24 kg and the standard deviation was 0.04 Kg. A
7. ## statistics/probability
QUESTION- machines are used to pack sugar into packets supposedly containing 1.20kg each.on testing a large number of packets over a long perodof time ,it was found that the mean weight of the packets was 1.24kg and the standard deviation was 0.04kg. a
8. ## Math
Elena bought 49 packets of red balloons, 66 packets of blue balloons, and 35 packets of yellow balloons. Each pack contained 12 balloons. She mixed them up and gave away some balloons. She then repacked the balance into packets of 25. How many balloons
9. ## Maths
Please help...I just can't figure it out.. Amanda, Belinda and mary share 9x packets of beads. Amanda gets x packets more than Belinda. mary gets 6 packets more than Belinda. (a) how many packets does belinda get in terms of x? (b) if amanda gets 3 packets
10. ## Maths Problem
Can anyone help me please? Amanda, Belinda and mary share 9x packets of beads. Amanda gets x packets more than Belinda. mary gets 6 packets more than Belinda. (a) how many packets does belinda get in terms of x? (b) if amanda gets 3 packets more than
11. ## Math
Pupils contributed some money to help needy people. They bought twenty four 2-kg packets of flour, thirty six 1-kg packets of flour and fifty kilogram bag of sugar. The flour was packed in 500 g packets and the sugar in 250 g packets. Hominy people got
12. ## math
Theta taa factroy produces equal number of 1/4kg packets and 1/2kg packets of tea. How many 1/4kg packets were produced from a totol of 45 tonnes of tea.
13. ## math
Theta taa factroy produces equal number of 1/4kg packets and 1/2kg packets of tea. How many 1/4kg packets were produced from a totol of 45 tonnes of tea. Pliz explain
14. ## math
Students contributed some money to help needy people. They bought twenty four 2-kg packets of flour, thirty six 1-kg packets of flour and a fifty kilogram bag of sugar. The flour was packed in 500 g packets and the sugar in 250 g packets. How many
15. ## math
Pupils contributed some money to help needy people. They bought twenty four 2-kg packets of flour, thirty six 1-kg packets of flour and a fifty kilogram bag of sugar. The flour was packed in 500g packets and the sugar in 250 g packets. How many people
16. ## Math
Pupils contributed some money to help needy people. They bought twenty four 2-kg packets of flour, thirty six 1-kg packets of flour and a 50 kilogram bag of sugar. The flour was packed in 500g packets and the sugar in 250 g packets. How many people got
17. ## math
Agus sold packets of sweets to Halim and Kairi in the ratio 8:13.He sold the sweets in 2-kg packets.What was the smallest possible mass of sweets Agus sold to both of them?
18. ## milli
elena bought 49 packets of red balloons, 66 packets of blue balloons, and 35 packets of yellow ballons. each packet contained 12 balloons. she mixed them up and gave away some balloons..she then repacket the balance into packet of 25
19. ## maths
An entire bag containing 4.8kg of coffee was repacked into smaller packets. each containing 250g. a) How many smaller packets were obtained? b) How much grams of coffee remained after the maximum amount of packets were obtained?
20. ## math
An entire bag containing 4.8kg of coffee was repacked into smaller packets. each containing 250g. a) How many smaller packets were obtained? b) How much grams of coffee remained after the maximum amount of packets were obtained?
21. ## math
Cathy and Judy bought some postcard packets.altogether there were 48 cards.Cathy bought three packets with 8 cards each.The packets Judy bought had 12 cardsin each.how many packets did Judy buy.
22. ## math
John and David had packets of tea to be packed into cartons. Each carton holds 46 packets. John packed 63 cartons and remained with 24 packets while David packed 54 cartons and remained with 19 packets. How many more packets of tea had John than David? =
23. ## aig math
A bookshop had 60 boxes of pencils. Each box contained 85 pencils. Half of the pencils were packed into packets of 25 pencils and the rest were packed into packets of 20. How many packets of pencils were there altogether? My answer is 229 am i right? How
24. ## math
The owner of a candy store packs all of the sweets from a box into packets of 9, 12, 18 sweets . a. what is the smallest number of sweets a box must contain if no sweets are to remain after packing? b. There are 180 sweets to be packed. what is the largest
25. ## math
A bookshop had 60 boxes of pencils. Each box contained 85 pencils. Half of the pencils were packed into packets of 25 pencils and the rest were packed into packets of 20. How many packets of pencils were there altogether? My answer is 229 am i right? How
26. ## Math
I am also having big trouble with this math problem. A teacher is making identical activity packets using 92 crayons and 23 sheets of paper. What is the greatest number of packets the teacher can make with no items left over? Please help me.
27. ## math
Twenty-four 5-decilitre packets of milk were emptied into a 50-litre container. How many more such packets of milk were needed to fill the container?
28. ## mathsP5
There are 2 packets of flour, Y & Z.The ratio of the mass of Y to Z is 4:1. If 26g is added to Z,the ratio of the mass of Y to Z is 7:5. Find the mass of the 2 packets at first. Pls show how to solve by model method as i have not learnt algebra. tks
29. ## Math
A rectangular container measures 4m by 3m by 500cm. The container is completely filled with 0. 5 long cubic packets. How many such packets are needed to filled 3/4 of the container? V = lwh = 3(4)(5) = 60
30. ## math
A factory produced 65, 160 sweets. The sweets were packed in packets each holding sweets. The cost of each packet was \$72. All the packets were then equally put in 15 cartons. What was the cost of the sweets in each carton?
31. ## AP Calculus BC Resources
Hi there, I was hoping if any one was familiar with any sites that have printable integration/differentiation packets? I'm hoping to use them for everyday practice. I skipped Calculus AB and need a little help. I love packets so if there are any out there
32. ## Maths
There are 2 packets of flour, Y & Z.The ratio of the mass of Y to Z is 4:1. If 26g is added to Z,the ratio of the mass of Y to Z is 7:5. Find the mass of the 2 packets at first.
33. ## Maths
There are 2 packets of flour, Y & Z.The ratio of the mass of Y to Z is 4:1. If 26g is added to Z,the ratio of the mass of Y to Z is 7:5. Find the mass of the 2 packets at first.
34. ## math
A box full of packets of biscuits each weighing 200 g weighs 12 1/2 kg.The empty box weighs 2 kg 500 g. calculate the number of packets of biscuits in the box.
35. ## Math
There are 10 packets of ham. Of the packets, 2/5 are turkey ham. Each packet of turkey ham weighs 1/3 pound. What is the total weight of the turkey ham?
36. ## statistics
question- a packaging device is set to fill detergent powder packets with a mean weight of 5 kg .the standered deviation is known to be 0.01kg.these are known to drift upward over a period of time due to machine fault,which is not tolerable.a random sample
37. ## Statistics
A packaging device is set to fill detergent power packets with a mean weight of 5 Kg. The standard deviation is known to be 0.01 Kg. These are known to drift upwards over a period of time due to machine fault, which is not tolerable. A random sample of 100
38. ## physics
A cannon is shooting emergency packets to people stranded on the roof of a flooded building of height H = 104 meters relative to the cannon, the corner of which is located a distance D = 53 meters from the cannon. It is desired that the incoming packets
39. ## marketing research
A packaging device is set to fill detergent power packets with a mean weight of 5 Kg. The standard deviation is known to be 0.01 Kg. These are known to drift upwards over a period of time due to machine fault, which is not tolerable. A random sample of 100
40. ## statistics
a packaging device is set to fill detergent powder packets with a mean weight of 5kg.The standard deviation is known to be .01kg.these are known to drift upwards over a period of time due to machine fault which is not tolerable. A random sample of 100
41. ## Math
Unable to derive any clues from the following. Please help! Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five,
42. ## math
i still don't get it ...Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six and have just one cookie
43. ## math
ok i got the multiples of 7 so what do i do next ...Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six
44. ## math
still don't get it.....Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six and have just one cookie left
45. ## math
Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six and have just one cookie left over in each case.If
46. ## math
Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six and have just one cookie left over in each case.If
47. ## physics
4 packets of charges are arranged in a circle. Each packet is 90 degrees from each other. Each packet consists of 22 elementary charges associated with the nucleus of two sodium atoms. A test charge is placed in the center of the circle of packets. The
48. ## physics
4 packets of charges are arranged in a circle. Each packet is 90 degrees from each other. Each packet consists of 22 elementary charges associated with the nucleus of two sodium atoms. A test charge is placed in the center of the circle of packets. The
49. ## algebra
Baking Cookies-Word Problem Our class planned a party for disadvantaged kids. Some of us baked cookies for the party. On the day of the party, we found we could divide the cookies into packets of two,three, four, five or six and have one cookie left over
50. ## Maths
A shopkeeper buys a number of blue pens and red pens. He pack 3 blue pens and 2 red pens into a packets for sale. If he wants to obtain 160 packets of the pens he needs another 25 blue pens. What is the total number of the pens bought by him at first?
51. ## Computer networks
Consider sending a file of 2Mbits from Host A to Host B. There are three links and two store-and-forward routers (or switches) between A and B. Assume that no link is congested (that is, no queueing delays), and the one-way propagation delay of each link
52. ## Statistics - Probability
Viruses reach networks by requests for communication that are identified as packets. A security system can detect them if at least one such packet is observed by a network sensor. Assume that the probability of observing a single packet sent from a new
53. ## Statistics/Probability
5) The product is sold in packets whose masses are normally distributes with mean of 1.42 kg and a standard deviation of 0.025 kg. (1) Find the probability that the mass of a packet, selected at random, lies between 1.37 kg and 1.45 kg
54. ## Algebra word problem
Baking Cookies-Word Problem Our class planned a party for disadvantaged kids. Some of us baked cookies for the party. On the day of the party, we found we could divide the cookies into packets of two,three, four, five or six and have one cookie left over
55. ## Math
please if anyone could help me with this question: 1. A class paid \$20 for a cake and \$4 per child for one slice of cheese pizza. they paid \$140. how many children are in the class? 2. Rajie want to divide 24 peanuts, 64 raisins, and 56 apricots evenly
56. ## STATISTICS
In the mass production of bolts it is found that 5% are defective. Bolts selected at random and put into packets of ten. A packet is selected at random. Find the probability that it contains (a) three defective bolts. (b) less than three are defective. Two
57. ## computer networks.
Question 1 (PIM) Assume you have a source S, and assume that some receivers have joined the shortestpath tree of S, while other receivers have joined only the shared tree. Prove that, if the network is in a steady state (no routing table changes, no
58. ## Maths
URGENTLY ANSWER RATIO MATHS !!! Please EXPLAIN the steps fully CLEARLY, I have AlREADY GOT ANSWER so no need to worry. This is for GCSE just so you know. I want explanation for both problems. Ratio Problem 1- Chantel makes muesli using wheat flakes and
59. ## Maths
URGENTLY ANSWER RATIO MATHS !!! Please EXPLAIN the steps fully CLEARLY, I have already got answer so no need to worry. This is for GCSE just so you know. Ratio Problem 1- Chantel makes muesli using wheat flakes and dried fruit in ratio 8:5. She has a 200g
60. ## Intro to Networking
While working with networks, does a larger number of hops have a effect on the delay of packets?
61. ## mathematics
seeds of type A and type B are sold in packets, each must contain 1) both type a and type b seeds 2) at least twice the amount of type B as there are type A 3) no more than 12 seeds state the minimum number in each paket of type A and of type B if here are
62. ## Mathematics
Anthony uses 1.2 metres of leather and one buckle when he makes a leather belt. He can make 15 belts per hour and he values his labour \$11 per hour. Anthony buys his supplies in bulk. The leather comes in rolls 50 metres long at a cost of \$180 per roll and
63. ## Math
Two shopkeepers, Juma and Wan bought some items from a wholesaler. Juma bought 18 loaves of bread, 40 packets of milk and 5 bars of soap while Wan bought 15 loaves of bread, 30 packets of milk and 6 bars of soap. The price of a loaf of bread, a packet of
64. ## Career Planning
Sarah needs to complete financial aid packets. During which school year would she do this? A.) freshman B.) sophomore C.) junior D.) senior I think that it is D...?
65. ## Math
Two shopkeepers, Juma and Wan bought some items from a wholesaler. Juma bought 18 loaves of bread, 40 packets of milk and 5 bars of soap while Wan bought 15 loaves of bread, 30 packets of milk and 6 bars of soap. The price of a loaf of bread, a packet of
66. ## Computer Science
Data packets travel from router on layers 3 and 4. As this happens their headers change. Explain why this is the case with reference to ip and mac adresses.
67. ## maths
5 packets of different sizes have a total mass of 800g find their average mass.
68. ## Math,Word problems(algebra)
Kai Ling has 4m kg of flour.She bought 2 more packets of flour, each of mass m kg.How much does she has now in terms of m? Please help me
69. ## maths
10 pencils in each packet 46 pencils how many packets are needed
70. ## mathematics
seeds of type a and b are sold in packets each containing both type a and b seeds at last twice the number of twice b seeds as there are in type a seeds and no more than 12 seeds a) using a scale of 1cm for each unit on both axis draw a graph n the axis to
71. ## Astronomy
Light is absorbed and emitted in the form of ________, which are “packets” of electromagnetic radiation that carry a specific amount of energy. photons ions excited atoms?? protons
72. ## math
Danny bought 6 packets of drink. Each packet contained 1/4 liter of drink. Find the total amount of drinks in liters.
73. ## algebra
mrs.Goh bought 4 bags of rice at \$5.28 per bag ,5 packets of biscuits at \$1.25 per packet and 2.8 kg of meat at \$7.40 kg per kg.Assuming that GST was absorbed by the retailer,calculate the total amount she spent ? What would be change if she paid for all
74. ## college math
A camper has 3 jackets, 6 sleeping bags, and 14 different packages of food. An "outfit" consists of 1 jacket, 1 sleeping bag, and 7 packets of food. How many different outfits are there?
75. ## chemistry
What does the fact that a blackbody can absorb light only in specific-sized "packets" say about light? A. It has no mass. B. It is composed of electrons. C. It is composed of atoms. D. It is quantized. I have no idea what the answer is
76. ## Finite
Linear programming A candy merchant sells two variety bags of cookies. Each pound of variety X contains 60 % chocolate chips and 40% raisin bran and sells for \$ 8.00 a pound. Each pound of variety Y contains 45% of chocolate chips and 55% raisin bran and
77. ## math
Assume that the camper has 3 jackets, 7 sleeping bags, and 9 different packages of food. An outfit consists of 1 jacket, 1 sleeping bag, and 6 packets of food. How many ways can the trip be outfitted?
78. ## Math
Eva is making bran muffin. She measures5/4 cups of bran, 1/4 cup of flour, and 7/4 cups of oats. What is the order of the ingredients from least to greatest amounts?
79. ## math
Students are baking muffins to sell to raise money. They bake muffins in trays of 24 muffins each. They want to divide the muffins on each tray into equal packets. Which packet size would result in muffins from each tray being left over? A.3 muffins per
80. ## Math 10
Cashew nuts are sold at 15 per kg. Walnuts are sold at 12 per kg. What quantities of each nut would a store owner put into a 100-kg barrel so that it could be sold for 13.20 per kg?
81. ## World History
So, last week Coach gave us 3 packets by Blackline Masters to do but I wasn't at school most of last so I didn't get them until Friday and he isn't giving me any extra time, and tneyre due tomorrow, and im kind of freaking out, does anyone know where I
82. ## algebra 2
A garden supply store sells two types of lawn mowers. The cashiers kept a tally chart for the total number of mowers sold. They tallied 30 mowers sold. Total sales of mowers for the year were \$8379.70. The small mowers cost \$249.99 and the large mowers
83. ## algebra 2
A garden supply store sells two types of lawn mowers. The cashiers kept a tally chart for the total number of mowers sold. They tallied 30 mowers sold. Total sales of mowers for the year were \$8379.70. The small mowers cost \$249.99 and the large mowers
84. ## Math
Assume that the camper has 3 jackets, 3 sleeping bags, and 11 different packages of food. An outfit consists of 1 jacket, 2 sleeping bags, and 4 packets of food.How many ways can the trip be outfitted?
85. ## math
Assume that the camper has 3 jackets, 3 sleeping bags, and 11 different packages of food. An outfit consists of 1 jacket, 2 sleeping bags, and 4 packets of food.
86. ## Math (again)
Sorry...I forgot to include all of the details for one of my previous questions the complete problem is: 1. Yesterday Josh sold some boxes of greeting cards. Today he sold seven boxes. If he sold 25 boxes in all, how many did he sell yesterday? I can solve
87. ## Economics
Could someone please help me to asnwer these tw economics questions? The following table shows the marginal benefits (MB) of consuming chocolate bars. Chocolate Bars (unit) 1 2 3 4 5 MB \$10 \$8 \$6 \$4 \$2 Suppose that the market price of chocolate bars is \$7
88. ## science
A class is to perform an experiment to study the effects of various plant foods on a type of indoor plant.The teacher provides three similar plants,a water supply,and three different packets of plant food.Explain how you might conduct a controlled
89. ## statistics
2 Packets of crisps from a production line have a weight distribution that is normal with a mean of 50.0 grams and a standard deviation of 2.0 grams. Samples are taken of size 36.
What is the probability that a sample mean is less than 49.0
90. ## Chemistry
One medium bran muffin connateness 4.2 g of fiber. How many ounces of fiber are obtained by eating three medium bran muffins? I need help understanding the conversion factor process
cassie and connor are selling peaches yesterday cassie sold 31 peachesconnor sold p fewer peaches than cassie. a. suppose connor sold 50 peaches today explain how to write an expression for the number of peaches cassie sold today if connor sold p fewer
92. ## internet
Consider sending a series of packets from a sending host to a receiving host over a fixed route. List the delay components in the end-to-end delay for a single packet. Which of these delays are constant and which are variable?
93. ## MATH
Assume that the camper has 4 jackets, 2 sleeping bags, and 11 different packages of food. An outfit consists of 2 jackets, 2 sleeping bags, and 7 packets of food. How many ways can the trip be outfitted?
94. ## math
Cashew nuts are sold at \$15.00/kg. Walnuts are sold at \$12.00/kg. What quantities of each nut would a store owner put into a 100-kg barrel so that it could be sold for \$13.20/kg? ---------------- 15.00c+12.00w= 31.20 C+w= 100 Are these equations right?
95. ## math,(fraction)
Becky bought 25kg of sweets.she kept 1kg of sweets for her use and packed the remaining sweets equally into some packets to distribute among some children.each packet contained 1/2kg of sweets.if each child received one packet of sweets,how many children
96. ## math
When winter jackets are sold for \$250 each, an outdoors store can sell 110 in a season. For every \$25 nicrease in the price, the number sold drops by 10. If the total sales revenue is \$22500, how many of these jackets were sold?
97. ## math metric unit
A box of hot cocoa packets has 30 grams of hot cocoa mix. How many boxes of mix can be made from 6 kilograms of hot cocoa mix. | 5,945 | 23,491 | {"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} | 3.59375 | 4 | CC-MAIN-2018-34 | latest | en | 0.954653 |
https://byjusexamprep.com/a-person-cannot-see-beyond-2-m-if-he-has-to-read-a-signboard-4-m-away-then-the-power-of-the-lens-required-is-i | 1,685,572,888,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647459.8/warc/CC-MAIN-20230531214247-20230601004247-00655.warc.gz | 179,510,811 | 55,266 | # A person cannot see beyond 2 m. If he has to read a signboard 4 m away, then the power of the lens required is
By Mandeep Kumar|Updated : May 23rd, 2023
A person cannot see beyond 2 m. If he has to read a signboard 4 m away, then the power of the lens required is -0.25 D. A lens is a transmissive optical device that, via the use of refraction, concentrates or disperses a light beam. While a compound lens is made up of numerous simple lenses that are often aligned along a common axis, a simple lens is made up of a single transparent piece.
## Power of the Lens if a Person cannot See beyond 2m & has to Read Signboard at 4m
The farthest distance that a person can see clearly is referred to as their far point, and the closest distance that they can see well is referred to as their near point. Many people experience vision issues as they age because of an inherited eye condition. To calculate the power of the lens, Lens formula has to be used which is 1/f = 1/v − 1/u or P = 1/f.
By putting the values in 1/f = 1/v − 1/u, we will get,
1/f = 1/(-2) - 1/(-4)
1/f = 1/(-4)
1/f = −0.25m−1
Therefore, the power of the lens will be -0.25 D.
Summary:
## A person cannot see beyond 2 m. If he has to read a signboard 4 m away, then the power of the lens required is
-0.25 D should be the power of the lens if a person cannot see beyond 2 m and he has to read a signboard 4 m away. A lens is a transmissive optical device that concentrates or disperses a light beam through the use of refraction.
Related Questions: | 419 | 1,530 | {"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} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.949276 |
https://elteoremadecuales.com/l-balance-theorem/ | 1,680,003,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00295.warc.gz | 276,874,789 | 9,373 | # L-balance theorem
L-balance theorem In mathematical finite group theory, the L-balance theorem was proved by Gorenstein & Walter (1975). The letter L stands for the layer of a group, and "balance" refers to the property discussed below.
Statement The L-balance theorem of Gorenstein and Walter states that if X is a finite group and T a 2-subgroup of X then {displaystyle L_{2'}(C_{X}(T))leq L_{2'}(X)} Here L2′(X) stands for the 2-layer of a group X, which is the product of all the 2-components of the group, the minimal subnormal subgroups of X mapping onto components of X/O(X).
A consequence is that if a and b are commuting involutions of a group G then {displaystyle L_{2'}(L_{2'}(C_{a})cap C_{b})=L_{2'}(L_{2'}(C_{b})cap C_{a})} This is the property called L-balance.
More generally similar results are true if the prime 2 is replaced by a prime p, and in this case the condition is called Lp-balance, but the proof of this requires the classification of finite simple groups (more precisely the Schreier conjecture).
References Gorenstein, D.; Walter, John H. (1975), "Balance and generation in finite groups", Journal of Algebra, 33: 224–287, doi:10.1016/0021-8693(75)90123-4, ISSN 0021-8693, MR 0357583 Categories: Theorems about finite groups
Si quieres conocer otros artículos parecidos a L-balance theorem puedes visitar la categoría Theorems about finite groups.
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Utilizamos cookies propias y de terceros para mejorar la experiencia de usuario Más información | 405 | 1,489 | {"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} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.705176 |
https://web2.0calc.com/questions/sqrt-x-sqrt-x-1-equals | 1,576,339,140,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00468.warc.gz | 596,915,334 | 5,870 | +0
sqrt(x)*sqrt(x+1) equals ?
0
631
1
sqrt(x)*sqrt(x+1) equals ?
Jul 8, 2015
#1
+423
+5
$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$
Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)
$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$
Now simplify since obviously the square of a square root is...
$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$
Expand out the equation inside the surd and you get:
$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$
.
Jul 9, 2015
#1
+423
+5
$${\sqrt{{\mathtt{x}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}$$
Let's combine the surds. To do this, square one of them (so put it in terms of it's square root)
$${\sqrt{{{\sqrt{{\mathtt{x}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$
Now simplify since obviously the square of a square root is...
$${\sqrt{{\mathtt{x}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}$$
Expand out the equation inside the surd and you get:
$${\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}}}$$
Sir-Emo-Chappington Jul 9, 2015 | 588 | 1,486 | {"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} | 4.6875 | 5 | CC-MAIN-2019-51 | latest | en | 0.615768 |
https://www.electrical4u.com/temperature-sensor-types-of-temperature-sensor/ | 1,579,745,946,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00360.warc.gz | 855,095,588 | 22,645 | # Temperature Sensor Temperature Measurement | Types of Temperature Sensor
In previous days the thermometer was only used for measuring human body temperature and atmospheric temperature, but now days due to drastic industrial developments, it is essential to measure temperature of various situations and conditions much accurately and precisely. To achieve required accuracy and precision of temperature measurement, different types of temperature sensor have been developed. Depending upon the working principle of temperature sensors, they have broadly classified as,
1. Liquid and gas thermometer
2. Bimetallic strip thermometer
3. Resistance thermometers, which include resistance temperature detector or RTD and thermistors
4. Thermocouple
5. Junction semiconductor sensor
Let’s us discuss all types of temperature sensor one by one in brief.
## Liquid and Gas Thermometer
This is the oldest process of temperature measurement. The method is very simple. In gas thermometer, one glass made bulb is filled with some specific gas and it is observed the different volumes of the gas inside the bulb at different temperatures. Here, Charles Law of Gas is applied, this law states that volume of gas is directly proportional to kelvin temperature of the gas when pressure is kept constant. In gas thermometer, the temperature is calibrated against volume of the gas inside the bulb, considering the pressure of the gas is constant as shown in the basic figure above.
The reading of this type of temperature sensors is not quite reliable as it is highly effected by atmospheric pressure. Another little bit improved version of gas thermometer was also introduced, where volume of the gas inside the bulb is kept constant at all temperatures by adjusting the level of capillary tube attached to the bulb and observed what is the change of head of the liquid in side the capillary tube. Here, also Charles Law of Gas is applied. As we know this law also states that pressure of an enclosed gas is directly proportional to its kelvin temperature when its volume is constant. Here, changed pressure of the gas is calibrated to its temperature. As both of these gas thermometers sense temperature, they can be considered as oldest and simplest version of temperature sensor.
## Bimetallic Strip Thermometer
This is another simple form of thermometer. Here two strips of different metals are joint together to form a bimetallic strip. The coefficient of expansion these two metal must be different enough. Due to these different coefficient of expansion, the metal strips will be expanded or contracted differently for the same temperature change in surrounding. Consequently, whole bimetallic strip assembly will bend or be deformed in shape. This mechanical deformation of the strip, is then caught by a pointer in an indication dial by means of lever – gear arrangement.
## Resistance Thermometers
We know that resistance of any material changes for change in temperature. There is a clear relation between resistance and temperature of a material. This relation can be utilized for temperature measurement. The rate of change of resistance per degree centigrade is known as temperature coefficient of resistance of the material. This coefficient can be either positive or negative. Generally, resistance of metal increases with increase of temperature – hence temperature coefficient of resistance of metal is referred as positive, where as resistance of semiconductor decreases with increase in temperature – hence temperature coefficient of resistance of semiconductor is referred as negative. Although, both metal and semiconductor can be used as temperature sensor. When metal is used as sensor the thermometer is referred as resistance temperature detector or RTD on the other hand when semiconductor is used for same purpose the thermometer is called thermistor.
## Thermocouple Thermometer
In a thermocouple system, two dissimilar metal conductors are connected together at their two ends. Due to this, two common junctions are created and if these two common junctions are exposed to two different temperatures, there will be a net thermal emf produced in the circuit. The value of produced emf depends upon the nature of these two dissimilar metals and the temperature difference between their common junctions. Hence, the temperature difference can easily be calibrated from the emf generated for that. This type of temperature sensors is mainly used in industrial purposes.
## Junction Semiconductor Sensor
A p-n junction diode and bipolar junction transistor have very strong dependency on temperature. If a diode or a BJT is connected a constant current source and if the forward biased voltage of these devices are measured, it would be observed that due to change in temperature there would be very much change in forward biased voltage. Hence this forward biased voltage can easily be calibrated in temperature such as the value of this voltage gives direct temperature measurement. these type of temperature sensors are very sensitive and mainly used in electronics purposes. | 925 | 5,105 | {"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} | 2.625 | 3 | CC-MAIN-2020-05 | longest | en | 0.941549 |
https://file.scirp.org/Html/65458_65458.htm | 1,679,745,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00356.warc.gz | 302,262,884 | 9,946 | On the Measurement of Lower Solution Bounds of the Discrete Algebraic Lyapunov Equation
Journal of Applied Mathematics and Physics
Vol.04 No.04(2016), Article ID:65458,7 pages
10.4236/jamp.2016.44075
On the Measurement of Lower Solution Bounds of the Discrete Algebraic Lyapunov Equation
Chien-Hua Lee
Department of Electrical Engineering, Cheng-Shiu University, Taiwan
Received 14 March 2016; accepted 6 April 2016; published 13 April 2016
ABSTRACT
In this paper, estimations of the lower solution bounds for the discrete algebraic Lyapunov Equation (the DALE) are addressed. By utilizing linear algebraic techniques, several new lower solution bounds of the DALE are presented. We also propose numerical algorithms to develop sharper solution bounds. The obtained bounds can give a supplement to those appeared in the literature.
Keywords:
Discrete Lyapunov Equation, Estimation, Lower Solution Bound, Linear Algebraic Technique
1. Introduction
It is known that the Lyapunov equation is widely used in various control systems. Furthermore, solution bounds of the above equation can also treat many control problems. For example, robust stability analysis for time-delay systems, robust root clustering for linear systems, determination of the size of the estimation error for multiplicative systems, and others can be solved by the mentioned solution bounds. Gajic and Qureshi [1] explained one motive for studying the solution bounds of the Lyapunov equation: sometimes we are simply interested in the general behavior of the underlying system, and this behavior can be determined by examining certain bounds on the parameters of the solution, rather than the full solution. During the past few decades, research on deriving solution bounds of the Lyapunov equation has become an attractive research topic, and a number of research approaches have been proposed to this problem [2]-[9]. Among those results, they focus on the evaluation for the bounds of single eigenvalues including the extreme ones, the trace, the determinant, as well as the bounds of solution matrix. In fact, it has been observed that all the aforementioned solution bounds can be defined by matrix bounds. Hence, the matrix bounds are the most general findings. In the literature, matrix bounds of the solution of the DALE have been studied in [3] [5]-[8]. Many good bounds have been presented. However, it seems that most of these approaches for the matrix bounds contain points of weakness. For example, those results proposed in [5]-[7] must assume that the matrix Q is positive definite. In many control problems, this matrix may be positive semi-definite. Bounds in [5]-[7] also have inverse matrices and the computation burden hence may become very heavy when the system dimension becomes large. The matrix A must be normal in one of the bound presented in [7] and must be diagonalizable in [6]. For another bound in [7], an extra Lyapunov equation must be solved. Furthermore, in [5], the lower solution bound has a free matrix and how to choose this matrix such that the obtained bound is the best is still an open problem. To give a supplement to those appeared in the literature is therefore the aim of this paper. A new approach for developing lower matrix bounds of the solution of the DALE then is proposed. In what follows, it is not necessary to assume that Q is positive definite and A is diagonalizable or normal. Several matrix bounds for the DALE (1) are developed by a simple approach. In addition, the presented results do not involve any free variable. Therefore, in comparison with existing literature on the subject, the proposed results are less restrictive and more easily calculated.
2. Main Results
Consider the discrete algebraic Lyapunov Equation (DALE) which are represented by
(1)
where, A is a stable matrix, Q denotes a given positive semi-definite matrix, and P is the unique positive semi-definite solution.
Before developing the main results, we review the following useful result.
Lemma 1 [10]: Let a real symmetric matrix U be defined as
(2)
where is a given positive semi-definite matrix. Then if and only if the real matrix is negative semi-definite then U is negative semi-definite.
Then, by utilizing lemma 1 and some linear algebraic techniques, new lower matrix bounds of the solution of the DALE (1) are derived as follows.
Theorem 1. The solution P of the DALE (1) has the following bounds.
(3)
and
(4)
where the positive semi-definite matrices and are defined, respectively, by
(5)
and
(6)
Proof. Let a positive semi-definite matrix R is defined as
. (7)
Then we have
(8)
which infers
. (9)
It is seen that by using the positive semi-definite matrix R, the DALE (1) can be transformed into a continuous-type Lyapunov Equation (9). Then, by (9), we rewrite the DALE (1) as
(10)
Since A is stable, we have and.Then
(11)
and
(12)
Then, according to Lemma 1, it is seen that the right-hand side of Equation (10) is negative semi-definite. Therefore, Equation (10) is a continuous Lyapunov equation and its solution is positive semi-definite. That is,
(13)
Define
(14)
The DALE (1) now can be rewritten as
(15)
Then, we have
(16)
Due to the facts that
(17)
and
(18)
the right-hand side of (16) then is negative semi-definite. Therefore, the solution of the Lyapunov Equation (16) is positive semi-definite. We have
(19)
Substituting and into (13) and (19), respectively, and from the definitions R and gives
(20)
and
(21)
Thus, the proof is completed.
Remark 1. According to the proof of Theorem 1, it is seen that if and are substituted into (19) and (13), respectively, we can also obtain the following results without proof.
Corollary 1. The solution P of the DALE (1) satisfies
(22)
and
(23)
where matrices and are defined by (5) and (6), respectively.
Remark 2.It is found that if then. This leads to and. Besides, it is seen that for. For this case, we have and. The tightness between bounds and, respectively, cannot be compared. Maybe they can give a supplement to each other. However, from Theorem 1, the following algorithms can be developed for obtaining tighter lower solution bound for the DALE (1).
Algorithm 1.
Step 1. Set.
Step 2. Compute
(24)
Then, comparing to, are tighter solution bounds for the DALE (1).
Proof. Let. From Step 1 and (24), we have
(25)
Now, we assume
.
Then the definition of yields
(26)
By the inductive method, one can conclude that.
Algorithm 2.
Step 1. Set.
Step 2. Compute
(27)
Then, solution bounds of the DALE (1) are tighter than.
Proof. Let. From Step 1 and (27), we have
. (28)
Now, we assume
.
Then the definition of yields
. (29)
By the inductive method, one can conclude that.
Remark 3. Surveying the literature, existing lower matrix bounds of the solution of the DALE (1) are summarized as follows.
[5] (30)
[3] (31)
, [7] (32)
, with [7] (33)
[8] (34)
[6] (35)
where
with (36)
(37)
where (38)
, with (39)
(40)
(41)
where (42)
the degree of the minimal polynomial of A. (43)
From the above conditions, it is seen that most of them contain points of weakness. The matrix Q in [5]-[7] must be positive definite. In many control problems, this matrix may be positive semi-definite. It is also seen that bounds in (30), (33), and (35) have inverse matrices. The computation burden hence may become very heavy. The matrix A must be normal in (33) and must be diagonalizable in (35). For bound (32), an extra Lyapunov equation must be solved. Furthermore, from (39), it is obvious that the computation of constants is very difficult. From the obtained results of this work, it is not necessary to assume that the matrix A is diagonalizable or normal. We also do not assume that the matrix Q is positive definite. Furthermore, the present bounds do not involve any inverse matrix and hence are easy to be evaluated. It is found the tightness of the obtained results and those appeared in the literature cannot be compared by any mathematical method. However, at least they can give a supplement to each other.
3. A Numerical Example
Example 1. Consider the DALE (1). Matrices A and Q are chosen as
and
where matrix A is diagonalizable and normal and Q is positive definite. In this case, we choose
.
Then, from the obtained results and (30)-(35), solution bounds of the DALE (1) for this case are shown below.
For this case, it is seen that, , and. However, the sharpness between, , and cannot be compared. It shows that the obtained results and those appeared in the literature can give a supplement to each other. By using Algorithm 1, we have
for.
Obviously our result now is tighter than the parallel ones in this case. This means that the presented algorithms indeed can work.
4. Conclusion
In this paper, the lower matrix bounds of the solution for the DALE have been discussed. By transform the DALE into a continuous-type Lyapunov equation, we have established several concise lower solution bounds of the DALE. All proposed bounds are new and less restrictive than the majority of those appeared in the literature. According to some of these results, iterative algorithms have also been developed for obtaining sharper lower matrix bounds. Finally, we give a numerical example to demonstrate the applicability of the presented schemes.
Acknowledgements
The author would like to thank the National Science Council for financial support of this research under the grant MOST 104-2221-E-230-009.
Cite this paper
Chien-Hua Lee, (2016) On the Measurement of Lower Solution Bounds of the Discrete Algebraic Lyapunov Equation. Journal of Applied Mathematics and Physics,04,655-661. doi: 10.4236/jamp.2016.44075
References
1. 1. Gajic, Z. and Qureshi, M. (1995) Lyapunov Matrix Equation in System Stability and Control. Academic Press, New York.
2. 2. Choi, H.H. and Kuc, T.Y. (2002) Lower Matrix Bounds for the Continuous Algebraic Riccati and Lyapunov Matrix Equations. Automatica, 38, 1147-1152. http://dx.doi.org/10.1016/S0005-1098(01)00304-1
3. 3. Lee, C.H. (1996) Upper and Lower Bounds of the Solution for the Discrete Lyapunov Equation. IEEE Transactions on Automatic Control, 41, 1338-1341. http://dx.doi.org/10.1109/9.536505
4. 4. Lee, C.H. and Chen, C.Y. (2009) Matrix Solution Bounds of the Continuous Lyapunov Equation. Journal of the Franklin Institute, 346, 741-751. http://dx.doi.org/10.1016/j.jfranklin.2009.07.001
5. 5. Lee, C.H. (2004) Solution Bounds of the Continuous and Discrete Lyapunov Matrix Equations. Journal of Optimization Theory and Applications, 120, 559-578. http://dx.doi.org/10.1023/B:JOTA.0000025710.59589.80
6. 6. Lee, C.H. and Kung, F.C. (1997) Upper and Lower Matrix Bounds of the Solutions for the Continuous and Discrete Lyapunov Equations. Journal of the Franklin Institute, 334B, 539-546. http://dx.doi.org/10.1016/S0016-0032(96)00106-8
7. 7. Mori, T., Fukuta, N. and Kuwahara, M. (1985) Eigenvalue Bounds for the Discrete Lyapunov Matrix Equation. IEEE Transactions on Automatic Control, 30, 925-926. http://dx.doi.org/10.1109/TAC.1985.1104092
8. 8. Troch, I. (1987) Improved Bounds for the Eigenvalues of Solutions of Lyapunov Equations. IEEE Transactions on Automatic Control, 32, 744-747. http://dx.doi.org/10.1109/TAC.1987.1104699
9. 9. Truhar, N. and Veselic, K. (2007) Bounds on the Trace of a Solution to the Lyapunov Equation with a General Stable Matrix. Systems & Control Letters, 56, 493-503. http://dx.doi.org/10.1016/j.sysconle.2007.02.003
10. 10. Ostrowski, A. and Schneider, H. (1961) Some Theorems on the Inertia of General Matrices. Journal of Mathematical Analysis and Applications, 4, 72-84. http://dx.doi.org/10.1016/0022-247X(62)90030-6 | 2,972 | 11,701 | {"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} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.924294 |
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Since 1985, pollution levels in Lake Thomas have dropped considerably primarily because of a state program to clean the lake water by means of a water refinery. Ironically, during this same period, the once-abundant population of sunfish in the lake has dwindled.
Which of the following, if true, would best explain why the sunfish population of Lake Thomas has dwindled at the same time that the lake water has become cleaner?
a. The life spans of sunfish are not diminished by high pollution levels, but the number of offspring they create during their lifetime is diminished.
b. Several artificial chemicals are introduced into the lake as a result of the refinement process, but these chemicals are known to have a benign effect on fish.
c. The water refinement process creates an environment extremely favorable to pike, a predator fish.
d. The heaviest concentrations of sunfish population in the lake are at its northern and northeastern shores, many miles away from the water refinery.
e. Ever since 1972, a strictly enforced state regulation has prevented anglers from over-fishing Lake Thomas.
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Re: Deja Vu CR : Lake Thomas [#permalink]
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17 Jul 2006, 13:18
C. The water refinement process creates an environment extremely favorable to pike, a predator fish.
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17 Jul 2006, 13:47
[quote="Praetorian"]Since 1985, pollution levels in Lake Thomas have dropped considerably primarily because of a state program to clean the lake water by means of a water refinery. Ironically, during this same period, the once-abundant population of sunfish in the lake has dwindled.
Which of the following, if true, would best explain why the sunfish population of Lake Thomas has dwindled at the same time that the lake water has become cleaner?
a. The life spans of sunfish are not diminished by high pollution levels, but the number of offspring they create during their lifetime is diminished. does not address passage
b. Several artificial chemicals are introduced into the lake as a result of the refinement process, but these chemicals are known to have a benign effect on fish. then out we should explain why less fish cleaner water
c. The water refinement process creates an environment extremely favorable to pike, a predator fish. is the right one
d. The heaviest concentrations of sunfish population in the lake are at its northern and northeastern shores, many miles away from the water refinery. but this does not justify the fact that the fish are declinging there should be another reason that could explain this paradox
e. Ever since 1972, a strictly enforced state regulation has prevented anglers from over-fishing Lake Thomas.then the nr of fish should increase
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17 Jul 2006, 14:41
c. The water refinement process creates an environment extremely favorable to pike, a predator fish.
A. This is cause decling even with pollution.
B. No affect on the population "benign"
C. Predator thrives sunfish population declines... correct answer.
D. Out of scope
E. Out of scope.
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17 Jul 2006, 15:38
no doubt here about C
we need to find alternative reason for lower population of this sunfish, that is not directly related to cleaning the pollution of the lake, which is C
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17 Jul 2006, 15:41
I think its C.
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17 Jul 2006, 17:45
Good Job.
Again, akamaibrah was gracious enough to explain:
"My choice is definitely C.
Remember, we are trying to explain the decrease in Sunfish...
(A) states that number of offspring is DIMINISHED by pollution. Hence, if we REDUCE pollution, offspring will no longer be diminished. This certainly does not explain the decrease in Sunfish -- more likely it inplies that an eventually increase will occur.
(B) states that the chemical do not harm fish. Hence, this does not explain the decrease in Sunfish.
(C) states that a predatory fish thrives in the environment created by the refinery. Nitpicking aside, since the subject of the argument is the "sunfish", it is reasonable to assume that the term "predatory" refers to the relationship of the pike to the sunfish. This would certanly explain a decrease in sunfish and IMO is the best choice.
(D). states that the heaviest concentration of pike are far away from the refinery. While this is consistent with (C), it does not in itself explain an overall decrease in the sunfish population.
(E) does nothing to explain the decrease in Sunfish since 1985 and is irrelavent. In fact, it removes a possible explaination for the decrease."
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17 Jul 2006, 23:11
Clear winner C.
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17 Jul 2006, 23:21
I will also go with C.
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19 Jul 2006, 05:06
Could you give us the OA Praetorian.
I have a problem with C - it mentions 'creates an environment extremely favorable to pike' - this does not imply that pikes exist in the lake as well. All it says is that the environment is pike friendly.
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19 Jul 2006, 07:44
C it is........
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21 Jul 2006, 02:00
C it is
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Studies of the human “sleep-wake cycleâ€
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C
"to" is needed to maintain parallelism
"staffing" is needed to keep it as a noun (from duty assign. to staffing)
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[quote="ritledge"]Studies of the human “sleep-wake cycleâ€
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From...to... is the idiomatic construction tested here.
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Section: LAPACK (3)
Updated: Tue Nov 14 2017
Page Index
cupmtr.f
## SYNOPSIS
### Functions/Subroutines
subroutine cupmtr (SIDE, UPLO, TRANS, M, N, AP, TAU, C, LDC, WORK, INFO)
CUPMTR
## Function/Subroutine Documentation
### subroutine cupmtr (character SIDE, character UPLO, character TRANS, integer M, integer N, complex, dimension( * ) AP, complex, dimension( * ) TAU, complex, dimension( ldc, * ) C, integer LDC, complex, dimension( * ) WORK, integer INFO)
CUPMTR
Purpose:
``` CUPMTR overwrites the general complex M-by-N matrix C with
SIDE = 'L' SIDE = 'R'
TRANS = 'N': Q * C C * Q
TRANS = 'C': Q**H * C C * Q**H
where Q is a complex unitary matrix of order nq, with nq = m if
SIDE = 'L' and nq = n if SIDE = 'R'. Q is defined as the product of
nq-1 elementary reflectors, as returned by CHPTRD using packed
storage:
if UPLO = 'U', Q = H(nq-1) . . . H(2) H(1);
if UPLO = 'L', Q = H(1) H(2) . . . H(nq-1).
```
Parameters:
SIDE
``` SIDE is CHARACTER*1
= 'L': apply Q or Q**H from the Left;
= 'R': apply Q or Q**H from the Right.
```
UPLO
``` UPLO is CHARACTER*1
= 'U': Upper triangular packed storage used in previous
call to CHPTRD;
= 'L': Lower triangular packed storage used in previous
call to CHPTRD.
```
TRANS
``` TRANS is CHARACTER*1
= 'N': No transpose, apply Q;
= 'C': Conjugate transpose, apply Q**H.
```
M
``` M is INTEGER
The number of rows of the matrix C. M >= 0.
```
N
``` N is INTEGER
The number of columns of the matrix C. N >= 0.
```
AP
``` AP is COMPLEX array, dimension
(M*(M+1)/2) if SIDE = 'L'
(N*(N+1)/2) if SIDE = 'R'
The vectors which define the elementary reflectors, as
returned by CHPTRD. AP is modified by the routine but
restored on exit.
```
TAU
``` TAU is COMPLEX array, dimension (M-1) if SIDE = 'L'
or (N-1) if SIDE = 'R'
TAU(i) must contain the scalar factor of the elementary
reflector H(i), as returned by CHPTRD.
```
C
``` C is COMPLEX array, dimension (LDC,N)
On entry, the M-by-N matrix C.
On exit, C is overwritten by Q*C or Q**H*C or C*Q**H or C*Q.
```
LDC
``` LDC is INTEGER
The leading dimension of the array C. LDC >= max(1,M).
```
WORK
``` WORK is COMPLEX array, dimension
(N) if SIDE = 'L'
(M) if SIDE = 'R'
```
INFO
``` INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
December 2016
Definition at line 152 of file cupmtr.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code. | 856 | 2,688 | {"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-2022-27 | latest | en | 0.785054 |
https://www.nagwa.com/en/videos/504156752457/ | 1,582,090,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00078.warc.gz | 826,630,056 | 6,260 | # Video: AQA GCSE Mathematics Foundation Tier Pack 1 β’ Paper 2 β’ Question 7
04:57
### Video Transcript
Adil is looking at some function machines. π₯ times two times three equals π¦. Adil says, βIf you know π¦, you can work out π₯ by either dividing by two first and then dividing by three or dividing by three first and then dividing by two.β Is Adil correct? You must give a reason for your answer.
According to Adil if we start with π¦, we could divide by two and then divide by three to get π₯. She also claims that this order doesnβt matter, that we could first divide by three and then divide by two. Why donβt we test it by plugging in some value for π₯? This function machine tells us that π₯ times two times three equals π¦. If π₯ equals one, one times two equals two, two times three equals π¦, six equals π¦. When π₯ equals one, π¦ equals six.
Now, we want to do the reverse π¦ divided by two divided by three should equal π₯. And according to Adil, π¦ divided by three divided by two should also equal π₯. In the first part, six divided by two divided by three: six divided by two equals three, three divided by three equals π₯, and π₯ equals one, which is correct. And what if we change the order of division? Is six divided by three divided by two also equal to the correct π₯-value? Six divided by three equals two, two divided by two equals π₯, and π₯ equals one.
Weβve shown that yes, Adil is correct. This is because divided by two and divided by three are the inverse operations of multiplied by two and multiplied by three, respectively and because the order of these operations does not matter. If the only operations weβre using are multiplication and division, then the order wonβt matter.
Now, Iβm gonna clear some space and consider part b), a second function machine.
Here is the second function machine. π minus two multiplied by three equals π. Adil says, βIf you know the value of π, you can work out the value of π by first dividing π by three and then adding two.β He says, βYou cannot add two first and then divide by three.β Is Adil correct? You must give a reason for your answer.
Letβs map out what Adil has said. He says if we start with π, we first divide by three and then we add two to get π. First thing we notice is that dividing by three is the inverse of multiplying by three and that adding two is the inverse of subtracting two. We found the inverse operation. But now we need to consider the question, βdoes order matter here?β
Weβre starting with π minus two and we multiply that by three to give us π. Can we switch these two operations and still come up with the same answer? Is π multiplied by three and subtract two the same value for π? Letβs try it if π equals two. Hereβs what we want to know: is two minus two multiplied by three the same thing as two times three minus two? Two minus two equals zero, zero times three equals zero, two times three equals six, six minus two equals four. Zero is not equal to four.
In this function machine, order matters And if order matters moving in the forward direction, then order will also matter moving in the reverse direction.
Adil is correct. Although divided by three and plus two are the inverse operations to multiply by three and subtract two, respectively, in this case, the order does matter. Given π, you cannot add two and then divide by three. You must divide by three and then add two to get π. | 936 | 3,541 | {"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} | 4.625 | 5 | CC-MAIN-2020-10 | latest | en | 0.955305 |
http://stackoverflow.com/questions/12958170/cut-function-split-by-factor-levels/12964916 | 1,430,048,811,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246654292.99/warc/CC-MAIN-20150417045734-00223-ip-10-235-10-82.ec2.internal.warc.gz | 247,565,891 | 18,953 | # Cut function split by factor levels
I have a problem with the cut function. I have this situation:
`````` codice
1 11GP2-0016
2 11GP2-0016
3 11GP2-0016
4 11OL2-074
5 11OL2-074
``````
and I would like to have a new variable "campione" splitted by variable "codice" like this:
`````` codice campione
1 11GP2-0016 [1,3]
2 11GP2-0016 [1,3]
3 11GP2-0016 [1,3]
4 11OL2-074 (4,5]
5 11OL2-074 (4,5]
``````
How can I use the cut function to split the "codice" creating a variable showing that from 1 to 3 i have the same code, from 4 to 5 same code and so on?
I need to solve another question. For the same issue I would like to obtain:
`````` codice campione
1 11GP2-0016 [11GP2-0016,11GP2-0016,11GP2-0016]
2 11GP2-0016 [11GP2-0016,11GP2-0016,11GP2-0016]
3 11GP2-0016 [11GP2-0016,11GP2-0016,11GP2-0016]
4 11OL2-074 (11OL2-074,11OL2-074]
5 11OL2-074 (11OL2-074,11OL2-074]
``````
Is there any solution to do this?
-
Do you just want to know the run length, or is it important to have the start and end position? (see `?rle`) – Brandon Bertelsen Oct 18 '12 at 15:37
is it important the position because 1 to 3 it means sample1 to sample3. Not the length. @hadley example is good but seems that the rank is following something different than the position. Nicola – Spigonico Oct 22 '12 at 10:12
This will do it. You can add brackets/parens, if you want.
``````dat <- read.table(text='codice
1 11GP2-0016
2 11GP2-0016
3 11GP2-0016
4 11OL2-074
within(dat,
campione <- with(rle(as.character(codice)), {
starts <- which(! duplicated(codice))
ends <- starts + lengths - 1
inverse.rle(list(values=paste(starts, ends, sep=','), lengths=lengths))
})
)
# codice campione
# 1 11GP2-0016 1,3
# 2 11GP2-0016 1,3
# 3 11GP2-0016 1,3
# 4 11OL2-074 4,5
# 5 11OL2-074 4,5
``````
-
Best way for me. Using @Gavin Simpson solution the intervals are not correct. No matter for me about brakets ;) – Spigonico Dec 5 '12 at 17:02
``````d <- read.table(text = "1 11GP2-0016
2 11GP2-0016
3 11GP2-0016
4 11OL2-074
5 11OL2-074", row.names = 1, stringsAsFactors = FALSE)
names(d) <- "codice"
``````
Here is a slightly convoluted example using `rle()`:
``````drle <- with(d, rle(codice))
``````
This gives us the run lengths of `codice`:
``````> drle
Run Length Encoding
lengths: int [1:2] 3 2
values : chr [1:2] "11GP2-0016" "11OL2-074"
``````
and it is the `\$lengths` component that I manipulate to create two indicates, the start (`ind1`) and the end (`ind2`) location:
``````ind1 <- with(drle, rep(seq_along(lengths), times = lengths) +
rep(c(0, head(lengths, -1) - 1), times = lengths))
ind2 <- ind1 + with(drle, rep(lengths- 1, times = lengths))
``````
Then I just paste these together:
``````d <- transform(d, campione = paste0("[", ind1, ",", ind2, "]"))
``````
Giving
``````> head(d)
codice campione
1 11GP2-0016 [1,3]
2 11GP2-0016 [1,3]
3 11GP2-0016 [1,3]
4 11OL2-074 [4,5]
5 11OL2-074 [4,5]
``````
-
Nice, although you're missing the rounded left parens on the second level of that factor... (Just kidding. Your notation's better.) – Josh O'Brien Oct 18 '12 at 15:54
Yeah, I know as I failed to even both using `cut()`! Bad me! ;-) Hope the OP doesn't mind - those are nicer but just labels that `cut()` produces... The OP can convert to a factor if the want... (And thanks by the way!) – Gavin Simpson Oct 18 '12 at 15:57
Well, strictly speaking, the interval `(4,5]` shouldn't include the 4th element at all, which is why yours is objectively better. – Josh O'Brien Oct 18 '12 at 15:59
Yes, I win!! ;-) – Gavin Simpson Oct 18 '12 at 16:05
An alternative approach is to use `rank`:
``````left <- rank(factor(d\$codice), ties.method = "min")
right <- rank(factor(d\$codice), ties.method = "max")
d\$campione <- paste("[", left, ",", right, "]", sep = "")
``````
-
Dear Hadley, I think that I'm doing something wrong because your way is easier for me than otherones but I have a problem because the rank doesen't start from 1 but it starts from 395 (my databse is around 900 cases). Thank you in advance – Spigonico Oct 22 '12 at 10:00
Please provide a reproducible example that illustrates the problem. – hadley Oct 22 '12 at 23:04
what I would like to do is to find duplicated rows in my data frame by some variables. In this case only one "Codice" but further I'd like to use more than one variable. – Spigonico Dec 3 '12 at 17:01 | 1,564 | 4,433 | {"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} | 3.09375 | 3 | CC-MAIN-2015-18 | latest | en | 0.733663 |
https://encyclopedia2.thefreedictionary.com/Unicursal+curve | 1,568,951,445,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00557.warc.gz | 461,200,482 | 10,860 | # Unicursal Curve
Also found in: Dictionary.
## Unicursal Curve
in mathematics, a plane curve that can be parametrized by a pair of rational functions Ф(t) and Ψ(t). The most important theorems on unicursal curves are the assertion that an algebraic curve with the maximal number of double points permitted by its order is unicursal and its converse, that is, the assertion that every unicursal curve is an algebraic curve with the maximal number of double points permitted by its order. Here it is assumed that, as usual, a point of higher multiplicity is counted in terms of its “double point equivalents,” so that, for example, a triple point is equivalent to three double points.
Figure 1
The maximal number of double points of an algebraic curve of order n is δ = (n – 1)(n – 2)/2. If the curve of order n has r double points, then the difference δ – r is called its defect, or genus. Therefore, a unicursal curve can be defined as an algebraic curve of genus zero. It is clear that lines and quadratic curves cannot have double points and so are unicursal. A curve of order 3 is unicursal if it has one double point, and a curve of order 4 is unicursal if it has three double points.
Figure 1 represents a third-order curve known as the folium of Descartes. It has one double point and so is unicursal. In fact, the folium of Descartes admits the parametrization
The parameter t is the tangent of the angle between the x-axis and the radius vector of the point (x,y).
In counting the number of double points, we must not rely on the actual graph of the curve, since we may leave out double points at infinity or complex double points. For example, the fourth-order curve known as the lemniscate of Bernoulli has just one real double point and two circular double points and so is unicursal.
Unicursal curves are important in the theory of integrals of algebraic functions. Every integral of the form
ʃ R(x,y) dx
where R is a rational function of two variables and x and y satisfy an equation F(x,y) = 0, which defines a unicursal curve, can be reduced to an integral of a rational function and expressed in terms of elementary functions.
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Open / Close | 504 | 2,189 | {"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} | 3.859375 | 4 | CC-MAIN-2019-39 | longest | en | 0.940491 |
https://www.jianshu.com/p/2931c8012723 | 1,610,933,112,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00079.warc.gz | 862,728,174 | 36,398 | # EP669
https://leetcode.com/problems/trim-a-binary-search-tree/
image.png
``````class Solution {
public:
TreeNode* trimBST(TreeNode* root, int L, int R) {
if (root == NULL) return NULL;
if (root->val < L) return trimBST(root->right, L, R);
if (root->val > R) return trimBST(root->left, L, R);
root->left = trimBST(root->left, L, R);
root->right = trimBST(root->right, L, R);
return root;
}
};
`````` | 126 | 403 | {"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-2021-04 | latest | en | 0.235887 |
http://forum.allaboutcircuits.com/threads/circuits-problems.15591/ | 1,477,185,691,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00067-ip-10-171-6-4.ec2.internal.warc.gz | 96,696,193 | 15,250 | # Circuits problems
Discussion in 'Homework Help' started by kirit, Oct 31, 2008.
1. ### kirit Thread Starter Active Member
Sep 9, 2008
35
0
Hello, I seem to be in need of some help with some homework problems. Cant seem to pinpoint why Im having so much trouble getting them done correctly. I'll post them one at a time so any help given will be easier to discuss. Thanks in advance!!!!!
So looks like an easy problem, not sure why I keep having to bang my head against a wall to get it right. I get an answer that is very close to that. I get 2^.5 * cos(40t - 135), so instead of +135 i get -135.
Here is my work:
Hope the pics are clear. If not I can type the work up. Thanks for the help again.
Feb 4, 2008
4,846
63
3. ### kirit Thread Starter Active Member
Sep 9, 2008
35
0
Thats part of me banging my head against the wall. I did and rechecked my signs. I even did it the first time using nodal analysis but couldnt get the answer.
4. ### Ratch New Member
Mar 20, 2007
1,068
3
kirit,
It appears you are right and the book is wrong. See the attached file. Interestingly, V1 appears to be zero.
Ratch
File size:
788.8 KB
Views:
18
5. ### The Electrician AAC Fanatic!
Oct 9, 2007
2,255
311
This is an interesting circuit. For purposes of discussion, eliminate Vs and number the nodes across the top of the circuit, left to right, V1, V2 and V3. The frequency for all calculations will be w=40.
Typically in a circuit with an unsymmetrically placed dependent source, the circuit is non-bilateral; for example, the equivalent circuit of a 3 terminal amplifying device such as a vacuum tube or a transistor.
But this one appears to be still bilateral.
The voltage transfer ratio (VTR; also known as voltage gain or loss) from from V1 to V3 (apply a voltage source to V1 and calculate V3/V1) is just unity. The VTR from V1 to V2 is zero.
Interestingly, the VTR from V3 to V1 is also unity, and from V3 to V2 is also zero.
But now, for those who wish to exercise their circuit analysis skills, try these:
Calculate the VTR from V2 to both V1 and V3.
Calculate the current transfer ratio (CTR; also known as current gain or loss), from V1 to V3. This is done by applying a current source to V1 and shorting V3 to ground. The CTR is the current in the short at V3 divided by the current applied to the input, V1; in other words, I3/I1, with a short at node V3.
Calculate the CTR from V3 to V1. Is the network bilateral for CTR?
Calculate the impedance from V1 to V3 as though measured by a floating AC impedance meter (at a frequency of w=40).
Calculate the impedance from V1 to V3 with the 2 ohm resistor removed.
6. ### Ratch New Member
Mar 20, 2007
1,068
3
The coil forms a series resonant circuit with the capacitors on each side of it. That is why the voltage on the high side of the resistor is the same as ground or zero, and the output voltage is the same as the input voltage. If Vs is removed, then no "i" is present, so the dependent current source outputs zero amps, and Vo is zero also.
Ratch | 814 | 3,028 | {"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} | 3.03125 | 3 | CC-MAIN-2016-44 | longest | en | 0.947257 |
https://math.stackexchange.com/questions/1261300/solving-the-recurrence-relation-obtained-from-the-power-series-method | 1,653,529,357,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00599.warc.gz | 432,125,124 | 65,997 | # Solving the recurrence relation obtained from the power series method
Assuming the solution to my differential equation is of the form $y=\sum_{n=0}^\infty a_nx^n$, I was able to get to the recurrence relation.
The recurrence relation is $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$
From this I get $$a_2 = \frac {-3}{2}a_0$$ $$a_3 = 0 \implies a_{\text{odd}} = 0 \tag{except maybe a_1} \\ a_4 = \frac 14a_2 = \frac {-3}8a_0 \\ a_6 = \frac 3{10}a_4 = \frac {-9}{80}a_0 \\ \vdots$$
Thus my solution should be of the form $$y= a_1(x) + a_0 (1-\frac 32x^2 -\frac 38x^4 -\frac 9{80}x^6 + \cdots)$$
How do I simplify this -- that is write the infinite series for the even terms? I've been trying for a while, but I'm not sure what the standard technique is for this.
• What is the differential equation that $y(x)$ solves?
– DVD
May 2, 2015 at 0:26
$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$ $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$ $$a_{2n+1} = 0$$
\begin{align} % a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)} \\ & \times \dots \\ & \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0} \end{align}
And canceling out terms:
\begin{align} % a_{n} &= a_0~3^{n/2}~\frac {1}{(n-1)(n-0)} \times \frac {1}{(n-2)} \times \frac {1}{(n-4)} \\ & \times \dots \\ & \times \frac {1}{6} \times \frac{1}{4}\times \frac{(-1)}{2} \end{align}
$$a_{n} = a_0~3^{n/2}~\frac{-1}{n-1}~\frac{1}{(n-0)(n-2)(n-4) \dots (4)(2)}$$
And rewriting it in various ways:
$$a_{n} = a_0~3^{n/2}~\frac{-1}{2(n/2)-1}~\frac{2^{-n/2}}{(n/2-0)(n/2-1)(n/2-2) \dots (2)(1)}$$
$$f_n(x) = a_{2n}x^{2n} = a_0~(3/2)^{n}~\frac{-1}{2n - 1}~\frac{x^{2n}}{(n-0)(n-1)(n-2) \dots (2)(1)}$$
$$f_n(x) = \frac{-a_0}{2n - 1}~\frac{\left(3x^2/2\right)^n}{n!}$$
$$-a_0~g_n\left( \frac{3x^2}{2} \right) = f_n(x) \iff \color{darkred}{g_n(x) = \frac{1}{2n - 1}~\frac{x^n}{n!}}$$
This $g$ doesn't look like any common taylor series, so checking wolfram online:
$$\sum_{n=0}^{\infty} \frac{1}{2n - 1}~\frac{x^n}{n!} = \sqrt{\pi x}~~ {\rm erfi}(\sqrt{x}) - e^{x}$$
where erfi is the imaginary error function. So while your recursion has the above closed form, your series does not have an elementary closed form.
• Hi Daniel. When you do \begin{align} a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)} \\ & \times \dots \\ & \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0} \end{align} how do you know what the last term in the multiplication is? Sep 4, 2021 at 0:12 | 1,179 | 2,615 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-21 | latest | en | 0.669544 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-8-rational-functions-8-3-graph-general-rational-functions-8-3-exercises-skill-practice-page-568/6 | 1,726,603,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00482.warc.gz | 729,910,075 | 15,014 | Algebra 2 (1st Edition)
In order to graph this function, it can be helpful to factor the numerator and the denominator, for the factors of the numerator will be the x-intercepts, while the factors of the denominator will be vertical asymptotes. The denominator must have factors of $\pm 2$, and the numerator must have no factors. Thus, A is correct. | 79 | 351 | {"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} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.920583 |
https://www.sawaal.com/time-and-distance-questions-and-answers/if-hema-walks-at-12-km-hr-instead-of-8-km-hr-she-would-have-walked-20-km-more-the-actual-distance-tr_12191 | 1,660,275,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00328.warc.gz | 845,997,093 | 14,337 | 6
Q:
If Hema walks at 12 km/hr instead of 8 km/hr, she would have walked 20 km more. The actual distance travelled by Hema is ?
A) 40 kms B) 30 kms C) 46 kms D) 32 kms
Explanation:
Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.
Q:
A train covers 60 km at a uniform speed. If the speed had been 8 km/h more, it would have taken 10 hours less for the same journey. What is the speed of the train (in km/h)?
A) 3 B) 2.5 C) 5 D) 4
Explanation:
1 233
Q:
A car covers 25 km at a uniform speed. If the speed had been 8 knv/h more, it would have taken 10 hours less for the same journey. What is the speed of the car (in km/h)?
A) 4 B) 3 C) 2.5 D) 2
Explanation:
2 230
Q:
A car covers 25 km at a uniform speed. If the speed had been 8 knv/h more, it would have taken 10 hours less for the same journey. What is the speed of the car (in km/h)?
A) 4 B) 3 C) 2.5 D) 2
Explanation:
2 223
Q:
The ratio between the speeds of two trains is 5 : 7. If the first train covers 300 km in 3 hours, then the speed (in km/h) of the second train is:
A) 100 B) 150 C) 140 D) 120
Explanation:
1 246
Q:
The ratio between the speeds of two trains is 2 : 5. If the first train runs 350 km in 5 h, then the difference between the speeds (in km/h) of both the trains is:
A) 350 B) 105 C) 165 D) 180
Explanation:
2 486
Q:
Given that the lengths of the paths of a ball thrown with different speeds by two boy sare the same, if they take 0.6 seconds and 1 second respectively to cover the said length, what is the average speed of travel for the first throw, if the same for the second is 96 km/h?
A) 160 km/h B) 200 km/h C) 150 km/h D) 100 km/h
Explanation:
1 489
Q:
A man divided his journey into three parts of distances of 18 km, 20 km and 27 km. He travelled the distances at the speeds of 6 km/h, 5 km/h and 9 kn/h,respectively. What was his average speed during the entire journey?
A) 6.5 km/h B) 4.5 km/h C) 7.5 km/h D) 5.5 km/h
Explanation:
2 745
Q:
The distance between two railway stations is 1176 km. To cover this distance, an express train takes 5 hours less than a passenger train while the average speed of the passenger train is 70 km/h less than that of the express train. The time taken by the passenger train to complete the travel is:
A) 12 hours B) 23 hours C) 17 hours D) 18 hours | 765 | 2,372 | {"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-2022-33 | latest | en | 0.912035 |
https://www.acmicpc.net/problem/21932 | 1,624,396,276,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00026.warc.gz | 533,015,568 | 8,885 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
2 초 512 MB 5 3 3 60.000%
## 문제
An undirected graph is given, each of its nodes associated with a positive integer value. Given a threshold, nodes of the graph are divided into two groups: one consisting of the nodes with values less than or equal to the threshold, and the other consisting of the rest of the nodes. Now, consider a subgraph of the original graph obtained by removing all the edges connecting two nodes belonging to different groups. When both of the node groups are non-empty, the resultant subgraph is disconnected, whether or not the given graph is connected.
Then a number of new edges are added to the subgraph to make it connected, but these edges must connect nodes in different groups, and each node can be incident with at most one new edge. The threshold is called feasible if neither of the groups is empty and the subgraph can be made connected by adding some new edges.
## 입력
The input consists of a single test case of the following format.
n m
l1 . . . ln
x1 y1
.
.
.
xm ym
The first line contains two integers n (2 ≤ n ≤ 105) and m (0 ≤ m ≤ min(105, n(n−1)/2)), the numbers of the nodes and the edges, respectively, of the graph. Nodes are numbered 1 through n. The second line contains n integers li (1 ≤ li ≤ 109), meaning that the value associated with the node i is li. Each of the following m lines contains two integers xj and yj (1 ≤ xj < yj ≤ n), meaning that an edge connects the nodes xj and yj. At most one edge exists between any two nodes.
## 출력
Output the minimum feasible threshold value. Output -1 if no threshold values are feasible.
## 예제 입력 1
4 2
10 20 30 40
1 2
3 4
## 예제 출력 1
20
## 예제 입력 2
2 1
3 5
1 2
## 예제 출력 2
3
## 예제 입력 3
3 0
9 2 8
## 예제 출력 3
-1
## 예제 입력 4
4 6
5 5 5 5
1 2
1 3
1 4
2 3
2 4
3 4
## 예제 출력 4
-1
## 예제 입력 5
7 6
3 1 4 1 5 9 2
2 3
3 5
5 6
1 4
1 7
4 7
## 예제 출력 5
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Quantitative Aptitude– Data Sufficiency- For Bank & Insurance
# Quantitative Aptitude– Data Sufficiency- For Bank & Insurance
Dear Readers, Expected Data Sufficiency questions for Bank/Insurance was given here.Aspirants those who are preparing for the examination can use this.
Quantitative Aptitude– Data Sufficiency – Set 10
Directions(1-3) Each of the questions below consist of a question & two statements numbered I & II given below it. You have to decide whether the data provided in the Statement are sufficient to answer the question. Read both the statements & give answer.
A)If the data in statement I alone are sufficient to answer the question , while the data in statement II alone are not sufficient to answer the question
B) If the data in statement II alone are sufficient to answer the question , while the data in statement I alone are not sufficient to answer the question
C)If the data either in statement I alone or in statement II alone are sufficient to answer the question.
D)If you cannot get the answer from the statement I & II together , but need even more data
E)If the data in both statement I & II together are necessary to answer the question
1)Is y an even number
I)Y is divisible by 2
II)Y + an odd number is an odd number
2)The ages of A & B are in the ratio of 6:5 . What is the age of B ?
I)The ages of A & C are in the ratio of 3:2
II) After 6 years, the ratio of C’s & B’s ages will be 6:7
3)In how many days 15 woman can complete a piece of work
I)If 16 children can complete the same piece of work in 48 days
II)If 9 woman can complete the same piece of work in 27 days
4)What is salary of “W” in a group of U,V,W,X,Y & Z whose average salary is Rs 60600 ?
I)Total of the salary of U & Y is Rs 64500
II)Total of the salary of V & Z is Rs 52600
5)What is the three digit number ?
I)one-third of that number is less by 27 of the half of that number
II)One-fifth of that number is 20 % of that number
Directions (6-10)You have to determine which statements(s) is /are sufficient /necessary to answer the questions.
6)What is the cost of painting the two adjacent walls of a hall which has no windows or doors ?
I)The area of the base of hall is 24 sq mt.
II)The breadth, length and the height of the hall are in the ratio of 4 : 6:5
III)Area of one wall is 30 Square metres
A)only I
B)Only II
C)Only III
D)Either I or III
7)A train crosses another train in 10 sec . Find out the lengths of the trains
I)Ratio between the lengths of second & first train is 4:5
II)Ratio between the speed of first and second train is 1:2
III)The speed of first train is 36 km/hr
A)Only I & II together
B)Only II & III together
C)Only I & III together
E)None of these
8)Find the average of present age of A,B & C
A)The average age of A,B & C 15 years ago was 40/3 years
B)The present age of A,B & C are in the ratio of 8 :9:12
C)The average age of A,B & C , 15 years hence is 190/3 years
A)A & B are sufficient
B)B & C are sufficient
C) Either A or B is sufficient
D)Either A or C is sufficient
E)All together are necessary
9)What is the area of circle ?
I)The breadth of a rectangle is three-fourths the radius of the circle.
II)The radius of the circle is equal to the side of the square whose area is 196 sq cm
III)The area of the rectangle is 252 sq cm
A)Only III is sufficient
B)Only I & II is sufficient
C)I , II & III all are required
D)Either I or III are sufficient
E)other than the given options
10)What is the cost of carpeting a rectangular hall ?
I)The perimeter of the hall is 60 m
II)The cost of carpeting the hall is Rs 120 per square metre
III)The area of the rectangular hall is 3/4 of the area of the square field
A)only II & III are sufficient
B)Only II is sufficient
C)Only III is sufficient
D)Question can’t be solved
E)Either II or III
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Ans
1.C)
2.E)
3.B)
4.D)
5.A)
6.E)cost price /unit is not given
7.D)can’t be answered because direction of the train are not given
8.D)
9.B)
10D)Since the area of the square field is not given , we can’t find the area of the rectangular hall.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
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https://stdkmd.net/nrr/cont/1/15551.htm | 1,660,154,861,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00087.warc.gz | 474,468,874 | 87,918 | 14×10152-419
c145
name 名前 JMB January 19, 2007 21:58:29 UTC 2007 年 1 月 20 日 (土) 6 時 58 分 29 秒 (日本時間) `1058693624412312835832221964029533520010813980653975666076772409870638354588713887513362292127898359918948926731302649916267202845985040887462393<145>` `847196941648551132106333324010269<33>` `1249642878021033284825168208794036075612473963907528490275936932677995158635356633750195668907903252864903462797<112>` ```Using B1=250000, B2=116469998, polynomial Dickson(3), sigma=3067084609 Step 1 took 3891ms Step 2 took 3047ms ********** Factor found in step 2: 847196941648551132106333324010269 Found probable prime factor of 33 digits: 847196941648551132106333324010269 Composite cofactor 1249642878021033284825168208794036075612473963907528490275936932677995158635356633750195668907903252864903462797 has 112 digits``` `GMP-ECM 6.0.1 [powered by GMP 4.2.1] [ECM]`
c112
name 名前 Tyler Cadigan January 27, 2007 07:27:26 UTC 2007 年 1 月 27 日 (土) 16 時 27 分 26 秒 (日本時間) `1249642878021033284825168208794036075612473963907528490275936932677995158635356633750195668907903252864903462797<112>` ```108123011159930669845571642420238883<36> 11557603368746529230971013510751174250574322894480256454543811212283540578959<77>``` ```Number: trial N=1249642878021033284825168208794036075612473963907528490275936932677995158635356633750195668907903252864903462797 ( 112 digits) SNFS difficulty: 153 digits. Divisors found: r1=108123011159930669845571642420238883 (pp36) r2=11557603368746529230971013510751174250574322894480256454543811212283540578959 (pp77) Version: GGNFS-0.77.1-20060722-pentium4 Total time: 47.63 hours. Scaled time: 26.15 units (timescale=0.549). Factorization parameters were as follows: n: 1249642878021033284825168208794036075612473963907528490275936932677995158635356633750195668907903252864903462797 m: 1000000000000000000000000000000 c5: 1400 c0: -41 skew: 1 type: snfsFactor base limits: 2400000/2400000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 48/48 Sieved algebraic special-q in [1200000, 2100001) Primes: RFBsize:176302, AFBsize:176059, largePrimes:5528339 encountered Relations: rels:5432457, finalFF:410172 Max relations in full relation-set: 0 Initial matrix: 352428 x 410172 with sparse part having weight 28109504. Pruned matrix : 318706 x 320532 with weight 20499291. Total sieving time: 39.98 hours. Total relation processing time: 0.37 hours. Matrix solve time: 6.94 hours. Time per square root: 0.34 hours. Prototype def-par.txt line would be: snfs,153,5,0,0,0,0,0,0,0,0,2400000,2400000,27,27,48,48,2.3,2.3,100000 total time: 47.63 hours. --------- CPU info (if available) ----------``` `GGNFS-0.77.1-20060722-pentium4` `Pentium 4 3.20 GHz, 1 Gig RAM, Windows XP and Cygwin`
14×10153-419
c119
name 名前 Tyler Cadigan October 8, 2006 17:47:47 UTC 2006 年 10 月 9 日 (月) 2 時 47 分 47 秒 (日本時間) `57445072003404701363203335515060261736047876565777623202277926783180283144761034643442564878604284961558202284453114149<119>` ```5869281371990934517896039538832558131<37> 24106869046755270050566332204861288659<38> 406000940317690411891991058441288457615883581<45>``` ```Number: test N=57445072003404701363203335515060261736047876565777623202277926783180283144761034643442564878604284961558202284453114149 ( 119 digits) SNFS difficulty: 154 digits. Divisors found: r1=5869281371990934517896039538832558131 (pp37) r2=24106869046755270050566332204861288659 (pp38) r3=406000940317690411891991058441288457615883581 (pp45) Version: GGNFS-0.77.1-20060722-pentium4 Total time: 34.66 hours. Scaled time: 29.67 units (timescale=0.856). Factorization parameters were as follows: n: 57445072003404701363203335515060261736047876565777623202277926783180283144761034643442564878604284961558202284453114149 m: 2000000000000000000000000000000 c5: 875 c0: -82 skew: 1 type: snfs Factor base limits: 2400000/2400000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 48/48 Sieved algebraic special-q in [1200000, 2200001) Primes: RFBsize:176302, AFBsize:176584, largePrimes:5473280 encountered Relations: rels:5306437, finalFF:400078 Max relations in full relation-set: 0 Initial matrix: 352952 x 400078 with sparse part having weight 33257430. Pruned matrix : 331945 x 333773 with weight 24776358. Total sieving time: 29.50 hours. Total relation processing time: 0.46 hours. Matrix solve time: 4.46 hours. Time per square root: 0.24 hours. Prototype def-par.txt line would be: snfs,154,5,0,0,0,0,0,0,0,0,2400000,2400000,27,27,48,48,2.3,2.3,100000 total time: 34.66 hours. --------- CPU info (if available) ----------``` `GGNFS-0.77.1-20060722-pentium4` `Pentium 4 3.20 GHz, Windows XP and Cygwin`
14×10157-419
c141
name 名前 Robert Backstrom April 15, 2007 07:51:35 UTC 2007 年 4 月 15 日 (日) 16 時 51 分 35 秒 (日本時間) `118588965978184191291589169425730928531988106038187106414487352675590392264054198803378536910232046817358432344625794912915634728702561481469<141>` ```1367055426296824724730956100096423270901258669<46> 86747738019244962078804322447103067350502553941506493682108511037049165854640129475050452781201<95>``` ```Number: n N=118588965978184191291589169425730928531988106038187106414487352675590392264054198803378536910232046817358432344625794912915634728702561481469 ( 141 digits) SNFS difficulty: 158 digits. Divisors found: r1=1367055426296824724730956100096423270901258669 (pp46) r2=86747738019244962078804322447103067350502553941506493682108511037049165854640129475050452781201 (pp95) Version: GGNFS-0.77.1-20060513-athlon-xp Total time: 30.51 hours. Scaled time: 41.55 units (timescale=1.362). Factorization parameters were as follows: name: KA_1_5_156_1 n: 118588965978184191291589169425730928531988106038187106414487352675590392264054198803378536910232046817358432344625794912915634728702561481469 skew: 0.49 deg: 5 c5: 1400 c0: -41 m: 10000000000000000000000000000000 type: snfs rlim: 3000000 alim: 3000000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3000000/3000000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved algebraic special-q in [100000, 1200001) Primes: RFBsize:216816, AFBsize:216347, largePrimes:6833902 encountered Relations: rels:6305635, finalFF:490090 Max relations in full relation-set: 28 Initial matrix: 433230 x 490090 with sparse part having weight 31304275. Pruned matrix : 384277 x 386507 with weight 20392536. Total sieving time: 27.27 hours. Total relation processing time: 0.20 hours. Matrix solve time: 2.93 hours. Total square root time: 0.10 hours, sqrts: 1. Prototype def-par.txt line would be: snfs,158,5,0,0,0,0,0,0,0,0,3000000,3000000,28,28,48,48,2.5,2.5,100000 total time: 30.51 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10158-419
c151
name 名前 Robert Backstrom April 24, 2007 16:43:17 UTC 2007 年 4 月 25 日 (水) 1 時 43 分 17 秒 (日本時間) `5387544406017014319327230238368785164959317286062833999131843397922409716848868223346056288825362813499140553902753137939643009070235018963983138109697<151>` ```16890701960639836030505223027599141408171380796039886568589214265562803591<74> 318965097991281408955651198675474281172186567241455132017352484669834254010167<78>``` ```Number: n N=5387544406017014319327230238368785164959317286062833999131843397922409716848868223346056288825362813499140553902753137939643009070235018963983138109697 ( 151 digits) SNFS difficulty: 159 digits. Divisors found: r1=16890701960639836030505223027599141408171380796039886568589214265562803591 (pp74) r2=318965097991281408955651198675474281172186567241455132017352484669834254010167 (pp78) Version: GGNFS-0.77.1-20060513-athlon-xp Total time: 33.67 hours. Scaled time: 43.06 units (timescale=1.279). Factorization parameters were as follows: name: KA_1_5_157_1 n: 5387544406017014319327230238368785164959317286062833999131843397922409716848868223346056288825362813499140553902753137939643009070235018963983138109697 skew: 0.62 deg: 5 c5: 875 c0: -82 m: 20000000000000000000000000000000 type: snfs rlim: 3000000 alim: 3000000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3000000/3000000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved algebraic special-q in [100000, 1500001) Primes: RFBsize:216816, AFBsize:217337, largePrimes:7010678 encountered Relations: rels:6493609, finalFF:497084 Max relations in full relation-set: 28 Initial matrix: 434219 x 497084 with sparse part having weight 34384792. Pruned matrix : 382432 x 384667 with weight 22554464. Total sieving time: 30.59 hours. Total relation processing time: 0.20 hours. Matrix solve time: 2.75 hours. Total square root time: 0.11 hours, sqrts: 1. Prototype def-par.txt line would be: snfs,159,5,0,0,0,0,0,0,0,0,3000000,3000000,28,28,48,48,2.5,2.5,100000 total time: 33.67 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10159-419
c150
name 名前 Robert Backstrom May 28, 2007 07:38:59 UTC 2007 年 5 月 28 日 (月) 16 時 38 分 59 秒 (日本時間) `395436898135329884560761905239269729221109523772888194939726864412661303745861547763429730960634179030786440951749566920750332476783936500229055425093<150>` ```6031459152120337884517237707176687<34> 65562393470958923411536676667369852579711360794342128936197039393056889008855481635744308817305722437678656721468939<116>``` ```Number: n N=395436898135329884560761905239269729221109523772888194939726864412661303745861547763429730960634179030786440951749566920750332476783936500229055425093 ( 150 digits) SNFS difficulty: 160 digits. Divisors found: r1=6031459152120337884517237707176687 (pp34) r2=65562393470958923411536676667369852579711360794342128936197039393056889008855481635744308817305722437678656721468939 (pp116) Version: GGNFS-0.77.1-20060513-athlon-xp Total time: 38.14 hours. Scaled time: 52.13 units (timescale=1.367). Factorization parameters were as follows: name: KA_1_5_158_1 n: 395436898135329884560761905239269729221109523772888194939726864412661303745861547763429730960634179030786440951749566920750332476783936500229055425093 skew: 1.96 deg: 5 c5: 7 c0: -205 m: 100000000000000000000000000000000 type: snfs rlim: 3500000 alim: 3500000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3500000/3500000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved algebraic special-q in [100000, 1600001) Primes: RFBsize:250150, AFBsize:250442, largePrimes:7155416 encountered Relations: rels:6677810, finalFF:563496 Max relations in full relation-set: 28 Initial matrix: 500657 x 563496 with sparse part having weight 34543217. Pruned matrix : 444914 x 447481 with weight 22819760. Total sieving time: 33.89 hours. Total relation processing time: 0.22 hours. Matrix solve time: 3.39 hours. Total square root time: 0.64 hours, sqrts: 5. Prototype def-par.txt line would be: snfs,160,5,0,0,0,0,0,0,0,0,3500000,3500000,28,28,48,48,2.5,2.5,100000 total time: 38.14 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10161-419
c131
name 名前 Robert Backstrom January 11, 2008 11:53:31 UTC 2008 年 1 月 11 日 (金) 20 時 53 分 31 秒 (日本時間) `98340968558953790892585585889392535669496255284578280374020149122636303507627134972206826846954930393394612505752654421559042963063<131>` ```1237731289012386762849485097499340001383842403737<49> 79452599632851028607355803437954089119988633926215410442037070045079459127725856399<83>``` ```Number: n N=98340968558953790892585585889392535669496255284578280374020149122636303507627134972206826846954930393394612505752654421559042963063 ( 131 digits) SNFS difficulty: 162 digits. Divisors found: Fri Jan 11 22:44:54 2008 prp49 factor: 1237731289012386762849485097499340001383842403737 Fri Jan 11 22:44:54 2008 prp83 factor: 79452599632851028607355803437954089119988633926215410442037070045079459127725856399 Fri Jan 11 22:44:54 2008 elapsed time 00:58:45 (Msieve 1.32) Version: GGNFS-0.77.1-20050930-k8 Total time: 40.38 hours. Scaled time: 33.84 units (timescale=0.838). Factorization parameters were as follows: name: KA_1_5_160_1 n: 98340968558953790892585585889392535669496255284578280374020149122636303507627134972206826846954930393394612505752654421559042963063 type: snfs deg: 5 c5: 140 c0: -41 skew: 0.78 m: 100000000000000000000000000000000 rlim: 3000000 alim: 3000000 lbpr: 28 lbpa: 28 mbpr: 48 mbpa: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3000000/3000000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 48/48 Sieved special-q in [100000, 3900001) Primes: RFBsize:216816, AFBsize:216922, largePrimes:5786060 encountered Relations: rels:5738816, finalFF:470992 Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Total sieving time: 40.25 hours. Total relation processing time: 0.13 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,162,5,0,0,0,0,0,0,0,0,3000000,3000000,27,27,48,48,2.5,2.5,100000 total time: 40.38 hours. --------- CPU info (if available) ---------- CPU0: AMD Athlon(tm) 64 X2 Dual Core Processor 6000+ stepping 03 CPU1: AMD Athlon(tm) 64 X2 Dual Core Processor 6000+ stepping 03 Memory: 2074672k/2097088k available (2382k kernel code, 21604k reserved, 681k data, 296k init, 1179584k highmem, 0k BadRAM) Calibrating delay loop... 5963.77 BogoMIPS (lpj=2981888) Calibrating delay loop... 6045.69 BogoMIPS (lpj=3022848) Total of 2 processors activated (12009.47 BogoMIPS).```
14×10162-419
c150
name 名前 Robert Backstrom June 19, 2007 22:01:42 UTC 2007 年 6 月 20 日 (水) 7 時 1 分 42 秒 (日本時間) `234452915565309717743310968590236453242416001444511606428126237948991274174971093497091887725636490945116371777925175942981720559775961753391568233761<150>` ```110021344519261749564662722162683047<36> 34504949858484862052592280714802542174189<41> 61758578210491351731564901830521726152917416595997540856609496672738529267<74>``` ```GMP-ECM 5.0 [powered by GMP 4.1.2] [ECM] Input number is 234452915565309717743310968590236453242416001444511606428126237948991274174971093497091887725636490945116371777925175942981720559775961753391568233761 (150 digits) Using B1=203500, B2=86869866, polynomial x^2, sigma=809407629 Step 1 took 4020ms Step 2 took 2820ms ********** Factor found in step 2: 110021344519261749564662722162683047 Found probable prime factor of 36 digits: 110021344519261749564662722162683047 Composite cofactor 2130976644484319852036608153111363410085013547135030736318682926017626785463045452028730619284253197320648388489463 has 115 digits GMP-ECM 5.0 [powered by GMP 4.1.2] [ECM] Input number is 2130976644484319852036608153111363410085013547135030736318682926017626785463045452028730619284253197320648388489463 (115 digits) Using B1=1024000, B2=868404622, polynomial Dickson(6), sigma=622108743 Step 1 took 13880ms Step 2 took 11310ms ********** Factor found in step 2: 34504949858484862052592280714802542174189 Found probable prime factor of 41 digits: 34504949858484862052592280714802542174189 Probable prime cofactor 61758578210491351731564901830521726152917416595997540856609496672738529267 has 74 digits```
14×10164-419
c155
name 名前 Robert Backstrom September 28, 2007 05:50:19 UTC 2007 年 9 月 28 日 (金) 14 時 50 分 19 秒 (日本時間) `21037291424252539446163109033236138674247346680037789377346270695525241546202958190719320217737047557792989848315711277798343615987894452010473997907897297<155>` ```241913548612846605274086927369517<33> 86962022362460490445785055865229214136443842893085986014021798676035510354833177041605490276301765124099002836589234876341<122>``` ```Number: n N=21037291424252539446163109033236138674247346680037789377346270695525241546202958190719320217737047557792989848315711277798343615987894452010473997907897297 ( 155 digits) SNFS difficulty: 165 digits. Divisors found: r1=241913548612846605274086927369517 (pp33) r2=86962022362460490445785055865229214136443842893085986014021798676035510354833177041605490276301765124099002836589234876341 (pp122) Version: GGNFS-0.77.1-20051202-athlon Total time: 74.48 hours. Scaled time: 98.54 units (timescale=1.323). Factorization parameters were as follows: name: KA_1_5_163_1 n: 21037291424252539446163109033236138674247346680037789377346270695525241546202958190719320217737047557792989848315711277798343615987894452010473997907897297 skew: 1.96 deg: 5 c5: 7 c0: -205 m: 1000000000000000000000000000000000 type: snfs rlim: 3500000 alim: 3500000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3500000/3500000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved algebraic special-q in [100000, 2800001) Primes: RFBsize:250150, AFBsize:250442, largePrimes:7581353 encountered Relations: rels:7099178, finalFF:586042 Max relations in full relation-set: 48 Initial matrix: 500657 x 586042 with sparse part having weight 50177668. Pruned matrix : 439767 x 442334 with weight 33392103. Total sieving time: 60.70 hours. Total relation processing time: 0.37 hours. Matrix solve time: 13.32 hours. Total square root time: 0.10 hours, sqrts: 1. Prototype def-par.txt line would be: snfs,165,5,0,0,0,0,0,0,0,0,3500000,3500000,28,28,48,48,2.5,2.5,100000 total time: 74.48 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10165-419
c136
name 名前 Robert Backstrom May 28, 2008 04:47:25 UTC 2008 年 5 月 28 日 (水) 13 時 47 分 25 秒 (日本時間) `2643961340767579204510392201562942474383505436157630831674023493387373870717733728070420140402288953838976680468196570012720491891696559<136>` ```145830821819731044721528535735047502903441900884043243247216827<63> 18130332859509736343289311799201732135771943135308657063773868356166511517<74>``` ```Number: n N=2643961340767579204510392201562942474383505436157630831674023493387373870717733728070420140402288953838976680468196570012720491891696559 ( 136 digits) SNFS difficulty: 166 digits. Divisors found: Wed May 28 10:17:40 2008 prp63 factor: 145830821819731044721528535735047502903441900884043243247216827 Wed May 28 10:17:40 2008 prp74 factor: 18130332859509736343289311799201732135771943135308657063773868356166511517 Wed May 28 10:17:40 2008 elapsed time 00:53:14 (Msieve 1.36) Version: GGNFS-0.77.1-20051202-athlon Total time: 35.95 hours. Scaled time: 65.76 units (timescale=1.829). Factorization parameters were as follows: name: KA_1_5_164_1 n: 2643961340767579204510392201562942474383505436157630831674023493387373870717733728070420140402288953838976680468196570012720491891696559 skew: 1.24 deg: 5 c5: 14 c0: -41 m: 1000000000000000000000000000000000 type: snfs rlim: 3300000 alim: 3300000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 3300000/3300000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved special-q in [100000, 2500339) Primes: RFBsize:236900, AFBsize:237378, largePrimes:7332565 encountered Relations: rels:6804891, finalFF:530984 Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Total sieving time: 35.79 hours. Total relation processing time: 0.17 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,166,5,0,0,0,0,0,0,0,0,3300000,3300000,28,28,48,48,2.5,2.5,100000 total time: 35.95 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 X2 6000+```
14×10166-419
c162
name 名前 Robert Backstrom November 18, 2007 23:30:08 UTC 2007 年 11 月 19 日 (月) 8 時 30 分 8 秒 (日本時間) `455013764166366032572485317680859846010341812839838405111754630577574971642892197488974041464754308817841739712626306945785109999577487218988374398325549347867773<162>` ```35081134283933574559653611257097<32> 2728630078335383137189177941861782066861<40> 126411714129835466690844912764467931579339687<45> 37602690897693034145575177417896418231525935887<47>``` ```GMP-ECM 6.0.1 [powered by GMP 4.1.4] [ECM] Input number is 455013764166366032572485317680859846010341812839838405111754630577574971642892197488974041464754308817841739712626306945785109999577487218988374398325549347867773 (162 digits) Using B1=396500, B2=224704516, polynomial Dickson(3), sigma=40752660 Step 1 took 9172ms Step 2 took 5188ms ********** Factor found in step 2: 35081134283933574559653611257097 Found probable prime factor of 32 digits: 35081134283933574559653611257097 Composite cofactor 12970326457624057370940093292765884024967210652070161540053382611780691229396033766335056809054710948583510049684525069811157738709 has 131 digits Number: n N=12970326457624057370940093292765884024967210652070161540053382611780691229396033766335056809054710948583510049684525069811157738709 ( 131 digits) SNFS difficulty: 167 digits. Divisors found: Mon Nov 19 08:29:02 2007 prp40 factor: 2728630078335383137189177941861782066861 Mon Nov 19 08:29:02 2007 prp45 factor: 126411714129835466690844912764467931579339687 Mon Nov 19 08:29:02 2007 prp47 factor: 37602690897693034145575177417896418231525935887 Mon Nov 19 08:29:02 2007 elapsed time 03:38:46 (Msieve 1.29) Version: GGNFS-0.77.1-20051202-athlon Total time: 212.15 hours. Scaled time: 254.37 units (timescale=1.199). Factorization parameters were as follows: name: KA_1_5_165_1 n: 12970326457624057370940093292765884024967210652070161540053382611780691229396033766335056809054710948583510049684525069811157738709 # n: 455013764166366032572485317680859846010341812839838405111754630577574971642892197488974041464754308817841739712626306945785109999577487218988374398325549347867773 type: snfs skew: 0.78 deg: 5 c5: 140 c0: -41 m: 1000000000000000000000000000000000 rlim: 3500000 alim: 3500000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 qintsize: 50000 Factor base limits: 3500000/3500000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved special-q in [100000, 3400001) Primes: RFBsize:250150, AFBsize:250097, largePrimes:7703762 encountered Relations: rels:7182168, finalFF:528137 Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Total sieving time: 211.32 hours. Total relation processing time: 0.83 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,167,5,0,0,0,0,0,0,0,0,3500000,3500000,28,28,48,48,2.6,2.6,100000 total time: 212.15 hours. --------- CPU info (if available) ---------- Cygwin on AMD XP 2700+```
14×10168-419
c154
name 名前 Robert Backstrom July 6, 2008 02:12:07 UTC 2008 年 7 月 6 日 (日) 11 時 12 分 7 秒 (日本時間) `1529960291681783151114531193400230721051917068342751517577310648254554333940389934727118286669463974028994922153019732058807777461380973311907897295786167<154>` ```16981226766391967119183913630606464339<38> 90097159217599722569381430542429490640646635539748578844942039627166323665820109863974364446625831381931678300595853<116>``` ```Number: n N=1529960291681783151114531193400230721051917068342751517577310648254554333940389934727118286669463974028994922153019732058807777461380973311907897295786167 ( 154 digits) SNFS difficulty: 169 digits. Divisors found: Sun Jul 06 09:20:10 2008 prp38 factor: 16981226766391967119183913630606464339 Sun Jul 06 09:20:10 2008 prp116 factor: 90097159217599722569381430542429490640646635539748578844942039627166323665820109863974364446625831381931678300595853 Sun Jul 06 09:20:10 2008 elapsed time 02:40:33 (Msieve 1.36) Version: GGNFS-0.77.1-20051202-athlon Total time: 81.62 hours. Scaled time: 142.92 units (timescale=1.751). Factorization parameters were as follows: name: KA_1_5_167_1 n: 1529960291681783151114531193400230721051917068342751517577310648254554333940389934727118286669463974028994922153019732058807777461380973311907897295786167 type: snfs skew: 0.62 deg: 5 c5: 875 c0: -82 m: 2000000000000000000000000000000000 rlim: 5800000 alim: 5800000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 qintsize: 100000 Factor base limits: 5800000/5800000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved special-q in [100000, 3800113) Primes: RFBsize:399993, AFBsize:400415, largePrimes:8020626 encountered Relations: rels:7627568, finalFF:810616 Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Total sieving time: 81.31 hours. Total relation processing time: 0.31 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,169,5,0,0,0,0,0,0,0,0,5800000,5800000,28,28,48,48,2.6,2.6,100000 total time: 81.62 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10170-419
c169
name 名前 Robert Backstrom June 3, 2007 23:08:55 UTC 2007 年 6 月 4 日 (月) 8 時 8 分 55 秒 (日本時間) `1747815230961298377028714107365792759051186017478152309612983770287141073657927590511860174781523096129837702871410736579275905118601747815230961298377028714107365792759<169>` ```23562871960702446576109209737209512980072163311<47> 74176663773255644917317906830538807621668868770340406914332112105118853831007147436387613726854741316452945435326628021369<122>``` ```Number: n N=1747815230961298377028714107365792759051186017478152309612983770287141073657927590511860174781523096129837702871410736579275905118601747815230961298377028714107365792759 ( 169 digits) SNFS difficulty: 171 digits. Divisors found: r1=23562871960702446576109209737209512980072163311 (pp47) r2=74176663773255644917317906830538807621668868770340406914332112105118853831007147436387613726854741316452945435326628021369 (pp122) Version: GGNFS-0.77.1-20051202-athlon Total time: 101.58 hours. Scaled time: 134.39 units (timescale=1.323). Factorization parameters were as follows: name: KA_1_5_169_1 n: 1747815230961298377028714107365792759051186017478152309612983770287141073657927590511860174781523096129837702871410736579275905118601747815230961298377028714107365792759 skew: 1.24 deg: 5 c5: 14 c0: -41 m: 10000000000000000000000000000000000 type: snfs rlim: 5000000 alim: 5000000 lpbr: 28 lpba: 28 mfbr: 48 mfba: 48 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 5000000/5000000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 48/48 Sieved algebraic special-q in [100000, 4000001) Primes: RFBsize:348513, AFBsize:349737, largePrimes:8140106 encountered Relations: rels:7766379, finalFF:797968 Max relations in full relation-set: 48 Initial matrix: 698316 x 797968 with sparse part having weight 47933037. Pruned matrix : 612389 x 615944 with weight 31910090. Total sieving time: 91.84 hours. Total relation processing time: 0.33 hours. Matrix solve time: 9.19 hours. Total square root time: 0.22 hours, sqrts: 2. Prototype def-par.txt line would be: snfs,171,5,0,0,0,0,0,0,0,0,5000000,5000000,28,28,48,48,2.5,2.5,100000 total time: 101.58 hours. --------- CPU info (if available) ---------- Cygwin on AMD 64 3200+```
14×10171-419
c162
name 名前 Robert Backstrom June 19, 2009 07:33:58 UTC 2009 年 6 月 19 日 (金) 16 時 33 分 58 秒 (日本時間) `326747806240499343230759596100293564963118613154761041676746005032714834420399700505602789363046603661020070633672157059323827132394838374455190425704435922805177<162>` ```8422572196437277344165892625074975797910517101552896422569<58> 38794301624236914293793808123357843854869719962248434313381337905654704362886863912964954781426002751633<104>``` ```Number: n N=326747806240499343230759596100293564963118613154761041676746005032714834420399700505602789363046603661020070633672157059323827132394838374455190425704435922805177 ( 162 digits) SNFS difficulty: 172 digits. Divisors found: Fri Jun 19 11:40:48 2009 prp58 factor: 8422572196437277344165892625074975797910517101552896422569 Fri Jun 19 11:40:48 2009 prp104 factor: 38794301624236914293793808123357843854869719962248434313381337905654704362886863912964954781426002751633 Fri Jun 19 11:40:48 2009 elapsed time 01:16:46 (Msieve 1.39 - dependency 1) Version: GGNFS-0.77.1-20060513-pentium-m Total time: 55.33 hours. Scaled time: 143.58 units (timescale=2.595). Factorization parameters were as follows: name: KA_1_5_170_1 n: 326747806240499343230759596100293564963118613154761041676746005032714834420399700505602789363046603661020070633672157059323827132394838374455190425704435922805177 m: 10000000000000000000000000000000000 deg: 5 c5: 140 c0: -41 skew: 0.78 type: snfs lss: 1 rlim: 5200000 alim: 5200000 lpbr: 28 lpba: 28 mfbr: 56 mfba: 56 rlambda: 2.4 alambda: 2.4 qintsize: 100000 Factor base limits: 5200000/5200000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 56/56 Sieved special-q in [2600000, 5682731) Primes: RFBsize:361407, AFBsize:361488, largePrimes:17700805 encountered Relations: rels:17570935, finalFF:657599 Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 2124084 hash collisions in 19715067 relations Msieve: matrix is 839295 x 839543 (222.4 MB) Total sieving time: 54.53 hours. Total relation processing time: 0.80 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,172,5,0,0,0,0,0,0,0,0,5200000,5200000,28,28,56,56,2.4,2.4,100000 total time: 55.33 hours. --------- CPU info (if available) ----------```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:00:41 UTC 2009 年 3 月 7 日 (土) 8 時 0 分 41 秒 (日本時間)
255e4230Andreas TeteJune 11, 2009 19:49:20 UTC 2009 年 6 月 12 日 (金) 4 時 49 分 20 秒 (日本時間)
3025e4412Andreas TeteJune 11, 2009 19:49:43 UTC 2009 年 6 月 12 日 (金) 4 時 49 分 43 秒 (日本時間)
351e61204Andreas TeteJune 11, 2009 19:50:29 UTC 2009 年 6 月 12 日 (金) 4 時 50 分 29 秒 (日本時間)
403e62500Dmitry DomanovJune 12, 2009 03:14:15 UTC 2009 年 6 月 12 日 (金) 12 時 14 分 15 秒 (日本時間)
14×10173-419
c153
name 名前 Markus Tervooren March 28, 2010 18:55:29 UTC 2010 年 3 月 29 日 (月) 3 時 55 分 29 秒 (日本時間) `347000135465724530503799425009385047835317539649822814252765478428183943593657226484694890368451833896352697067016831821349402963395720095424908569483431<153>` ```23080100935767937552807262322325734205629659558792838583<56> 15034602163631261212715734728010325972403035090978675640483615735970342078121418262955553225997457<98>``` ```Msieve v. 1.44 Sun Mar 28 19:02:01 2010 random seeds: 2ef1c838 a1074f0b factoring 347000135465724530503799425009385047835317539649822814252765478428183943593657226484694890368451833896352697067016831821349402963395720095424908569483431 (153 digits) searching for 15-digit factors commencing number field sieve (153-digit input) R0: -10000000000000000000000000000000000 R1: 1 A0: -41 A1: 0 A2: 0 A3: 0 A4: 0 A5: 14000 skew 0.31, size 3.969596e-12, alpha -1.807517, combined = 1.907255e-10 commencing relation filtering estimated available RAM is 8010.2 MB commencing duplicate removal, pass 1 read 10M relations read 20M relations found 2930406 hash collisions in 29459129 relations added 65379 free relations commencing duplicate removal, pass 2 found 1795359 duplicates and 27729149 unique relations memory use: 98.6 MB reading ideals above 720000 commencing singleton removal, initial pass memory use: 753.0 MB reading all ideals from disk memory use: 830.2 MB keeping 31582432 ideals with weight <= 200, target excess is 189188 commencing in-memory singleton removal begin with 27729149 relations and 31582432 unique ideals reduce to 7611355 relations and 7011020 ideals in 22 passes max relations containing the same ideal: 85 removing 1615921 relations and 1425483 ideals in 190438 cliques commencing in-memory singleton removal begin with 5995434 relations and 7011020 unique ideals reduce to 5734823 relations and 5315884 ideals in 13 passes max relations containing the same ideal: 69 removing 1180158 relations and 989720 ideals in 190438 cliques commencing in-memory singleton removal begin with 4554665 relations and 5315884 unique ideals reduce to 4361573 relations and 4126304 ideals in 11 passes max relations containing the same ideal: 59 removing 163334 relations and 147524 ideals in 15810 cliques commencing in-memory singleton removal begin with 4198239 relations and 4126304 unique ideals reduce to 4193351 relations and 3973878 ideals in 6 passes max relations containing the same ideal: 57 relations with 0 large ideals: 4511 relations with 1 large ideals: 20662 relations with 2 large ideals: 131937 relations with 3 large ideals: 462031 relations with 4 large ideals: 953084 relations with 5 large ideals: 1189296 relations with 6 large ideals: 926241 relations with 7+ large ideals: 505589 commencing 2-way merge reduce to 2338368 relation sets and 2118894 unique ideals commencing full merge memory use: 226.7 MB found 1159051 cycles, need 1131094 weight of 1131094 cycles is about 79262281 (70.08/cycle) distribution of cycle lengths: 1 relations: 140964 2 relations: 127141 3 relations: 122671 4 relations: 111120 5 relations: 102342 6 relations: 87553 7 relations: 79222 8 relations: 68328 9 relations: 58346 10+ relations: 233407 heaviest cycle: 21 relations commencing cycle optimization start with 6851942 relations pruned 126648 relations memory use: 234.4 MB distribution of cycle lengths: 1 relations: 140964 2 relations: 129373 3 relations: 126021 4 relations: 112998 5 relations: 104147 6 relations: 88605 7 relations: 79862 8 relations: 68559 9 relations: 58338 10+ relations: 222227 heaviest cycle: 21 relations RelProcTime: 620 elapsed time 00:10:21 -> Doing matrix solving step... =>nice -n 19 "../msieve" -s 172.dat -l ggnfs.log -i 172.ini -v -nf 172.fb -t 4 -nc2 Msieve v. 1.44 Sun Mar 28 19:12:23 2010 random seeds: 6c55cfa6 6a3af0e3 factoring 347000135465724530503799425009385047835317539649822814252765478428183943593657226484694890368451833896352697067016831821349402963395720095424908569483431 (153 digits) searching for 15-digit factors commencing number field sieve (153-digit input) R0: -10000000000000000000000000000000000 R1: 1 A0: -41 A1: 0 A2: 0 A3: 0 A4: 0 A5: 14000 skew 0.31, size 3.969596e-12, alpha -1.807517, combined = 1.907255e-10 commencing linear algebra read 1131094 cycles cycles contain 3988639 unique relations read 3988639 relations using 20 quadratic characters above 536868782 building initial matrix memory use: 483.9 MB read 1131094 cycles matrix is 1130907 x 1131094 (340.3 MB) with weight 104092949 (92.03/col) sparse part has weight 76775126 (67.88/col) filtering completed in 2 passes matrix is 1125929 x 1126115 (339.8 MB) with weight 103890318 (92.26/col) sparse part has weight 76697372 (68.11/col) read 1126115 cycles matrix is 1125929 x 1126115 (339.8 MB) with weight 103890318 (92.26/col) sparse part has weight 76697372 (68.11/col) saving the first 48 matrix rows for later matrix is 1125881 x 1126115 (324.7 MB) with weight 82559648 (73.31/col) sparse part has weight 73854808 (65.58/col) matrix includes 64 packed rows using block size 65536 for processor cache size 4096 kB commencing Lanczos iteration (4 threads) memory use: 343.0 MB linear algebra at 0.1%, ETA 1h13m126115 dimensions (0.1%, ETA 1h13m) linear algebra completed 1125728 of 1126115 dimensions (100.0%, ETA 0h 0m) lanczos halted after 17808 iterations (dim = 1125881) recovered 39 nontrivial dependencies BLanczosTime: 5136 elapsed time 01:25:37 =>nice -n 19 "../msieve" -s 172.dat -l ggnfs.log -i 172.ini -v -nf 172.fb -t 4 -nc3 Msieve v. 1.44 Sun Mar 28 20:38:00 2010 random seeds: 11f1e36c ded5e247 factoring 347000135465724530503799425009385047835317539649822814252765478428183943593657226484694890368451833896352697067016831821349402963395720095424908569483431 (153 digits) searching for 15-digit factors commencing number field sieve (153-digit input) R0: -10000000000000000000000000000000000 R1: 1 A0: -41 A1: 0 A2: 0 A3: 0 A4: 0 A5: 14000 skew 0.31, size 3.969596e-12, alpha -1.807517, combined = 1.907255e-10 commencing square root phase reading relations for dependency 1 read 562704 cycles cycles contain 1992760 unique relations read 1992760 relations multiplying 1992760 relations multiply complete, coefficients have about 69.44 million bits initial square root is modulo 96581 reading relations for dependency 2 read 562741 cycles cycles contain 1991774 unique relations read 1991774 relations multiplying 1991774 relations multiply complete, coefficients have about 69.41 million bits initial square root is modulo 96211 reading relations for dependency 3 read 563713 cycles cycles contain 1995770 unique relations read 1995770 relations multiplying 1995770 relations multiply complete, coefficients have about 69.55 million bits initial square root is modulo 98221 reading relations for dependency 4 read 562322 cycles cycles contain 1990392 unique relations read 1990392 relations multiplying 1990392 relations multiply complete, coefficients have about 69.36 million bits initial square root is modulo 95191 sqrtTime: 988 prp56 factor: 23080100935767937552807262322325734205629659558792838583 prp98 factor: 15034602163631261212715734728010325972403035090978675640483615735970342078121418262955553225997457 elapsed time 00:16:29```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:01:12 UTC 2009 年 3 月 7 日 (土) 8 時 1 分 12 秒 (日本時間)
255e40--
3025e4430Erik BrangerSeptember 11, 2009 16:44:34 UTC 2009 年 9 月 12 日 (土) 1 時 44 分 34 秒 (日本時間)
351e60--
403e62336Wataru SakaiSeptember 15, 2009 05:21:34 UTC 2009 年 9 月 15 日 (火) 14 時 21 分 34 秒 (日本時間)
4511e60--
5043e61140 / 7465yoyo@homeJanuary 15, 2010 06:50:12 UTC 2010 年 1 月 15 日 (金) 15 時 50 分 12 秒 (日本時間)
14×10174-419
c157
name 名前 Ignacio Santos September 2, 2010 11:18:08 UTC 2010 年 9 月 2 日 (木) 20 時 18 分 8 秒 (日本時間) `2852369260900869362316423549810699947802201450091696271793506865217669614989771609502170881030039380641982448315685948161717739940431982431044867466172920491<157>` ```972903877050095322201729952570985360045293<42> 2931809943598363724639189406254713941279726209214477444932991429078165303110719735541116020713833296683821829860087<115>``` ```Number: 9 N=2852369260900869362316423549810699947802201450091696271793506865217669614989771609502170881030039380641982448315685948161717739940431982431044867466172920491 ( 157 digits) SNFS difficulty: 175 digits. Divisors found: r1=972903877050095322201729952570985360045293 (pp42) r2=2931809943598363724639189406254713941279726209214477444932991429078165303110719735541116020713833296683821829860087 (pp115) Version: Msieve-1.40 Total time: 71.46 hours. Scaled time: 124.62 units (timescale=1.744). Factorization parameters were as follows: n: 2852369260900869362316423549810699947802201450091696271793506865217669614989771609502170881030039380641982448315685948161717739940431982431044867466172920491 m: 20000000000000000000000000000000000 deg: 5 c5: 4375 c0: -41 skew: 0.39 type: snfs lss: 1 rlim: 5800000 alim: 5800000 lpbr: 28 lpba: 28 mfbr: 52 mfba: 52 rlambda: 2.5 alambda: 2.5 Factor base limits: 5800000/5800000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 52/52 Sieved rational special-q in [2900000, 6400001) Primes: , , Relations: relations Max relations in full relation-set: Initial matrix: Pruned matrix : 1176517 x 1176742 Total sieving time: 68.89 hours. Total relation processing time: 0.16 hours. Matrix solve time: 1.89 hours. Time per square root: 0.51 hours. Prototype def-par.txt line would be: snfs,175.000,5,0,0,0,0,0,0,0,0,5800000,5800000,28,28,52,52,2.5,2.5,100000 total time: 71.46 hours. --------- CPU info (if available) ----------``` `GGNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:01:48 UTC 2009 年 3 月 7 日 (土) 8 時 1 分 48 秒 (日本時間)
255e4204Ignacio SantosMarch 14, 2010 20:39:16 UTC 2010 年 3 月 15 日 (月) 5 時 39 分 16 秒 (日本時間)
3025e4403Ignacio SantosMarch 14, 2010 21:44:23 UTC 2010 年 3 月 15 日 (月) 6 時 44 分 23 秒 (日本時間)
351e6868458Ignacio SantosMarch 18, 2010 19:54:45 UTC 2010 年 3 月 19 日 (金) 4 時 54 分 45 秒 (日本時間)
410Ignacio SantosMarch 29, 2010 08:50:44 UTC 2010 年 3 月 29 日 (月) 17 時 50 分 44 秒 (日本時間)
403e62100150Ignacio SantosMarch 29, 2010 08:50:44 UTC 2010 年 3 月 29 日 (月) 17 時 50 分 44 秒 (日本時間)
1950Wataru SakaiMay 26, 2010 06:41:40 UTC 2010 年 5 月 26 日 (水) 15 時 41 分 40 秒 (日本時間)
14×10175-419
c173
name 名前 Sinkiti Sibata January 29, 2007 05:56:53 UTC 2007 年 1 月 29 日 (月) 14 時 56 分 53 秒 (日本時間) `32887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887<173>` ```1971848954400133691125566917937107287858416479<46> 16678260045126227226173832334535265219726130696522534014568395028983269564908150799866271813714347853920132775969907902330680553<128>``` ```Number: 15551_175 N=32887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887 ( 173 digits) SNFS difficulty: 176 digits. Divisors found: r1=1971848954400133691125566917937107287858416479 (pp46) r2=16678260045126227226173832334535265219726130696522534014568395028983269564908150799866271813714347853920132775969907902330680553 (pp128) Version: GGNFS-0.77.1-20060722-pentium4 Total time: 390.89 hours. Scaled time: 238.83 units (timescale=0.611). Factorization parameters were as follows: name: 15551_175 n: 32887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887009631195677707305614282358468404980032887 m: 100000000000000000000000000000000000 c5: 14 c0: -41 skew: 4 type: snfs Factor base limits: 7400000/7400000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 48/48 Sieved algebraic special-q in [3700000, 11300001) Primes: RFBsize:501962, AFBsize:502607, largePrimes:6474994 encountered Relations: rels:6944147, finalFF:1127040 Max relations in full relation-set: 0 Initial matrix: 1004635 x 1127040 with sparse part having weight 69379171. Pruned matrix : 898670 x 903757 with weight 54223156. Total sieving time: 335.51 hours. Total relation processing time: 1.06 hours. Matrix solve time: 53.90 hours. Time per square root: 0.42 hours. Prototype def-par.txt line would be: snfs,176,5,0,0,0,0,0,0,0,0,7400000,7400000,27,27,48,48,2.6,2.6,100000 total time: 390.89 hours. --------- CPU info (if available) ----------``` `Pentium 4 2.4GHz, Windows XP and Cygwin`
14×10176-419
c144
name 名前 Serge Batalov August 8, 2008 01:48:43 UTC 2008 年 8 月 8 日 (金) 10 時 48 分 43 秒 (日本時間) `853033207546563461451718362889740780953837235756093028082603839348423212464125183593320212885364289524600134366427205672362581025957408927915491<144>` `6253487942641704767889353221<28>` `136409187220118099594259543876309212197184154651522471682143115643170104578299575205818479783053844257159831548060871<117>` ```Using B1=1000000, B2=2139490090, polynomial Dickson(6), sigma=1412173085 Step 1 took 4788ms Step 2 took 3348ms ********** Factor found in step 2: 6253487942641704767889353221 Found probable prime factor of 28 digits: 6253487942641704767889353221 Composite cofactor 136409187220118099594259543876309212197184154651522471682143115643170104578299575205818479783053844257159831548060871 has 117 digits``` `GMP-ECM 6.2.1`
c117
name 名前 Serge Batalov August 10, 2008 19:08:23 UTC 2008 年 8 月 11 日 (月) 4 時 8 分 23 秒 (日本時間) `136409187220118099594259543876309212197184154651522471682143115643170104578299575205818479783053844257159831548060871<117>` ```1019051608417910918330923759093085376287866529<46> 133858958754694373843982955055938024720353027482661337116522058059690599<72>``` ```Number: 15551_176 N=136409187220118099594259543876309212197184154651522471682143115643170104578299575205818479783053844257159831548060871 ( 117 digits) Divisors found: r1=1019051608417910918330923759093085376287866529 r2=133858958754694373843982955055938024720353027482661337116522058059690599 Version: Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.314). Factorization parameters were as follows: name: 15551_176 n: 136409187220118099594259543876309212197184154651522471682143115643170104578299575205818479783053844257159831548060871 skew: 49132.51 # norm 2.08e+16 c5: 6660 c4: 447743516 c3: -609459037292731 c2: -187213174297973853 c1: 85240638334503841045441 c0: 142345279227926981946484747 # alpha -5.80 Y1: 173069743693 Y0: -28991710782521877328584 # Murphy_E 4.58e-10 # M 111726850663256429068149510231893237790161040770684032330899062146596737783001767146726624576283412255547250616765441 type: gnfs rlim: 4500000 alim: 4500000 lpbr: 27 lpba: 27 mfbr: 50 mfba: 50 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 4500000/4500000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 50/50 Sieved algebraic special-q in [2250000, 5550001) Primes: rational ideals filtering, algebraic ideals filtering, Relations: relations Max relations in full relation-set: Initial matrix: Pruned matrix : 624752 x 624980 Total sieving time: 0.00 hours. Total relation processing time: 0.00 hours. Matrix solve time: 1.9 hours. Time per square root: 0.33 hours. Prototype def-par.txt line would be: gnfs,116,5,maxs1,maxskew,goodScore,efrac,j0,j1,eStepSize,maxTime,4500000,4500000,27,27,50,50,2.5,2.5,100000 total time: 31 hours.``` `Msieve-1.36` `Opteron-2.2GHz; Linux x86_64`
14×10179-419
c126
name 名前 Robert Backstrom January 28, 2008 02:58:30 UTC 2008 年 1 月 28 日 (月) 11 時 58 分 30 秒 (日本時間) `612214321443508783027241882722763189436929423776438294876205034862636149584748928187912109677383254425120789930651974479517241<126>` ```20107693569497065346641840555374563<35> 30446770005100168171382797846562982573356796889366415814698830439233930405861605026830253107<92>``` ```GMP-ECM 6.0 [powered by GMP 4.1.4] [ECM] Input number is 612214321443508783027241882722763189436929423776438294876205034862636149584748928187912109677383254425120789930651974479517241 (126 digits) Using B1=1086000, B2=944273532, polynomial Dickson(3), sigma=616903093 Step 1 took 9992ms Step 2 took 4663ms ********** Factor found in step 2: 20107693569497065346641840555374563 Found probable prime factor of 35 digits: 20107693569497065346641840555374563 Probable prime cofactor 30446770005100168171382797846562982573356796889366415814698830439233930405861605026830253107 has 92 digits```
14×10180-419
c143
name 名前 Ignacio Santos September 25, 2010 22:58:12 UTC 2010 年 9 月 26 日 (日) 7 時 58 分 12 秒 (日本時間) `37620770792961260169000776314768279227423125480790131521738298307291451823488282089158009025505982963394152884192179599243092310473533977711963<143>` ```377792798387183521521767477672641<33> 99580433913950252411989798540909150574762318664033411196238101031586466398615937104293186061775420154935452443<110>``` ```Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=1059588015 Step 1 took 71417ms Step 2 took 34024ms ********** Factor found in step 2: 377792798387183521521767477672641 Found probable prime factor of 33 digits: 377792798387183521521767477672641 Probable prime cofactor 99580433913950252411989798540909150574762318664033411196238101031586466398615937104293186061775420154935452443 has 110 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:02:24 UTC 2009 年 3 月 7 日 (土) 8 時 2 分 24 秒 (日本時間)
255e40--
3025e460Rich DickersonSeptember 25, 2010 02:03:30 UTC 2010 年 9 月 25 日 (土) 11 時 3 分 30 秒 (日本時間)
351e6100 / 837Rich DickersonSeptember 25, 2010 02:03:30 UTC 2010 年 9 月 25 日 (土) 11 時 3 分 30 秒 (日本時間)
403e620 / 2321Rich DickersonSeptember 25, 2010 02:21:28 UTC 2010 年 9 月 25 日 (土) 11 時 21 分 28 秒 (日本時間)
14×10181-419
c162
name 名前 Dmitry Domanov October 29, 2012 08:10:29 UTC 2012 年 10 月 29 日 (月) 17 時 10 分 29 秒 (日本時間) `151178013085473247123592318029129285052854439704018850606634858275707709881500257531613762346524228710900082390786886759631587156448782668276806855854282508097509<162>` ```45675311350624366066456907382414462817794805137268093686527314438719307<71> 3309840888110480166167154996700718752489184274272528640641948515171493403268046966540185487<91>``` ```N=151178013085473247123592318029129285052854439704018850606634858275707709881500257531613762346524228710900082390786886759631587156448782668276806855854282508097509 ( 162 digits) SNFS difficulty: 182 digits. Divisors found: r1=45675311350624366066456907382414462817794805137268093686527314438719307 (pp71) r2=3309840888110480166167154996700718752489184274272528640641948515171493403268046966540185487 (pp91) Version: Msieve-1.40 Total time: 176.87 hours. Scaled time: 344.19 units (timescale=1.946). Factorization parameters were as follows: n: 151178013085473247123592318029129285052854439704018850606634858275707709881500257531613762346524228710900082390786886759631587156448782668276806855854282508097509 m: 1000000000000000000000000000000000000 deg: 5 c5: 140 c0: -41 skew: 0.78 # Murphy_E = 8.921e-11 type: snfs lss: 1 rlim: 7600000 alim: 7600000 lpbr: 28 lpba: 28 mfbr: 53 mfba: 53 rlambda: 2.5 alambda: 2.5 qintsize: 400000 Factor base limits: 7600000/7600000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 53/53 Sieved rational special-q in [3800000, 6600001) Primes: , , Relations: relations Max relations in full relation-set: Initial matrix: Pruned matrix : 1129707 x 1129932 Total sieving time: 175.22 hours. Total relation processing time: 0.19 hours. Matrix solve time: 1.33 hours. Time per square root: 0.12 hours. Prototype def-par.txt line would be: snfs,182.000,5,0,0,0,0,0,0,0,0,7600000,7600000,28,28,53,53,2.5,2.5,100000 total time: 176.87 hours. --------- CPU info (if available) ----------```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:03:05 UTC 2009 年 3 月 7 日 (土) 8 時 3 分 5 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 25, 2010 23:46:07 UTC 2010 年 9 月 26 日 (日) 8 時 46 分 7 秒 (日本時間)
351e6300Ignacio SantosSeptember 25, 2010 23:56:54 UTC 2010 年 9 月 26 日 (日) 8 時 56 分 54 秒 (日本時間)
403e62144110Ignacio SantosSeptember 25, 2010 23:56:54 UTC 2010 年 9 月 26 日 (日) 8 時 56 分 54 秒 (日本時間)
2034apoOctober 15, 2012 07:46:00 UTC 2012 年 10 月 15 日 (月) 16 時 46 分 0 秒 (日本時間)
4511e632 / 3991Ignacio SantosSeptember 25, 2010 23:56:54 UTC 2010 年 9 月 26 日 (日) 8 時 56 分 54 秒 (日本時間)
14×10186-419
c163
name 名前 Serge Batalov November 14, 2008 05:54:39 UTC 2008 年 11 月 14 日 (金) 14 時 54 分 39 秒 (日本時間) `1761802176713554260439600271227320073560053489692942860792504699397952438152606637476421940803327426096191842876451365120475029111493672266456440951198876595108569<163>` ```32588984944879249360239874831<29> 54061278057400452561143167961063501861872785250750941089820142488320368986133543167855544550220331678923941602572569248399536665621399<134>``` ```Using B1=3000000, B2=11414255590, polynomial Dickson(12), sigma=319754870 Step 1 took 19249ms Step 2 took 12557ms ********** Factor found in step 2: 32588984944879249360239874831 Found probable prime factor of 29 digits: 32588984944879249360239874831 Probable prime cofactor 54061278057400452561143167961063501861872785250750941089820142488320368986133543167855544550220331678923941602572569248399536665621399 has 134 digits``` `GMP-ECM 6.2.1`
14×10187-419
c178
name 名前 Ignacio Santos September 26, 2010 08:31:45 UTC 2010 年 9 月 26 日 (日) 17 時 31 分 45 秒 (日本時間) `1706318429313935541446795563874087520169621852577261483930470700731604913124723544047723436393570817299749045585134873042367967564674490440365375501443098951264542648050307437051<178>` ```3689731686791057511182493254095616953<37> 462450544960333618227622623208059861475009706651116695284225242264765051852021894512774787633648215514948788708010551378932007874192315440467<141>``` ```Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=478754712 Step 1 took 97064ms Step 2 took 43587ms ********** Factor found in step 2: 3689731686791057511182493254095616953 Found probable prime factor of 37 digits: 3689731686791057511182493254095616953 Probable prime cofactor 462450544960333618227622623208059861475009706651116695284225242264765051852021894512774787633648215514948788708010551378932007874192315440467 has 141 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:03:37 UTC 2009 年 3 月 7 日 (土) 8 時 3 分 37 秒 (日本時間)
14×10188-419
c168
name 名前 apo November 4, 2012 02:36:56 UTC 2012 年 11 月 4 日 (日) 11 時 36 分 56 秒 (日本時間) `102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527<168>` ```131456931873872480857924922991230434748690395524856259<54> 779946789311843779118706893173113999653152352396773863425234903444320946522719838551521514375849686893027305254253<114>``` ```-> makeJobFile(): Adjusted to q0=4950000, q1=5130000. -> client 1 q0: 4950000 [10/31 19:40:35] GGNFS-0.77.1-VC8(UTE) : makefb [10/31 19:40:45] name: [10/31 19:40:45] n=102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527 (168 digits) [10/31 19:40:45] c0: -41 [10/31 19:40:45] c1: 0 [10/31 19:40:45] c2: 0 [10/31 19:40:45] c3: 0 [10/31 19:40:45] c4: 0 [10/31 19:40:45] c5: 14000 [10/31 19:40:45] RFBsize: 658445 (upto 9899999) [10/31 19:40:45] AFBsize: 658173 (upto 9899999) [10/31 19:40:45] maxNumLargeRatPrimes: 3 [10/31 19:40:45] maxLargeRatPrime: 268435456 [10/31 19:40:45] maxNumLargeAlgPrimes: 3 [10/31 19:40:45] maxLargeAlgPrime: 268435456 LatSieveTime: 4393 LatSieveTime: 4411 LatSieveTime: 4412 LatSieveTime: 4423 LatSieveTime: 4427 LatSieveTime: 4471 -> makeJobFile(): Adjusted to q0=5130001, q1=5310000. -> client 1 q0: 5130001 [10/31 20:29:27] GGNFS-0.77.1-VC8(UTE) : procrels [10/31 20:30:00] There were 1938/452123 duplicates. [10/31 20:30:00] RelProcTime: 32.2 [10/31 20:30:01] largePrimes: 1108754 , relations: 450185 [10/31 20:30:33] GGNFS-0.77.1-VC8(UTE) : matbuild [10/31 20:30:34] largePrimes: 1108754 , relations: 450185 [10/31 20:30:34] Heap stats for matbuild run. [10/31 20:30:34] Max heap usage: 0 MB [10/31 20:30:34] malloc/realloc errors: 0 [10/31 20:30:34] total malloc's : 52 [10/31 20:30:34] total realloc's: 20 [10/31 20:30:34] rels:450185, initialFF:0, finalFF:7995 LatSieveTime: 3491 LatSieveTime: 3505 LatSieveTime: 3517 LatSieveTime: 3519 LatSieveTime: 3529 LatSieveTime: 3553 -> makeJobFile(): Adjusted to q0=5310001, q1=5490000. -> client 1 q0: 5310001 LatSieveTime: 4388 LatSieveTime: 4394 LatSieveTime: 4409 LatSieveTime: 4433 LatSieveTime: 4436 LatSieveTime: 4483 -> makeJobFile(): Adjusted to q0=5490001, q1=5670000. -> client 1 q0: 5490001 LatSieveTime: 3094 LatSieveTime: 3137 LatSieveTime: 3153 LatSieveTime: 3156 LatSieveTime: 3159 LatSieveTime: 3162 -> makeJobFile(): Adjusted to q0=5670001, q1=5850000. -> client 1 q0: 5670001 LatSieveTime: 3475 LatSieveTime: 3531 LatSieveTime: 3556 LatSieveTime: 3579 LatSieveTime: 3592 LatSieveTime: 3618 -> makeJobFile(): Adjusted to q0=5850001, q1=6030000. -> client 1 q0: 5850001 [11/01 00:46:28] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 00:48:45] There were 43873/1802364 duplicates. [11/01 00:48:45] RelProcTime: 134.1 [11/01 00:48:47] largePrimes: 4724201 , relations: 2208675 [11/01 01:02:07] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 01:02:13] largePrimes: 4724201 , relations: 2208675 [11/01 01:02:14] Heap stats for matbuild run. [11/01 01:02:14] Max heap usage: 0 MB [11/01 01:02:14] malloc/realloc errors: 0 [11/01 01:02:14] total malloc's : 71 [11/01 01:02:14] total realloc's: 44 [11/01 01:02:14] rels:2208675, initialFF:0, finalFF:39248 LatSieveTime: 4122 LatSieveTime: 4133 LatSieveTime: 4134 LatSieveTime: 4208 LatSieveTime: 4209 LatSieveTime: 4214 -> makeJobFile(): Adjusted to q0=6030001, q1=6210000. -> client 1 q0: 6030001 [11/01 01:23:56] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 01:24:34] There were 19316/450948 duplicates. [11/01 01:24:34] RelProcTime: 33.5 [11/01 01:24:39] largePrimes: 5473973 , relations: 2640307 [11/01 01:24:42] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 01:24:52] largePrimes: 5473973 , relations: 2640307 [11/01 01:24:53] Heap stats for matbuild run. [11/01 01:24:53] Max heap usage: 0 MB [11/01 01:24:53] malloc/realloc errors: 0 [11/01 01:24:53] total malloc's : 89 [11/01 01:24:53] total realloc's: 47 [11/01 01:24:53] rels:2640307, initialFF:0, finalFF:47137 LatSieveTime: 3123 LatSieveTime: 3138 LatSieveTime: 3176 LatSieveTime: 3192 LatSieveTime: 3202 LatSieveTime: 3215 -> makeJobFile(): Adjusted to q0=6210001, q1=6390000. -> client 1 q0: 6210001 LatSieveTime: 4297 LatSieveTime: 4333 LatSieveTime: 4383 LatSieveTime: 4404 LatSieveTime: 4435 LatSieveTime: 4440 -> makeJobFile(): Adjusted to q0=6390001, q1=6570000. -> client 1 q0: 6390001 LatSieveTime: 3162 LatSieveTime: 3207 LatSieveTime: 3212 LatSieveTime: 3219 LatSieveTime: 3223 LatSieveTime: 3232 -> makeJobFile(): Adjusted to q0=6570001, q1=6750000. -> client 1 q0: 6570001 LatSieveTime: 3138 LatSieveTime: 3176 LatSieveTime: 3226 LatSieveTime: 3255 LatSieveTime: 3256 LatSieveTime: 3289 -> makeJobFile(): Adjusted to q0=6750001, q1=6930000. -> client 1 q0: 6750001 LatSieveTime: 3192 LatSieveTime: 3221 LatSieveTime: 3234 LatSieveTime: 3237 LatSieveTime: 3254 LatSieveTime: 3311 -> makeJobFile(): Adjusted to q0=6930001, q1=7110000. -> client 1 q0: 6930001 LatSieveTime: 3219 LatSieveTime: 3225 LatSieveTime: 3246 LatSieveTime: 3255 LatSieveTime: 3285 LatSieveTime: 3330 -> makeJobFile(): Adjusted to q0=7110001, q1=7290000. -> client 1 q0: 7110001 LatSieveTime: 3143 LatSieveTime: 3183 LatSieveTime: 3226 LatSieveTime: 3238 LatSieveTime: 3277 LatSieveTime: 3307 -> makeJobFile(): Adjusted to q0=7290001, q1=7470000. -> client 1 q0: 7290001 LatSieveTime: 3209 LatSieveTime: 3221 LatSieveTime: 3249 LatSieveTime: 3253 LatSieveTime: 3258 LatSieveTime: 3303 -> makeJobFile(): Adjusted to q0=7470001, q1=7650000. -> client 1 q0: 7470001 LatSieveTime: 3199 LatSieveTime: 3231 LatSieveTime: 3266 LatSieveTime: 3270 LatSieveTime: 3298 LatSieveTime: 3347 -> makeJobFile(): Adjusted to q0=7650001, q1=7830000. -> client 1 q0: 7650001 LatSieveTime: 3166 LatSieveTime: 3265 LatSieveTime: 3274 LatSieveTime: 3286 LatSieveTime: 3288 LatSieveTime: 3300 -> makeJobFile(): Adjusted to q0=7830001, q1=8010000. -> client 1 q0: 7830001 LatSieveTime: 3213 LatSieveTime: 3230 LatSieveTime: 3270 LatSieveTime: 3326 LatSieveTime: 3345 LatSieveTime: 3360 -> makeJobFile(): Adjusted to q0=8010001, q1=8190000. -> client 1 q0: 8010001 LatSieveTime: 3222 LatSieveTime: 3223 LatSieveTime: 3279 LatSieveTime: 3299 LatSieveTime: 3326 LatSieveTime: 3352 -> makeJobFile(): Adjusted to q0=8190001, q1=8370000. -> client 1 q0: 8190001 LatSieveTime: 3256 LatSieveTime: 3296 LatSieveTime: 3299 LatSieveTime: 3303 LatSieveTime: 3305 LatSieveTime: 3326 -> makeJobFile(): Adjusted to q0=8370001, q1=8550000. -> client 1 q0: 8370001 [11/01 13:39:38] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 13:47:18] There were 467957/5721346 duplicates. [11/01 13:47:18] RelProcTime: 453.7 [11/01 13:47:37] largePrimes: 12047020 , relations: 7893696 [11/01 13:47:48] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 13:48:23] largePrimes: 12047020 , relations: 7893696 [11/01 13:48:28] Heap stats for matbuild run. [11/01 13:48:28] Max heap usage: 0 MB [11/01 13:48:28] malloc/realloc errors: 0 [11/01 13:48:28] total malloc's : 152 [11/01 13:48:28] total realloc's: 120 [11/01 13:48:28] rels:7893696, initialFF:0, finalFF:149361 LatSieveTime: 3406 LatSieveTime: 3435 LatSieveTime: 3442 LatSieveTime: 3449 LatSieveTime: 3459 LatSieveTime: 3497 -> makeJobFile(): Adjusted to q0=8550001, q1=8730000. -> client 1 q0: 8550001 [11/01 14:44:33] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 14:45:54] There were 49639/430835 duplicates. [11/01 14:45:54] RelProcTime: 49.6 [11/01 14:46:18] largePrimes: 12400076 , relations: 8274892 [11/01 14:46:32] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 14:47:12] largePrimes: 12400076 , relations: 8274892 [11/01 14:47:17] Heap stats for matbuild run. [11/01 14:47:17] Max heap usage: 0 MB [11/01 14:47:17] malloc/realloc errors: 0 [11/01 14:47:17] total malloc's : 176 [11/01 14:47:17] total realloc's: 119 [11/01 14:47:17] rels:8274892, initialFF:0, finalFF:157923 LatSieveTime: 3365 LatSieveTime: 3391 LatSieveTime: 3393 LatSieveTime: 3393 LatSieveTime: 3397 LatSieveTime: 3432 -> makeJobFile(): Adjusted to q0=8730001, q1=8910000. -> client 1 q0: 8730001 LatSieveTime: 3287 LatSieveTime: 3311 LatSieveTime: 3315 LatSieveTime: 3324 LatSieveTime: 3327 LatSieveTime: 3370 -> makeJobFile(): Adjusted to q0=8910001, q1=9090000. -> client 1 q0: 8910001 LatSieveTime: 3130 LatSieveTime: 3253 LatSieveTime: 3260 LatSieveTime: 3325 LatSieveTime: 3335 LatSieveTime: 3346 -> makeJobFile(): Adjusted to q0=9090001, q1=9270000. -> client 1 q0: 9090001 LatSieveTime: 3289 LatSieveTime: 3303 LatSieveTime: 3316 LatSieveTime: 3331 LatSieveTime: 3335 LatSieveTime: 3343 -> makeJobFile(): Adjusted to q0=9270001, q1=9450000. -> client 1 q0: 9270001 LatSieveTime: 3219 LatSieveTime: 3269 LatSieveTime: 3272 LatSieveTime: 3276 LatSieveTime: 3279 LatSieveTime: 3296 -> makeJobFile(): Adjusted to q0=9450001, q1=9630000. -> client 1 q0: 9450001 [11/01 19:52:14] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 19:56:11] There were 270653/2142596 duplicates. [11/01 19:56:11] RelProcTime: 190.0 [11/01 19:56:49] largePrimes: 13973175 , relations: 10146833 [11/01 19:59:14] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 20:00:10] largePrimes: 13973175 , relations: 10146833 [11/01 20:00:17] Heap stats for matbuild run. [11/01 20:00:17] Max heap usage: 0 MB [11/01 20:00:17] malloc/realloc errors: 0 [11/01 20:00:17] total malloc's : 236 [11/01 20:00:17] total realloc's: 148 [11/01 20:00:17] rels:10146833, initialFF:0, finalFF:202985 LatSieveTime: 3888 LatSieveTime: 3906 LatSieveTime: 3917 LatSieveTime: 3924 LatSieveTime: 3929 LatSieveTime: 4031 -> makeJobFile(): Adjusted to q0=9630001, q1=9810000. -> client 1 q0: 9630001 [11/01 20:57:03] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 20:58:42] There were 58626/425571 duplicates. [11/01 20:58:42] RelProcTime: 41.8 [11/01 20:59:27] largePrimes: 14254065 , relations: 10513778 [11/01 20:59:40] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 21:00:45] largePrimes: 14254065 , relations: 10513778 [11/01 21:00:52] Heap stats for matbuild run. [11/01 21:00:52] Max heap usage: 0 MB [11/01 21:00:52] malloc/realloc errors: 0 [11/01 21:00:52] total malloc's : 264 [11/01 21:00:52] total realloc's: 155 [11/01 21:00:52] rels:10513778, initialFF:0, finalFF:212531 LatSieveTime: 3472 LatSieveTime: 3479 LatSieveTime: 3502 LatSieveTime: 3551 LatSieveTime: 3565 LatSieveTime: 3568 -> makeJobFile(): Adjusted to q0=9810001, q1=9990000. -> client 1 q0: 9810001 -> makeJobFile(): Adjusted to q0=9810001, q1=9850000. -> client 1 q0: 9810001 [11/01 21:57:41] GGNFS-0.77.1-VC8(UTE) : procrels [11/01 21:58:54] There were 59750/425558 duplicates. [11/01 21:58:54] RelProcTime: 72.0 [11/01 21:59:43] largePrimes: 14526637 , relations: 10879586 [11/01 22:00:11] GGNFS-0.77.1-VC8(UTE) : matbuild [11/01 22:01:22] largePrimes: 14526637 , relations: 10879586 [11/01 22:01:30] Heap stats for matbuild run. [11/01 22:01:30] Max heap usage: 0 MB [11/01 22:01:30] malloc/realloc errors: 0 [11/01 22:01:30] total malloc's : 288 [11/01 22:01:30] total realloc's: 165 [11/01 22:01:30] rels:10879586, initialFF:0, finalFF:222614 LatSieveTime: 683 LatSieveTime: 687 LatSieveTime: 693 LatSieveTime: 713 LatSieveTime: 726 LatSieveTime: 742 LatSieveTime: 752 LatSieveTime: 776 -> makeJobFile(): Adjusted to q0=9850001, q1=10030000. -> client 1 q0: 9850001 LatSieveTime: 3200 LatSieveTime: 3202 LatSieveTime: 3208 LatSieveTime: 3231 LatSieveTime: 3231 LatSieveTime: 3248 LatSieveTime: 3256 LatSieveTime: 3321 -> makeJobFile(): Adjusted to q0=10030001, q1=10210000. -> client 1 q0: 10030001 LatSieveTime: 3193 LatSieveTime: 3234 LatSieveTime: 3263 LatSieveTime: 3263 LatSieveTime: 3290 LatSieveTime: 3292 LatSieveTime: 3308 LatSieveTime: 3310 -> makeJobFile(): Adjusted to q0=10210001, q1=10390000. -> client 1 q0: 10210001 [11/02 00:22:57] GGNFS-0.77.1-VC8(UTE) : makefb [11/02 00:23:08] name: [11/02 00:23:08] n=102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527 (168 digits) [11/02 00:23:08] c0: -41 [11/02 00:23:08] c1: 0 [11/02 00:23:08] c2: 0 [11/02 00:23:08] c3: 0 [11/02 00:23:08] c4: 0 [11/02 00:23:08] c5: 14000 [11/02 00:23:08] RFBsize: 818703 (upto 12499999) [11/02 00:23:08] AFBsize: 658173 (upto 9899999) [11/02 00:23:08] maxNumLargeRatPrimes: 3 [11/02 00:23:08] maxLargeRatPrime: 268435456 [11/02 00:23:08] maxNumLargeAlgPrimes: 3 [11/02 00:23:08] maxLargeAlgPrime: 268435456 [11/02 00:23:52] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 00:25:42] There were 561212/1460896 duplicates. [11/02 00:25:42] RelProcTime: 106.6 [11/02 00:26:35] largePrimes: 15251644 , relations: 11779270 [11/02 00:27:16] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 00:28:07] There were 1938/452123 duplicates. [11/02 00:28:07] RelProcTime: 45.4 [11/02 00:28:08] largePrimes: 1075108 , relations: 450185 [11/02 00:29:20] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 00:32:26] There were 43873/1802364 duplicates. [11/02 00:32:26] RelProcTime: 171.7 [11/02 00:32:30] largePrimes: 4613089 , relations: 2208675 [11/02 00:34:40] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 00:35:27] There were 19316/450948 duplicates. [11/02 00:35:27] RelProcTime: 35.7 [11/02 00:35:32] largePrimes: 5352573 , relations: 2640307 [11/02 00:38:06] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 00:47:40] There were 467957/5721346 duplicates. [11/02 00:47:40] RelProcTime: 533.3 [11/02 00:48:02] largePrimes: 11888709 , relations: 7893696 LatSieveTime: 3206 LatSieveTime: 3212 LatSieveTime: 3254 LatSieveTime: 3267 LatSieveTime: 3287 LatSieveTime: 3387 LatSieveTime: 3403 [11/02 00:56:58] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 3420 -> makeJobFile(): Adjusted to q0=10390001, q1=10570000. -> client 1 q0: 10390001 [11/02 00:58:24] There were 49639/430835 duplicates. [11/02 00:58:24] RelProcTime: 37.2 [11/02 00:58:48] largePrimes: 12241394 , relations: 8274892 [11/02 01:04:43] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 01:08:42] There were 270653/2142596 duplicates. [11/02 01:08:42] RelProcTime: 194.0 [11/02 01:09:22] largePrimes: 13813464 , relations: 10146833 [11/02 01:21:43] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 01:23:37] There were 58626/425571 duplicates. [11/02 01:23:37] RelProcTime: 46.8 [11/02 01:24:24] largePrimes: 14094253 , relations: 10513778 [11/02 01:48:42] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 01:49:36] There were 59750/425558 duplicates. [11/02 01:49:36] RelProcTime: 52.5 [11/02 01:50:27] largePrimes: 14366727 , relations: 10879586 LatSieveTime: 3209 LatSieveTime: 3210 LatSieveTime: 3224 LatSieveTime: 3228 LatSieveTime: 3233 LatSieveTime: 3244 LatSieveTime: 3315 LatSieveTime: 3346 -> makeJobFile(): Adjusted to q0=10570001, q1=10750000. -> client 1 q0: 10570001 [11/02 01:55:14] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 01:57:02] There were 561212/1460896 duplicates. [11/02 01:57:02] RelProcTime: 104.9 [11/02 01:57:55] largePrimes: 15091734 , relations: 11779270 [11/02 02:03:04] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 02:04:28] largePrimes: 15091734 , relations: 11779270 [11/02 02:04:40] Heap stats for matbuild run. [11/02 02:04:40] Max heap usage: 0 MB [11/02 02:04:40] malloc/realloc errors: 0 [11/02 02:04:40] total malloc's : 324 [11/02 02:04:40] total realloc's: 182 [11/02 02:04:40] rels:11779270, initialFF:0, finalFF:317397 LatSieveTime: 3180 LatSieveTime: 3205 LatSieveTime: 3210 LatSieveTime: 3213 LatSieveTime: 3258 LatSieveTime: 3276 LatSieveTime: 3289 LatSieveTime: 3327 -> makeJobFile(): Adjusted to q0=10750001, q1=10930000. -> client 1 q0: 10750001 [11/02 03:10:02] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 03:14:33] There were 195977/1400843 duplicates. [11/02 03:14:33] RelProcTime: 173.2 [11/02 03:15:49] largePrimes: 15988759 , relations: 12984136 [11/02 03:15:55] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 03:17:38] largePrimes: 15988759 , relations: 12984136 [11/02 03:17:57] Heap stats for matbuild run. [11/02 03:17:57] Max heap usage: 0 MB [11/02 03:17:57] malloc/realloc errors: 0 [11/02 03:17:57] total malloc's : 405 [11/02 03:17:57] total realloc's: 217 [11/02 03:17:57] rels:12984136, initialFF:0, finalFF:369413 LatSieveTime: 3168 LatSieveTime: 3225 LatSieveTime: 3239 LatSieveTime: 3245 LatSieveTime: 3254 LatSieveTime: 3261 LatSieveTime: 3263 LatSieveTime: 3396 -> makeJobFile(): Adjusted to q0=10930001, q1=11110000. -> client 1 q0: 10930001 LatSieveTime: 3099 LatSieveTime: 3153 LatSieveTime: 3175 LatSieveTime: 3175 LatSieveTime: 3176 LatSieveTime: 3183 LatSieveTime: 3203 LatSieveTime: 3211 -> makeJobFile(): Adjusted to q0=11110001, q1=11290000. -> client 1 q0: 11110001 LatSieveTime: 3122 LatSieveTime: 3124 LatSieveTime: 3162 LatSieveTime: 3172 LatSieveTime: 3176 LatSieveTime: 3184 LatSieveTime: 3187 LatSieveTime: 3270 -> makeJobFile(): Adjusted to q0=11290001, q1=11470000. -> client 1 q0: 11290001 LatSieveTime: 3134 LatSieveTime: 3152 LatSieveTime: 3166 LatSieveTime: 3189 LatSieveTime: 3189 LatSieveTime: 3193 LatSieveTime: 3200 LatSieveTime: 3229 -> makeJobFile(): Adjusted to q0=11470001, q1=11650000. -> client 1 q0: 11470001 LatSieveTime: 3101 LatSieveTime: 3178 LatSieveTime: 3181 LatSieveTime: 3191 LatSieveTime: 3213 LatSieveTime: 3223 LatSieveTime: 3234 LatSieveTime: 3334 -> makeJobFile(): Adjusted to q0=11650001, q1=11830000. -> client 1 q0: 11650001 LatSieveTime: 3163 LatSieveTime: 3187 LatSieveTime: 3206 LatSieveTime: 3206 LatSieveTime: 3206 LatSieveTime: 3209 LatSieveTime: 3223 LatSieveTime: 3226 -> makeJobFile(): Adjusted to q0=11830001, q1=12010000. -> client 1 q0: 11830001 LatSieveTime: 3146 LatSieveTime: 3161 LatSieveTime: 3196 LatSieveTime: 3220 LatSieveTime: 3235 LatSieveTime: 3242 LatSieveTime: 3245 LatSieveTime: 3320 -> makeJobFile(): Adjusted to q0=12010001, q1=12190000. -> client 1 q0: 12010001 LatSieveTime: 3125 LatSieveTime: 3184 LatSieveTime: 3207 LatSieveTime: 3215 LatSieveTime: 3215 LatSieveTime: 3218 LatSieveTime: 3237 LatSieveTime: 3261 -> makeJobFile(): Adjusted to q0=12190001, q1=12370000. -> client 1 q0: 12190001 LatSieveTime: 3139 LatSieveTime: 3160 LatSieveTime: 3233 LatSieveTime: 3234 LatSieveTime: 3240 LatSieveTime: 3271 LatSieveTime: 3288 LatSieveTime: 3331 -> makeJobFile(): Adjusted to q0=12370001, q1=12550000. -> client 1 q0: 12370001 LatSieveTime: 3126 LatSieveTime: 3169 LatSieveTime: 3187 LatSieveTime: 3214 LatSieveTime: 3221 LatSieveTime: 3232 LatSieveTime: 3256 LatSieveTime: 3269 -> makeJobFile(): Adjusted to q0=12550001, q1=12730000. -> client 1 q0: 12550001 [11/02 12:25:44] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 12:34:13] There were 716052/4574574 duplicates. [11/02 12:34:13] RelProcTime: 502.9 [11/02 12:35:42] largePrimes: 18402143 , relations: 16842658 [11/02 12:35:54] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 12:37:53] largePrimes: 18402143 , relations: 16842658 [11/02 12:39:40] Heap stats for matbuild run. [11/02 12:39:40] Max heap usage: 0 MB [11/02 12:39:40] malloc/realloc errors: 0 [11/02 12:39:40] total malloc's : 1318 [11/02 12:39:40] total realloc's: 698 [11/02 12:39:40] rels:16842658, initialFF:0, finalFF:730327 [11/02 12:46:06] GGNFS-0.77.1-VC8(UTE) : makefb [11/02 12:46:19] name: [11/02 12:46:19] n=102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527 (168 digits) [11/02 12:46:19] c0: -41 [11/02 12:46:19] c1: 0 [11/02 12:46:19] c2: 0 [11/02 12:46:19] c3: 0 [11/02 12:46:19] c4: 0 [11/02 12:46:19] c5: 14000 [11/02 12:46:19] RFBsize: 876616 (upto 13449979) [11/02 12:46:19] AFBsize: 658173 (upto 9899999) [11/02 12:46:19] maxNumLargeRatPrimes: 3 [11/02 12:46:19] maxLargeRatPrime: 268435456 [11/02 12:46:19] maxNumLargeAlgPrimes: 3 [11/02 12:46:19] maxLargeAlgPrime: 268435456 [11/02 12:47:02] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 12:47:48] There were 1938/452123 duplicates. [11/02 12:47:48] RelProcTime: 40.1 [11/02 12:47:49] largePrimes: 1064318 , relations: 450185 [11/02 12:47:57] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 12:50:53] There were 43873/1802364 duplicates. [11/02 12:50:53] RelProcTime: 161.6 [11/02 12:50:58] largePrimes: 4575829 , relations: 2208675 LatSieveTime: 3232 [11/02 12:51:30] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 3265 LatSieveTime: 3267 [11/02 12:52:18] There were 19316/450948 duplicates. [11/02 12:52:18] RelProcTime: 32.7 [11/02 12:52:23] largePrimes: 5311377 , relations: 2640307 LatSieveTime: 3293 LatSieveTime: 3301 LatSieveTime: 3317 LatSieveTime: 3521 LatSieveTime: 3592 -> makeJobFile(): Adjusted to q0=12730001, q1=12910000. -> client 1 q0: 12730001 [11/02 13:00:16] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:08:52] There were 467957/5721346 duplicates. [11/02 13:08:52] RelProcTime: 473.5 [11/02 13:09:09] largePrimes: 11831980 , relations: 7893696 [11/02 13:15:54] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:17:18] There were 49639/430835 duplicates. [11/02 13:17:18] RelProcTime: 42.1 [11/02 13:17:44] largePrimes: 12184448 , relations: 8274892 [11/02 13:17:50] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:23:08] There were 270653/2142596 duplicates. [11/02 13:23:08] RelProcTime: 243.8 [11/02 13:23:53] largePrimes: 13755893 , relations: 10146833 [11/02 13:28:46] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:31:31] There were 58626/425571 duplicates. [11/02 13:31:31] RelProcTime: 72.6 [11/02 13:32:25] largePrimes: 14036609 , relations: 10513778 [11/02 13:32:32] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:33:20] There were 59750/425558 duplicates. [11/02 13:33:20] RelProcTime: 47.2 [11/02 13:34:07] largePrimes: 14309047 , relations: 10879586 [11/02 13:34:17] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:36:12] There were 561212/1460896 duplicates. [11/02 13:36:12] RelProcTime: 104.0 [11/02 13:37:02] largePrimes: 15033982 , relations: 11779270 [11/02 13:37:18] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 13:41:03] There were 195977/1400843 duplicates. [11/02 13:41:03] RelProcTime: 143.4 [11/02 13:42:08] largePrimes: 15930935 , relations: 12984136 [11/02 13:42:35] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 3231 [11/02 13:51:36] There were 716052/4574574 duplicates. [11/02 13:51:36] RelProcTime: 534.4 LatSieveTime: 3299 LatSieveTime: 3301 [11/02 13:52:56] largePrimes: 18344241 , relations: 16842658 LatSieveTime: 3332 [11/02 13:53:07] GGNFS-0.77.1-VC8(UTE) : matbuild LatSieveTime: 3356 LatSieveTime: 3356 [11/02 13:54:35] largePrimes: 18344241 , relations: 16842658 [11/02 13:55:31] Heap stats for matbuild run. [11/02 13:55:31] Max heap usage: 0 MB [11/02 13:55:31] malloc/realloc errors: 0 [11/02 13:55:31] total malloc's : 1150 [11/02 13:55:31] total realloc's: 617 [11/02 13:55:31] rels:16842658, initialFF:0, finalFF:794609 LatSieveTime: 3719 LatSieveTime: 3830 -> makeJobFile(): Adjusted to q0=12910001, q1=13090000. -> client 1 q0: 12910001 [11/02 14:02:26] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 14:08:26] There were 154241/902253 duplicates. [11/02 14:08:26] RelProcTime: 217.9 [11/02 14:11:05] largePrimes: 18743185 , relations: 17590670 [11/02 14:11:12] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 14:13:45] largePrimes: 18743185 , relations: 17590670 [11/02 14:15:41] Heap stats for matbuild run. [11/02 14:15:41] Max heap usage: 0 MB [11/02 14:15:41] malloc/realloc errors: 0 [11/02 14:15:41] total malloc's : 1147 [11/02 14:15:41] total realloc's: 604 [11/02 14:15:41] rels:17590670, initialFF:0, finalFF:943832 LatSieveTime: 3236 LatSieveTime: 3276 LatSieveTime: 3295 LatSieveTime: 3317 LatSieveTime: 3344 LatSieveTime: 3353 LatSieveTime: 3367 LatSieveTime: 3381 -> makeJobFile(): Adjusted to q0=13090001, q1=13270000. -> client 1 q0: 13090001 LatSieveTime: 3233 LatSieveTime: 3240 LatSieveTime: 3269 LatSieveTime: 3275 LatSieveTime: 3283 LatSieveTime: 3292 LatSieveTime: 3300 LatSieveTime: 3310 -> makeJobFile(): Adjusted to q0=13270001, q1=13450000. -> client 1 q0: 13270001 LatSieveTime: 3203 LatSieveTime: 3207 LatSieveTime: 3209 LatSieveTime: 3213 LatSieveTime: 3228 LatSieveTime: 3229 LatSieveTime: 3250 LatSieveTime: 3258 -> makeJobFile(): Adjusted to q0=13450001, q1=13630000. -> client 1 q0: 13450001 LatSieveTime: 3170 LatSieveTime: 3174 LatSieveTime: 3181 LatSieveTime: 3199 LatSieveTime: 3221 LatSieveTime: 3226 LatSieveTime: 3228 LatSieveTime: 3280 -> makeJobFile(): Adjusted to q0=13630001, q1=13810000. -> client 1 q0: 13630001 [11/02 17:48:19] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 17:56:41] There were 317344/1771732 duplicates. [11/02 17:56:41] RelProcTime: 234.4 [11/02 17:58:34] largePrimes: 19466088 , relations: 19045058 [11/02 17:58:48] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 18:01:22] largePrimes: 19466088 , relations: 19045058 [11/02 18:04:55] Heap stats for matbuild run. [11/02 18:04:55] Max heap usage: 0 MB [11/02 18:04:55] malloc/realloc errors: 0 [11/02 18:04:55] total malloc's : 1117 [11/02 18:04:55] total realloc's: 577 [11/02 18:04:55] rels:19045058, initialFF:0, finalFF:1304133 LatSieveTime: 3359 LatSieveTime: 3416 LatSieveTime: 3418 LatSieveTime: 3419 LatSieveTime: 3463 LatSieveTime: 3502 LatSieveTime: 3635 LatSieveTime: 3658 -> makeJobFile(): Adjusted to q0=13810001, q1=13990000. -> client 1 q0: 13810001 [11/02 18:45:22] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 18:46:53] There were 81623/440305 duplicates. [11/02 18:46:53] RelProcTime: 89.5 [11/02 18:48:49] largePrimes: 19634014 , relations: 19403740 [11/02 18:49:34] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 18:52:16] largePrimes: 19634014 , relations: 19403740 [11/02 18:55:58] Heap stats for matbuild run. [11/02 18:55:58] Max heap usage: 0 MB [11/02 18:55:58] malloc/realloc errors: 0 [11/02 18:55:58] total malloc's : 1190 [11/02 18:55:58] total realloc's: 614 [11/02 18:55:58] rels:19403740, initialFF:0, finalFF:1398147 LatSieveTime: 3205 LatSieveTime: 3219 LatSieveTime: 3265 LatSieveTime: 3274 LatSieveTime: 3297 LatSieveTime: 3317 LatSieveTime: 3419 LatSieveTime: 3541 -> makeJobFile(): Adjusted to q0=13990001, q1=14170000. -> client 1 q0: 13990001 [11/02 19:43:04] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 19:43:58] There were 81713/435756 duplicates. [11/02 19:43:58] RelProcTime: 53.6 [11/02 19:45:55] largePrimes: 19795613 , relations: 19757781 [11/02 19:50:54] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 19:53:33] largePrimes: 19795613 , relations: 19757781 [11/02 19:57:33] Heap stats for matbuild run. [11/02 19:57:33] Max heap usage: 0 MB [11/02 19:57:33] malloc/realloc errors: 0 [11/02 19:57:33] total malloc's : 1188 [11/02 19:57:33] total realloc's: 616 [11/02 19:57:33] rels:19757781, initialFF:0, finalFF:1496854 LatSieveTime: 3197 LatSieveTime: 3311 LatSieveTime: 3313 LatSieveTime: 3320 LatSieveTime: 3328 LatSieveTime: 3339 LatSieveTime: 3449 LatSieveTime: 3517 -> makeJobFile(): Adjusted to q0=14170001, q1=14350000. -> client 1 q0: 14170001 [11/02 20:41:37] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 20:42:50] There were 82486/434614 duplicates. [11/02 20:42:50] RelProcTime: 71.5 [11/02 20:44:57] largePrimes: 19952377 , relations: 20109909 [11/02 20:45:04] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 20:47:44] largePrimes: 19952377 , relations: 20109909 [11/02 20:56:09] reduceRelSets dropped relation-set weight from 28715321 to 23045276. [11/02 20:58:58] reduceRelSets dropped relation-set weight from 23045276 to 23002957. [11/02 20:58:58] After removing heavy rel-sets, weight is 17538593. [11/02 20:58:59] Heap stats for matbuild run. [11/02 20:58:59] Max heap usage: 0 MB [11/02 20:58:59] malloc/realloc errors: 0 [11/02 20:58:59] total malloc's : 1220 [11/02 20:58:59] total realloc's: 631 [11/02 20:58:59] rels:20109909, initialFF:0, finalFF:1488875 [11/02 21:00:02] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 21:00:32] There were 25685/133917 duplicates. [11/02 21:00:32] RelProcTime: 30.1 [11/02 21:02:35] largePrimes: 19999773 , relations: 20218133 [11/02 21:02:41] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 21:05:28] largePrimes: 19999773 , relations: 20218133 [11/02 21:15:14] reduceRelSets dropped relation-set weight from 29580674 to 23493815. [11/02 21:18:15] reduceRelSets dropped relation-set weight from 23493815 to 23448522. [11/02 21:18:15] After removing heavy rel-sets, weight is 18052014. [11/02 21:18:15] Heap stats for matbuild run. [11/02 21:18:15] Max heap usage: 0 MB [11/02 21:18:15] malloc/realloc errors: 0 [11/02 21:18:15] total malloc's : 1288 [11/02 21:18:15] total realloc's: 660 [11/02 21:18:15] rels:20218133, initialFF:0, finalFF:1517994 [11/02 21:20:13] GGNFS-0.77.1-VC8(UTE) : procrels [11/02 21:25:07] There were 161279/276118 duplicates. [11/02 21:25:07] RelProcTime: 35.3 [11/02 21:27:23] largePrimes: 20049711 , relations: 20332965 [11/02 21:27:44] GGNFS-0.77.1-VC8(UTE) : matbuild [11/02 21:30:43] largePrimes: 20049711 , relations: 20332965 LatSieveTime: 3266 LatSieveTime: 3319 LatSieveTime: 3341 LatSieveTime: 3343 LatSieveTime: 3365 LatSieveTime: 3406 [11/02 21:39:34] reduceRelSets dropped relation-set weight from 30643854 to 24128173. [11/02 21:41:39] reduceRelSets dropped relation-set weight from 24128173 to 24079804. [11/02 21:41:39] After removing heavy rel-sets, weight is 18652516. [11/02 21:45:26] Heap stats for matbuild run. [11/02 21:45:26] Max heap usage: 0 MB [11/02 21:45:26] malloc/realloc errors: 0 [11/02 21:45:26] total malloc's : 1370 [11/02 21:45:26] total realloc's: 645 [11/02 21:45:26] rels:20332965, initialFF:0, finalFF:1550875 [11/02 21:45:26] depinf file written. Run matprune. LatSieveTime: 3949 LatSieveTime: 3980 -> makeJobFile(): Adjusted to q0=14350001, q1=14530000. -> client 1 q0: 14350001 [11/02 21:48:05] GGNFS-0.77.1-VC8(UTE) : matprune [11/02 21:48:17] Pruning matrix with wt=0.600 [11/02 21:48:17] Initial matrix is 1534856 x 1550875 with sparse part having weight 191125734. [11/02 21:48:17] (total weight is 296992299) [11/02 21:52:58] Matrix pruned to 1529548 x 1537286 with weight 187652461. [11/02 21:53:09] Matrix is pruned. Run matsolve. [11/02 21:53:09] Heap stats for matprune run: [11/02 21:53:09] Max heap usage: 0 MB [11/02 21:53:09] malloc/realloc errors: 0 [11/02 21:53:09] total malloc's : 14 [11/02 21:53:09] total realloc's: 0 [11/02 21:53:47] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=1611432705) [11/03 21:17:37] BLanczosTime: 84227.6 [11/03 21:17:38] Heap stats for matsolve run: [11/03 21:17:38] Max heap usage: 0 MB [11/03 21:17:38] malloc/realloc errors: 0 [11/03 21:17:38] total malloc's : 12 [11/03 21:17:38] total realloc's: 0 [11/04 01:00:04] GGNFS-0.77.1-VC8(UTE) : sqrt [11/04 01:00:06] bmultiplier=200 [11/04 01:21:10] From dependence 0, sqrt obtained: [11/04 01:21:10] r1=1 [11/04 01:21:10] r2=102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527 [11/04 01:21:10] sqrtTime: 1266.1 [11/04 01:21:51] GGNFS-0.77.1-VC8(UTE) : sqrt [11/04 01:21:52] bmultiplier=200 [11/04 01:39:35] From dependence 1, sqrt obtained: [11/04 01:39:35] r1=102529411947812620867501298653740028164848653681797955505966360422920206878151314364154626371607815725282005047296362581058360535593403888179804061389898620533273419527 [11/04 01:39:35] r2=1 [11/04 01:39:35] sqrtTime: 1064.4 [11/04 02:10:57] GGNFS-0.77.1-VC8(UTE) : sqrt [11/04 02:10:59] bmultiplier=200 [11/04 02:34:37] From dependence 2, sqrt obtained: [11/04 02:34:37] r1=131456931873872480857924922991230434748690395524856259 (pp54) [11/04 02:34:37] r2=779946789311843779118706893173113999653152352396773863425234903444320946522719838551521514375849686893027305254253 (pp114) [11/04 02:34:37] (pp=probable prime, c=composite) [11/04 02:34:37] sqrtTime: 1420.3```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:04:05 UTC 2009 年 3 月 7 日 (土) 8 時 4 分 5 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 08:33:08 UTC 2010 年 9 月 26 日 (日) 17 時 33 分 8 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 08:33:23 UTC 2010 年 9 月 26 日 (日) 17 時 33 分 23 秒 (日本時間)
403e62144110Ignacio SantosSeptember 26, 2010 08:33:23 UTC 2010 年 9 月 26 日 (日) 17 時 33 分 23 秒 (日本時間)
2034apoOctober 11, 2012 22:51:30 UTC 2012 年 10 月 12 日 (金) 7 時 51 分 30 秒 (日本時間)
4511e6399232Ignacio SantosSeptember 26, 2010 08:33:23 UTC 2010 年 9 月 26 日 (日) 17 時 33 分 23 秒 (日本時間)
3960apoOctober 24, 2012 16:49:54 UTC 2012 年 10 月 25 日 (木) 1 時 49 分 54 秒 (日本時間)
14×10189-419
c181
name 名前 Dmitry Domanov June 19, 2010 10:00:06 UTC 2010 年 6 月 19 日 (土) 19 時 0 分 6 秒 (日本時間) `8540802527712489900569642428864096513098833923437816893323429735952006102222829082845255599111757269186689139982748664796056436703136680433946194907239878040683018397920251626586233<181>` `239520538881060197905744372643280941<36>` `35657912960665285623816019575561369109492179973770108769108354949812805491773330613114678787974465052096451753581411492556939864057675922912374013<146>` `Factor=239520538881060197905744372643280941 Method=ECM B1=11000000 Sigma=324657124`
c146
name 名前 apo October 24, 2012 20:12:10 UTC 2012 年 10 月 25 日 (木) 5 時 12 分 10 秒 (日本時間) `35657912960665285623816019575561369109492179973770108769108354949812805491773330613114678787974465052096451753581411492556939864057675922912374013<146>` ```2002403664239885757046178465165412553631040634691159646468491732802599<70> 17807554789009568483347101192272315973838652529738302758061791566306292912187<77>``` ```-> makeJobFile(): Adjusted to q0=5300000, q1=5480000. -> client 1 q0: 5300000 -> makeJobFile(): Adjusted to q0=5300000, q1=5480000. -> client 1 q0: 5300000 LatSieveTime: 2592 LatSieveTime: 2594 LatSieveTime: 2605 LatSieveTime: 2610 LatSieveTime: 2621 LatSieveTime: 2636 LatSieveTime: 2641 LatSieveTime: 2645 -> makeJobFile(): Adjusted to q0=5480001, q1=5660000. -> client 1 q0: 5480001 LatSieveTime: 2607 LatSieveTime: 2637 LatSieveTime: 2641 LatSieveTime: 2655 LatSieveTime: 2660 LatSieveTime: 2675 LatSieveTime: 2679 LatSieveTime: 2693 -> makeJobFile(): Adjusted to q0=5660001, q1=5840000. -> client 1 q0: 5660001 LatSieveTime: 2650 LatSieveTime: 2658 LatSieveTime: 2661 LatSieveTime: 2668 LatSieveTime: 2669 LatSieveTime: 2676 LatSieveTime: 2699 LatSieveTime: 2739 -> makeJobFile(): Adjusted to q0=5840001, q1=6020000. -> client 1 q0: 5840001 LatSieveTime: 2619 LatSieveTime: 2628 LatSieveTime: 2658 LatSieveTime: 2670 LatSieveTime: 2672 LatSieveTime: 2687 LatSieveTime: 2695 LatSieveTime: 2703 -> makeJobFile(): Adjusted to q0=6020001, q1=6200000. -> client 1 q0: 6020001 LatSieveTime: 2722 LatSieveTime: 2726 LatSieveTime: 2732 LatSieveTime: 2733 LatSieveTime: 2735 LatSieveTime: 2739 LatSieveTime: 2753 LatSieveTime: 2795 -> makeJobFile(): Adjusted to q0=6200001, q1=6380000. -> client 1 q0: 6200001 LatSieveTime: 2647 LatSieveTime: 2702 LatSieveTime: 2703 LatSieveTime: 2713 LatSieveTime: 2741 LatSieveTime: 2743 LatSieveTime: 2750 LatSieveTime: 2760 -> makeJobFile(): Adjusted to q0=6380001, q1=6560000. -> client 1 q0: 6380001 LatSieveTime: 2682 LatSieveTime: 2716 LatSieveTime: 2724 LatSieveTime: 2730 LatSieveTime: 2731 LatSieveTime: 2739 LatSieveTime: 2741 LatSieveTime: 2745 -> makeJobFile(): Adjusted to q0=6560001, q1=6740000. -> client 1 q0: 6560001 LatSieveTime: 2698 LatSieveTime: 2708 LatSieveTime: 2711 LatSieveTime: 2715 LatSieveTime: 2734 LatSieveTime: 2735 LatSieveTime: 2748 LatSieveTime: 2775 -> makeJobFile(): Adjusted to q0=6740001, q1=6920000. -> client 1 q0: 6740001 LatSieveTime: 2817 LatSieveTime: 2830 LatSieveTime: 2836 LatSieveTime: 2844 LatSieveTime: 2846 LatSieveTime: 2852 LatSieveTime: 2874 LatSieveTime: 2895 -> makeJobFile(): Adjusted to q0=6920001, q1=7100000. -> client 1 q0: 6920001 LatSieveTime: 2762 LatSieveTime: 2781 LatSieveTime: 2813 LatSieveTime: 2816 LatSieveTime: 2820 LatSieveTime: 2835 LatSieveTime: 2851 LatSieveTime: 2856 -> makeJobFile(): Adjusted to q0=7100001, q1=7280000. -> client 1 q0: 7100001 LatSieveTime: 2826 LatSieveTime: 2844 LatSieveTime: 2874 LatSieveTime: 2881 LatSieveTime: 2884 LatSieveTime: 2888 LatSieveTime: 2892 LatSieveTime: 2896 -> makeJobFile(): Adjusted to q0=7280001, q1=7460000. -> client 1 q0: 7280001 LatSieveTime: 3337 LatSieveTime: 3396 LatSieveTime: 3409 LatSieveTime: 3412 LatSieveTime: 3421 LatSieveTime: 3437 LatSieveTime: 3436 LatSieveTime: 3493 -> makeJobFile(): Adjusted to q0=7460001, q1=7640000. -> client 1 q0: 7460001 LatSieveTime: 2935 LatSieveTime: 2959 LatSieveTime: 2963 LatSieveTime: 2972 LatSieveTime: 2981 LatSieveTime: 2980 LatSieveTime: 3010 LatSieveTime: 3042 -> makeJobFile(): Adjusted to q0=7640001, q1=7820000. -> client 1 q0: 7640001 LatSieveTime: 2876 LatSieveTime: 2902 LatSieveTime: 2920 LatSieveTime: 2943 LatSieveTime: 2954 LatSieveTime: 2953 LatSieveTime: 2958 LatSieveTime: 2993 -> makeJobFile(): Adjusted to q0=7820001, q1=8000000. -> client 1 q0: 7820001 LatSieveTime: 3031 LatSieveTime: 3036 LatSieveTime: 3067 LatSieveTime: 3072 LatSieveTime: 3073 LatSieveTime: 3090 LatSieveTime: 3111 LatSieveTime: 3144 -> makeJobFile(): Adjusted to q0=8000001, q1=8180000. -> client 1 q0: 8000001 LatSieveTime: 3138 LatSieveTime: 3147 LatSieveTime: 3169 LatSieveTime: 3173 LatSieveTime: 3187 LatSieveTime: 3192 LatSieveTime: 3203 LatSieveTime: 3225 -> makeJobFile(): Adjusted to q0=8180001, q1=8360000. -> client 1 q0: 8180001 LatSieveTime: 3100 LatSieveTime: 3105 LatSieveTime: 3129 LatSieveTime: 3132 LatSieveTime: 3134 LatSieveTime: 3139 LatSieveTime: 3141 LatSieveTime: 3164 -> makeJobFile(): Adjusted to q0=8360001, q1=8540000. -> client 1 q0: 8360001 LatSieveTime: 3455 LatSieveTime: 3526 LatSieveTime: 3537 LatSieveTime: 3537 LatSieveTime: 3544 LatSieveTime: 3554 LatSieveTime: 3566 LatSieveTime: 3621 -> makeJobFile(): Adjusted to q0=8540001, q1=8720000. -> client 1 q0: 8540001 LatSieveTime: 3390 LatSieveTime: 3404 LatSieveTime: 3404 LatSieveTime: 3408 LatSieveTime: 3410 LatSieveTime: 3422 LatSieveTime: 3437 LatSieveTime: 3450 -> makeJobFile(): Adjusted to q0=8720001, q1=8900000. -> client 1 q0: 8720001 [10/21 00:47:32] GGNFS-0.77.1-VC8(UTE) : makefb [10/21 00:47:52] name: [10/21 00:47:52] n=35657912960665285623816019575561369109492179973770108769108354949812805491773330613114678787974465052096451753581411492556939864057675922912374013 (146 digits) [10/21 00:47:52] c0: -205 [10/21 00:47:52] c1: 0 [10/21 00:47:52] c2: 0 [10/21 00:47:52] c3: 0 [10/21 00:47:52] c4: 0 [10/21 00:47:52] c5: 7 [10/21 00:47:52] RFBsize: 701795 (upto 10599977) [10/21 00:47:52] AFBsize: 1007238 (upto 15599959) [10/21 00:47:52] maxNumLargeRatPrimes: 3 [10/21 00:47:52] maxLargeRatPrime: 1073741824 [10/21 00:47:52] maxNumLargeAlgPrimes: 3 [10/21 00:47:52] maxLargeAlgPrime: 1073741824 [10/21 00:50:42] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:02:26] There were 309650/11176096 duplicates. [10/21 01:02:26] RelProcTime: 692.4 [10/21 01:02:54] largePrimes: 20331304 , relations: 10866445 [10/21 01:05:14] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:07:34] There were 11176095/11176096 duplicates. [10/21 01:07:34] RelProcTime: 83.0 [10/21 01:08:10] largePrimes: 20331304 , relations: 10866445 [10/21 01:08:13] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:10:13] There were 11176095/11176096 duplicates. [10/21 01:10:13] RelProcTime: 87.4 [10/21 01:10:54] largePrimes: 20331304 , relations: 10866445 [10/21 01:11:27] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 01:12:31] largePrimes: 20331304 , relations: 10866445 [10/21 01:12:37] Heap stats for matbuild run. [10/21 01:12:37] Max heap usage: 0 MB [10/21 01:12:37] malloc/realloc errors: 0 [10/21 01:12:37] total malloc's : 128 [10/21 01:12:37] total realloc's: 131 [10/21 01:12:37] rels:10866445, initialFF:0, finalFF:97359 [10/21 01:13:52] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:16:00] There were 11176095/11176096 duplicates. [10/21 01:16:00] RelProcTime: 88.5 [10/21 01:16:38] largePrimes: 20331304 , relations: 10866445 [10/21 01:17:30] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:19:44] There were 11176095/11176096 duplicates. [10/21 01:19:44] RelProcTime: 90.0 [10/21 01:20:31] largePrimes: 20331304 , relations: 10866445 [10/21 01:21:40] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:41:24] There were 789089/11151454 duplicates. [10/21 01:41:24] RelProcTime: 1169.5 [10/21 01:43:39] largePrimes: 32193353 , relations: 21228809 [10/21 01:43:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 01:47:31] There were 11151453/11151454 duplicates. [10/21 01:47:31] RelProcTime: 109.8 [10/21 01:50:02] largePrimes: 32193353 , relations: 21228809 [10/21 10:22:35] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 10:26:49] largePrimes: 32193353 , relations: 21228809 [10/21 10:27:06] Heap stats for matbuild run. [10/21 10:27:06] Max heap usage: 0 MB [10/21 10:27:06] malloc/realloc errors: 0 [10/21 10:27:06] total malloc's : 252 [10/21 10:27:06] total realloc's: 239 [10/21 10:27:06] rels:21228809, initialFF:0, finalFF:202859 [10/21 10:28:05] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 10:32:59] There were 11151453/11151454 duplicates. [10/21 10:32:59] RelProcTime: 111.8 [10/21 10:35:31] largePrimes: 32193353 , relations: 21228809 -> makeJobFile(): Adjusted to q0=8720001, q1=8900000. -> client 1 q0: 8720001 LatSieveTime: 2684 LatSieveTime: 2706 LatSieveTime: 2729 LatSieveTime: 2740 LatSieveTime: 2749 LatSieveTime: 2751 LatSieveTime: 2752 LatSieveTime: 2811 -> makeJobFile(): Adjusted to q0=8900001, q1=9080000. -> client 1 q0: 8900001 [10/21 11:58:39] GGNFS-0.77.1-VC8(UTE) : makefb [10/21 11:58:57] name: [10/21 11:58:57] n=35657912960665285623816019575561369109492179973770108769108354949812805491773330613114678787974465052096451753581411492556939864057675922912374013 (146 digits) [10/21 11:58:57] c0: -205 [10/21 11:58:57] c1: 0 [10/21 11:58:57] c2: 0 [10/21 11:58:57] c3: 0 [10/21 11:58:57] c4: 0 [10/21 11:58:57] c5: 7 [10/21 11:58:57] RFBsize: 701795 (upto 10599977) [10/21 11:58:57] AFBsize: 1007238 (upto 15599959) [10/21 11:58:57] maxNumLargeRatPrimes: 3 [10/21 11:58:57] maxLargeRatPrime: 1073741824 [10/21 11:58:57] maxNumLargeAlgPrimes: 3 [10/21 11:58:57] maxLargeAlgPrime: 1073741824 [10/21 11:59:32] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 12:18:34] There were 309650/11176096 duplicates. [10/21 12:18:34] RelProcTime: 1102.9 [10/21 12:19:44] largePrimes: 20331304 , relations: 10866445 [10/21 12:20:27] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 12:23:35] There were 11176095/11176096 duplicates. [10/21 12:23:35] RelProcTime: 79.1 LatSieveTime: 3603 [10/21 12:24:08] largePrimes: 20331304 , relations: 10866445 LatSieveTime: 3610 [10/21 12:24:11] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 3617 LatSieveTime: 3628 LatSieveTime: 3657 LatSieveTime: 3661 [10/21 12:25:59] There were 11176095/11176096 duplicates. [10/21 12:25:59] RelProcTime: 48.2 LatSieveTime: 3726 [10/21 12:26:23] largePrimes: 20331304 , relations: 10866445 LatSieveTime: 3784 -> makeJobFile(): Adjusted to q0=9080001, q1=9260000. -> client 1 q0: 9080001 [10/21 12:30:29] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 12:36:52] There were 113081/2312143 duplicates. [10/21 12:36:52] RelProcTime: 333.9 [10/21 12:38:12] largePrimes: 23223834 , relations: 13065507 [10/21 12:39:28] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 12:40:51] There were 2312143/2312143 duplicates. [10/21 12:40:51] RelProcTime: 25.6 [10/21 12:42:14] largePrimes: 23223834 , relations: 13065507 [10/21 12:51:14] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 12:52:57] There were 2312143/2312143 duplicates. [10/21 12:52:57] RelProcTime: 28.1 [10/21 12:54:30] largePrimes: 23223834 , relations: 13065507 [10/21 12:54:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 13:18:31] There were 886550/11151454 duplicates. [10/21 13:18:31] RelProcTime: 1373.1 [10/21 13:20:58] largePrimes: 34187884 , relations: 23330410 [10/21 13:23:55] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 13:27:06] There were 2312143/2312143 duplicates. [10/21 13:27:06] RelProcTime: 32.6 [10/21 13:30:13] largePrimes: 34187884 , relations: 23330410 LatSieveTime: 3820 LatSieveTime: 3861 LatSieveTime: 3945 LatSieveTime: 3961 LatSieveTime: 3974 LatSieveTime: 3981 LatSieveTime: 3999 [10/21 13:34:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 13:37:37] There were 2312143/2312143 duplicates. [10/21 13:37:37] RelProcTime: 26.9 LatSieveTime: 4237 -> makeJobFile(): Adjusted to q0=9260001, q1=9440000. -> client 1 q0: 9260001 [10/21 13:40:07] largePrimes: 34187884 , relations: 23330410 [10/21 13:43:15] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 13:47:30] There were 2312143/2312143 duplicates. [10/21 13:47:30] RelProcTime: 35.1 [10/21 13:51:12] largePrimes: 34187884 , relations: 23330410 [10/21 13:54:01] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 13:58:22] largePrimes: 34187884 , relations: 23330410 [10/21 13:58:44] Heap stats for matbuild run. [10/21 13:58:44] Max heap usage: 0 MB [10/21 13:58:44] malloc/realloc errors: 0 [10/21 13:58:44] total malloc's : 358 [10/21 13:58:44] total realloc's: 268 [10/21 13:58:44] rels:23330410, initialFF:0, finalFF:227391 [10/21 14:00:34] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 14:03:32] There were 111265/1163621 duplicates. [10/21 14:03:32] RelProcTime: 176.7 [10/21 14:07:29] largePrimes: 35147120 , relations: 24382766 [10/21 14:07:36] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 14:12:18] largePrimes: 35147120 , relations: 24382766 [10/21 14:12:44] Heap stats for matbuild run. [10/21 14:12:44] Max heap usage: 0 MB [10/21 14:12:44] malloc/realloc errors: 0 [10/21 14:12:44] total malloc's : 360 [10/21 14:12:44] total realloc's: 277 [10/21 14:12:44] rels:24382766, initialFF:0, finalFF:240458 LatSieveTime: 3742 LatSieveTime: 3749 LatSieveTime: 3761 LatSieveTime: 3774 LatSieveTime: 3775 LatSieveTime: 3799 LatSieveTime: 3827 LatSieveTime: 3840 -> makeJobFile(): Adjusted to q0=9440001, q1=9620000. -> client 1 q0: 9440001 [10/21 14:45:41] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 14:48:40] There were 114247/1150097 duplicates. [10/21 14:48:40] RelProcTime: 175.6 [10/21 14:54:15] largePrimes: 36069322 , relations: 25418616 [10/21 14:59:30] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 15:04:13] largePrimes: 36069322 , relations: 25418616 [10/21 15:04:37] Heap stats for matbuild run. [10/21 15:04:37] Max heap usage: 0 MB [10/21 15:04:37] malloc/realloc errors: 0 [10/21 15:04:37] total malloc's : 366 [10/21 15:04:37] total realloc's: 291 [10/21 15:04:37] rels:25418616, initialFF:0, finalFF:253746 LatSieveTime: 3662 LatSieveTime: 3698 LatSieveTime: 3709 LatSieveTime: 3733 LatSieveTime: 3757 LatSieveTime: 3774 LatSieveTime: 3804 LatSieveTime: 3836 -> makeJobFile(): Adjusted to q0=9620001, q1=9800000. -> client 1 q0: 9620001 [10/21 15:57:27] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 16:05:18] There were 117415/1155140 duplicates. [10/21 16:05:18] RelProcTime: 195.3 [10/21 16:10:09] largePrimes: 36970695 , relations: 26456341 [10/21 16:40:57] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 16:46:31] largePrimes: 36970695 , relations: 26456341 [10/21 16:46:56] Heap stats for matbuild run. [10/21 16:46:56] Max heap usage: 0 MB [10/21 16:46:56] malloc/realloc errors: 0 [10/21 16:46:56] total malloc's : 407 [10/21 16:46:56] total realloc's: 306 [10/21 16:46:56] rels:26456341, initialFF:0, finalFF:267729 LatSieveTime: 3768 LatSieveTime: 3788 LatSieveTime: 3808 LatSieveTime: 3808 LatSieveTime: 3829 LatSieveTime: 3836 LatSieveTime: 3848 LatSieveTime: 3857 -> makeJobFile(): Adjusted to q0=9800001, q1=9980000. -> client 1 q0: 9800001 [10/21 16:58:34] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 17:02:33] There were 119915/1154429 duplicates. [10/21 17:02:33] RelProcTime: 237.3 [10/21 17:07:22] largePrimes: 37847979 , relations: 27490855 [10/21 17:23:37] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 17:31:05] largePrimes: 37847979 , relations: 27490855 [10/21 17:31:39] Heap stats for matbuild run. [10/21 17:31:39] Max heap usage: 0 MB [10/21 17:31:39] malloc/realloc errors: 0 [10/21 17:31:39] total malloc's : 418 [10/21 17:31:39] total realloc's: 316 [10/21 17:31:39] rels:27490855, initialFF:0, finalFF:282266 LatSieveTime: 3745 LatSieveTime: 3762 LatSieveTime: 3775 LatSieveTime: 3779 LatSieveTime: 3797 LatSieveTime: 3811 LatSieveTime: 3813 LatSieveTime: 3982 -> makeJobFile(): Adjusted to q0=9980001, q1=10160000. -> client 1 q0: 9980001 LatSieveTime: 3675 LatSieveTime: 3737 LatSieveTime: 3739 LatSieveTime: 3747 LatSieveTime: 3761 LatSieveTime: 3768 LatSieveTime: 3768 LatSieveTime: 3775 -> makeJobFile(): Adjusted to q0=10160001, q1=10340000. -> client 1 q0: 10160001 [10/21 19:02:13] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 19:18:31] There were 249997/2290268 duplicates. [10/21 19:18:31] RelProcTime: 417.1 [10/21 19:24:07] largePrimes: 39518992 , relations: 29531126 [10/21 19:44:52] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 19:54:23] largePrimes: 39518992 , relations: 29531126 [10/21 19:54:56] Heap stats for matbuild run. [10/21 19:54:56] Max heap usage: 0 MB [10/21 19:54:56] malloc/realloc errors: 0 [10/21 19:54:56] total malloc's : 455 [10/21 19:54:56] total realloc's: 339 [10/21 19:54:56] rels:29531126, initialFF:0, finalFF:313114 LatSieveTime: 3778 LatSieveTime: 3838 LatSieveTime: 3839 LatSieveTime: 3854 LatSieveTime: 3859 LatSieveTime: 3861 LatSieveTime: 3883 LatSieveTime: 3905 -> makeJobFile(): Adjusted to q0=10340001, q1=10520000. -> client 1 q0: 10340001 [10/21 20:13:38] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 20:17:54] There were 128562/1140585 duplicates. [10/21 20:17:54] RelProcTime: 254.7 [10/21 20:21:51] largePrimes: 40322848 , relations: 30543149 [10/21 20:25:16] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 20:33:04] largePrimes: 40322848 , relations: 30543149 [10/21 20:33:32] Heap stats for matbuild run. [10/21 20:33:32] Max heap usage: 0 MB [10/21 20:33:32] malloc/realloc errors: 0 [10/21 20:33:32] total malloc's : 480 [10/21 20:33:32] total realloc's: 359 [10/21 20:33:32] rels:30543149, initialFF:0, finalFF:329856 LatSieveTime: 3183 LatSieveTime: 3203 LatSieveTime: 3221 LatSieveTime: 3240 LatSieveTime: 3243 LatSieveTime: 3248 LatSieveTime: 3354 LatSieveTime: 3399 -> makeJobFile(): Adjusted to q0=10520001, q1=10700000. -> client 1 q0: 10520001 [10/21 21:04:10] GGNFS-0.77.1-VC8(UTE) : makefb [10/21 21:04:27] name: [10/21 21:04:27] n=35657912960665285623816019575561369109492179973770108769108354949812805491773330613114678787974465052096451753581411492556939864057675922912374013 (146 digits) [10/21 21:04:27] c0: -205 [10/21 21:04:27] c1: 0 [10/21 21:04:27] c2: 0 [10/21 21:04:27] c3: 0 [10/21 21:04:27] c4: 0 [10/21 21:04:27] c5: 7 [10/21 21:04:27] RFBsize: 1006966 (upto 15599981) [10/21 21:04:27] AFBsize: 702306 (upto 10599977) [10/21 21:04:27] maxNumLargeRatPrimes: 3 [10/21 21:04:27] maxLargeRatPrime: 1073741824 [10/21 21:04:27] maxNumLargeAlgPrimes: 3 [10/21 21:04:27] maxLargeAlgPrime: 1073741824 [10/21 21:04:39] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 21:07:48] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 21:09:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 21:23:22] There were 309650/11176096 duplicates. [10/21 21:23:22] RelProcTime: 794.6 [10/21 21:23:56] largePrimes: 20307907 , relations: 10848549 [10/21 21:24:00] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 21:27:14] There were 11158199/11176096 duplicates. [10/21 21:27:14] RelProcTime: 80.0 [10/21 21:27:48] largePrimes: 20307907 , relations: 10848549 LatSieveTime: 3605 LatSieveTime: 3612 LatSieveTime: 3631 LatSieveTime: 3634 LatSieveTime: 3664 LatSieveTime: 3734 LatSieveTime: 3907 LatSieveTime: 3940 -> makeJobFile(): Adjusted to q0=10700001, q1=10880000. -> client 1 q0: 10700001 [10/21 22:11:23] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 22:12:42] largePrimes: 20307907 , relations: 10848549 [10/21 22:12:50] Heap stats for matbuild run. [10/21 22:12:50] Max heap usage: 0 MB [10/21 22:12:50] malloc/realloc errors: 0 [10/21 22:12:50] total malloc's : 105 [10/21 22:12:50] total realloc's: 133 [10/21 22:12:50] rels:10848549, initialFF:0, finalFF:99604 [10/21 22:13:37] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:18:40] There were 112925/2312143 duplicates. [10/21 22:18:40] RelProcTime: 238.5 [10/21 22:19:34] largePrimes: 23194935 , relations: 13043557 [10/21 22:21:47] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 22:23:30] largePrimes: 23194935 , relations: 13043557 [10/21 22:23:39] Heap stats for matbuild run. [10/21 22:23:39] Max heap usage: 0 MB [10/21 22:23:39] malloc/realloc errors: 0 [10/21 22:23:39] total malloc's : 152 [10/21 22:23:39] total realloc's: 158 [10/21 22:23:39] rels:13043557, initialFF:0, finalFF:119789 [10/21 22:24:40] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:28:02] There were 63273/1163621 duplicates. [10/21 22:28:02] RelProcTime: 117.5 [10/21 22:29:14] largePrimes: 24552029 , relations: 14141672 [10/21 22:29:25] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:33:30] There were 67005/1150097 duplicates. [10/21 22:33:30] RelProcTime: 146.5 [10/21 22:35:10] largePrimes: 25838344 , relations: 15222565 [10/21 22:35:19] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:41:37] There were 70883/1155140 duplicates. [10/21 22:41:37] RelProcTime: 247.4 [10/21 22:45:13] largePrimes: 27079768 , relations: 16304651 [10/21 22:45:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:50:16] There were 74468/1154429 duplicates. [10/21 22:50:16] RelProcTime: 137.5 [10/21 22:52:25] largePrimes: 28274088 , relations: 17382468 [10/21 22:52:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 22:58:00] There were 160127/2290268 duplicates. [10/21 22:58:00] RelProcTime: 296.0 [10/21 23:00:25] largePrimes: 30518554 , relations: 19508194 [10/21 23:00:39] GGNFS-0.77.1-VC8(UTE) : procrels [10/21 23:06:29] There were 84850/1140585 duplicates. [10/21 23:06:29] RelProcTime: 156.1 [10/21 23:09:52] largePrimes: 31582436 , relations: 20561740 [10/21 23:10:05] GGNFS-0.77.1-VC8(UTE) : matbuild LatSieveTime: 3824 [10/21 23:13:58] largePrimes: 31582436 , relations: 20561740 [10/21 23:14:14] Heap stats for matbuild run. [10/21 23:14:14] Max heap usage: 0 MB [10/21 23:14:14] malloc/realloc errors: 0 [10/21 23:14:14] total malloc's : 301 [10/21 23:14:14] total realloc's: 236 [10/21 23:14:14] rels:20561740, initialFF:0, finalFF:195464 LatSieveTime: 3896 LatSieveTime: 3979 [10/21 23:16:23] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 4017 LatSieveTime: 4056 LatSieveTime: 4079 LatSieveTime: 4108 LatSieveTime: 4306 -> makeJobFile(): Adjusted to q0=10880001, q1=11060000. -> client 1 q0: 10880001 [10/21 23:39:58] There were 1205160/11151454 duplicates. [10/21 23:39:58] RelProcTime: 1399.7 [10/21 23:45:26] largePrimes: 40276895 , relations: 30488512 [10/21 23:48:14] GGNFS-0.77.1-VC8(UTE) : matbuild [10/21 23:56:16] largePrimes: 40276895 , relations: 30488512 [10/21 23:56:52] Heap stats for matbuild run. [10/21 23:56:52] Max heap usage: 0 MB [10/21 23:56:52] malloc/realloc errors: 0 [10/21 23:56:52] total malloc's : 415 [10/21 23:56:52] total realloc's: 360 [10/21 23:56:52] rels:30488512, initialFF:0, finalFF:334561 [10/21 23:59:00] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 00:23:06] There were 403158/3425958 duplicates. [10/22 00:23:06] RelProcTime: 932.8 LatSieveTime: 3827 LatSieveTime: 3897 LatSieveTime: 3936 LatSieveTime: 3956 LatSieveTime: 3962 LatSieveTime: 4004 LatSieveTime: 4015 [10/22 00:29:42] largePrimes: 42575900 , relations: 33504717 [10/22 00:30:13] GGNFS-0.77.1-VC8(UTE) : matbuild LatSieveTime: 4389 -> makeJobFile(): Adjusted to q0=11060001, q1=11240000. -> client 1 q0: 11060001 [10/22 00:37:01] largePrimes: 42575900 , relations: 33504717 [10/22 00:37:30] Heap stats for matbuild run. [10/22 00:37:30] Max heap usage: 0 MB [10/22 00:37:30] malloc/realloc errors: 0 [10/22 00:37:30] total malloc's : 518 [10/22 00:37:30] total realloc's: 418 [10/22 00:37:30] rels:33504717, initialFF:0, finalFF:391359 [10/22 00:38:48] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 00:49:16] There were 3558192/4556201 duplicates. [10/22 00:49:16] RelProcTime: 227.0 [10/22 00:55:53] largePrimes: 43298744 , relations: 34493943 [10/22 00:56:25] GGNFS-0.77.1-VC8(UTE) : matbuild [10/22 01:04:30] largePrimes: 43298744 , relations: 34493943 -> makeJobFile(): Adjusted to q0=11060001, q1=11220000. -> client 1 q0: 11060001 [10/22 01:05:06] Heap stats for matbuild run. [10/22 01:05:06] Max heap usage: 0 MB [10/22 01:05:06] malloc/realloc errors: 0 [10/22 01:05:06] total malloc's : 641 [10/22 01:05:06] total realloc's: 473 [10/22 01:05:06] rels:34493943, initialFF:0, finalFF:413639 LatSieveTime: 4216 LatSieveTime: 4303 LatSieveTime: 4316 LatSieveTime: 4346 LatSieveTime: 4348 LatSieveTime: 4383 LatSieveTime: 4384 LatSieveTime: 4432 -> makeJobFile(): Adjusted to q0=11220001, q1=11400000. -> client 1 q0: 11220001 LatSieveTime: 4798 LatSieveTime: 4813 LatSieveTime: 4815 LatSieveTime: 4833 LatSieveTime: 4847 LatSieveTime: 4850 LatSieveTime: 4858 LatSieveTime: 4880 -> makeJobFile(): Adjusted to q0=11400001, q1=11580000. -> client 1 q0: 11400001 LatSieveTime: 4975 LatSieveTime: 5007 LatSieveTime: 5015 LatSieveTime: 5032 LatSieveTime: 5042 LatSieveTime: 5051 LatSieveTime: 5071 LatSieveTime: 5104 -> makeJobFile(): Adjusted to q0=11580001, q1=11760000. -> client 1 q0: 11580001 LatSieveTime: 4848 LatSieveTime: 4971 LatSieveTime: 4973 LatSieveTime: 4984 LatSieveTime: 4988 LatSieveTime: 5008 LatSieveTime: 5029 LatSieveTime: 5146 -> makeJobFile(): Adjusted to q0=11760001, q1=11940000. -> client 1 q0: 11760001 LatSieveTime: 4940 LatSieveTime: 4965 LatSieveTime: 4976 LatSieveTime: 4994 LatSieveTime: 5004 LatSieveTime: 5024 LatSieveTime: 5030 LatSieveTime: 5138 -> makeJobFile(): Adjusted to q0=11940001, q1=12120000. -> client 1 q0: 11940001 [10/22 08:03:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 08:33:29] There were 680123/6851644 duplicates. [10/22 08:33:29] RelProcTime: 1179.0 [10/22 08:42:54] largePrimes: 48907200 , relations: 40665464 LatSieveTime: 5313 LatSieveTime: 5407 LatSieveTime: 5425 LatSieveTime: 5454 LatSieveTime: 5462 LatSieveTime: 5513 LatSieveTime: 5633 LatSieveTime: 5696 -> makeJobFile(): Adjusted to q0=12120001, q1=12300000. -> client 1 q0: 12120001 LatSieveTime: 5342 LatSieveTime: 5373 LatSieveTime: 5379 LatSieveTime: 5435 LatSieveTime: 5443 LatSieveTime: 5460 LatSieveTime: 5475 LatSieveTime: 5494 -> makeJobFile(): Adjusted to q0=12300001, q1=12480000. -> client 1 q0: 12300001 LatSieveTime: 5147 LatSieveTime: 5269 LatSieveTime: 5281 LatSieveTime: 5320 LatSieveTime: 5340 LatSieveTime: 5379 LatSieveTime: 5382 LatSieveTime: 5448 -> makeJobFile(): Adjusted to q0=12480001, q1=12660000. -> client 1 q0: 12480001 LatSieveTime: 5149 LatSieveTime: 5191 LatSieveTime: 5217 LatSieveTime: 5220 LatSieveTime: 5259 LatSieveTime: 5265 LatSieveTime: 5282 LatSieveTime: 5335 -> makeJobFile(): Adjusted to q0=12660001, q1=12840000. -> client 1 q0: 12660001 LatSieveTime: 5284 LatSieveTime: 5296 LatSieveTime: 5316 LatSieveTime: 5330 LatSieveTime: 5345 LatSieveTime: 5353 LatSieveTime: 5356 LatSieveTime: 5367 -> makeJobFile(): Adjusted to q0=12840001, q1=13020000. -> client 1 q0: 12840001 LatSieveTime: 5719 LatSieveTime: 5748 LatSieveTime: 5749 LatSieveTime: 5749 LatSieveTime: 5830 LatSieveTime: 5849 LatSieveTime: 6095 LatSieveTime: 6299 -> makeJobFile(): Adjusted to q0=13020001, q1=13200000. -> client 1 q0: 13020001 [10/22 18:00:43] GGNFS-0.77.1-VC8(UTE) : matbuild [10/22 18:16:17] largePrimes: 48907200 , relations: 40665464 [10/22 18:19:52] Heap stats for matbuild run. [10/22 18:19:52] Max heap usage: 0 MB [10/22 18:19:52] malloc/realloc errors: 0 [10/22 18:19:52] total malloc's : 3218 [10/22 18:19:52] total realloc's: 1800 [10/22 18:19:52] rels:40665464, initialFF:0, finalFF:580506 [10/22 18:23:28] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 5697 LatSieveTime: 5767 LatSieveTime: 5785 LatSieveTime: 5798 LatSieveTime: 5849 LatSieveTime: 5885 LatSieveTime: 5956 LatSieveTime: 6042 -> makeJobFile(): Adjusted to q0=13200001, q1=13380000. -> client 1 q0: 13200001 [10/22 19:11:17] There were 916513/8348367 duplicates. [10/22 19:11:17] RelProcTime: 2034.1 [10/22 19:24:31] largePrimes: 54860882 , relations: 48097318 [10/22 19:25:44] GGNFS-0.77.1-VC8(UTE) : matbuild [10/22 19:46:39] largePrimes: 54860882 , relations: 48097318 [10/22 20:00:06] Heap stats for matbuild run. [10/22 20:00:06] Max heap usage: 0 MB [10/22 20:00:06] malloc/realloc errors: 0 [10/22 20:00:06] total malloc's : 2484 [10/22 20:00:06] total realloc's: 1454 [10/22 20:00:06] rels:48097318, initialFF:0, finalFF:1042402 LatSieveTime: 6185 LatSieveTime: 6438 LatSieveTime: 6572 LatSieveTime: 6631 LatSieveTime: 6653 LatSieveTime: 6674 LatSieveTime: 6680 LatSieveTime: 6992 -> makeJobFile(): Adjusted to q0=13380001, q1=13560000. -> client 1 q0: 13380001 [10/22 21:00:54] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 21:35:07] There were 322368/2758901 duplicates. [10/22 21:35:07] RelProcTime: 905.2 [10/22 21:51:45] largePrimes: 56646106 , relations: 50533851 [10/22 21:52:42] GGNFS-0.77.1-VC8(UTE) : matbuild [10/22 22:21:21] largePrimes: 56646106 , relations: 50533851 LatSieveTime: 5836 LatSieveTime: 5930 LatSieveTime: 6076 LatSieveTime: 6090 LatSieveTime: 6113 [10/22 22:40:32] Heap stats for matbuild run. [10/22 22:40:32] Max heap usage: 0 MB [10/22 22:40:32] malloc/realloc errors: 0 [10/22 22:40:32] total malloc's : 2400 [10/22 22:40:32] total realloc's: 1413 [10/22 22:40:32] rels:50533851, initialFF:0, finalFF:1284228 LatSieveTime: 6190 LatSieveTime: 6289 LatSieveTime: 6965 -> makeJobFile(): Adjusted to q0=13560001, q1=13740000. -> client 1 q0: 13560001 [10/22 22:55:17] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 22:55:24] GGNFS-0.77.1-VC8(UTE) : procrels [10/22 23:15:05] There were 162438/1371033 duplicates. [10/22 23:15:05] RelProcTime: 517.3 -> makeJobFile(): Adjusted to q0=13560001, q1=13740000. -> client 1 q0: 13560001 [10/22 23:27:46] largePrimes: 57504451 , relations: 51742446 [10/22 23:27:55] GGNFS-0.77.1-VC8(UTE) : matbuild [10/22 23:49:29] largePrimes: 57504451 , relations: 51742446 [10/23 00:04:54] Heap stats for matbuild run. [10/23 00:04:54] Max heap usage: 0 MB [10/23 00:04:54] malloc/realloc errors: 0 [10/23 00:04:54] total malloc's : 2460 [10/23 00:04:54] total realloc's: 1439 [10/23 00:04:54] rels:51742446, initialFF:0, finalFF:1420553 LatSieveTime: 5451 LatSieveTime: 5516 LatSieveTime: 5517 LatSieveTime: 5525 LatSieveTime: 5560 LatSieveTime: 5596 LatSieveTime: 5843 LatSieveTime: 5851 -> makeJobFile(): Adjusted to q0=13740001, q1=13920000. -> client 1 q0: 13740001 LatSieveTime: 5114 LatSieveTime: 5167 LatSieveTime: 5172 LatSieveTime: 5207 LatSieveTime: 5218 LatSieveTime: 5227 LatSieveTime: 5228 LatSieveTime: 5241 -> makeJobFile(): Adjusted to q0=13920001, q1=14100000. -> client 1 q0: 13920001 LatSieveTime: 5040 LatSieveTime: 5062 LatSieveTime: 5094 LatSieveTime: 5095 LatSieveTime: 5097 LatSieveTime: 5126 LatSieveTime: 5146 LatSieveTime: 5149 -> makeJobFile(): Adjusted to q0=14100001, q1=14280000. -> client 1 q0: 14100001 LatSieveTime: 5069 LatSieveTime: 5090 LatSieveTime: 5143 LatSieveTime: 5177 LatSieveTime: 5188 LatSieveTime: 5190 LatSieveTime: 5192 LatSieveTime: 5193 -> makeJobFile(): Adjusted to q0=14280001, q1=14460000. -> client 1 q0: 14280001 LatSieveTime: 5085 LatSieveTime: 5188 LatSieveTime: 5192 LatSieveTime: 5199 LatSieveTime: 5216 LatSieveTime: 5247 LatSieveTime: 5268 LatSieveTime: 5319 -> makeJobFile(): Adjusted to q0=14460001, q1=14640000. -> client 1 q0: 14460001 [10/23 07:33:36] GGNFS-0.77.1-VC8(UTE) : procrels [10/23 08:19:48] There were 852493/6882783 duplicates. [10/23 08:19:48] RelProcTime: 1652.8 [10/23 08:34:03] largePrimes: 61531660 , relations: 57772736 [10/23 17:18:44] GGNFS-0.77.1-VC8(UTE) : matbuild [10/23 17:40:52] largePrimes: 61531660 , relations: 57772736 [10/23 18:10:58] reduceRelSets dropped relation-set weight from 55240320 to 40816273. [10/23 18:16:47] reduceRelSets dropped relation-set weight from 40816273 to 40750915. [10/23 18:16:47] After removing heavy rel-sets, weight is 29732497. [10/23 18:33:33] Heap stats for matbuild run. [10/23 18:33:33] Max heap usage: 0 MB [10/23 18:33:33] malloc/realloc errors: 0 [10/23 18:33:33] total malloc's : 3041 [10/23 18:33:33] total realloc's: 1653 [10/23 18:33:33] rels:57772736, initialFF:0, finalFF:2045533 [10/23 18:33:33] depinf file written. Run matprune. [10/23 18:34:45] GGNFS-0.77.1-VC8(UTE) : matprune [10/23 18:34:56] Pruning matrix with wt=0.700 [10/23 18:34:56] Initial matrix is 1709337 x 2045532 with sparse part having weight 301338639. [10/23 18:34:56] (total weight is 448543908) [10/23 19:02:41] Matrix pruned to 1605042 x 1613652 with weight 202308874. [10/23 19:02:51] Matrix is pruned. Run matsolve. [10/23 19:02:51] Heap stats for matprune run: [10/23 19:02:51] Max heap usage: 0 MB [10/23 19:02:51] malloc/realloc errors: 0 [10/23 19:02:51] total malloc's : 14 [10/23 19:02:51] total realloc's: 0 [10/23 19:13:11] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=4046353774) [10/24 20:25:47] BLanczosTime: 90751.1 [10/24 20:25:48] Heap stats for matsolve run: [10/24 20:25:48] Max heap usage: 0 MB [10/24 20:25:48] malloc/realloc errors: 0 [10/24 20:25:48] total malloc's : 12 [10/24 20:25:48] total realloc's: 0 [10/24 20:26:46] GGNFS-0.77.1-VC8(UTE) : sqrt [10/24 20:26:49] bmultiplier=1 [10/24 21:07:47] From dependence 0, sqrt obtained: [10/24 21:07:47] r1=17807554789009568483347101192272315973838652529738302758061791566306292912187 (pp77) [10/24 21:07:47] r2=2002403664239885757046178465165412553631040634691159646468491732802599 (pp70) [10/24 21:07:47] (pp=probable prime, c=composite) [10/24 21:07:47] sqrtTime: 2459.8```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:04:44 UTC 2009 年 3 月 7 日 (土) 8 時 4 分 44 秒 (日本時間)
255e40--
3025e40--
351e6120Serge BatalovJune 18, 2010 07:33:22 UTC 2010 年 6 月 18 日 (金) 16 時 33 分 22 秒 (日本時間)
403e61907Pawel ApostolOctober 8, 2012 07:57:11 UTC 2012 年 10 月 8 日 (月) 16 時 57 分 11 秒 (日本時間)
4511e64053119Dmitry DomanovJune 19, 2010 09:59:47 UTC 2010 年 6 月 19 日 (土) 18 時 59 分 47 秒 (日本時間)
3934apoOctober 19, 2012 18:37:52 UTC 2012 年 10 月 20 日 (土) 3 時 37 分 52 秒 (日本時間)
14×10190-419
c191
name 名前 Sinkiti Sibata August 4, 2007 19:24:16 UTC 2007 年 8 月 5 日 (日) 4 時 24 分 16 秒 (日本時間) `15555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551<191>` ```35979232514979028691658608275491778123813<41> 432348176106296870587027279656162097057836779149603113487443591832639183538536761174932277931864748203116574693057107906194995991781437958621511650227<150>``` ```Number: 15551_190 N=15555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551 ( 191 digits) SNFS difficulty: 191 digits. Divisors found: r1=35979232514979028691658608275491778123813 (pp41) r2=432348176106296870587027279656162097057836779149603113487443591832639183538536761174932277931864748203116574693057107906194995991781437958621511650227 (pp150) Version: GGNFS-0.77.1-20060513-k8 Total time: 1349.01 hours. Scaled time: 2702.07 units (timescale=2.003). Factorization parameters were as follows: name: 15551_190 n: 15555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551 m: 100000000000000000000000000000000000000 c5: 14 c0: -41 skew: 1.24 type: snfs Factor base limits: 7400000/7400000 Large primes per side: 3 Large prime bits: 27/27 Max factor residue bits: 48/48 Sieved algebraic special-q in [3700000, 19800001) Primes: RFBsize:501962, AFBsize:502607, largePrimes:7128971 encountered Relations: rels:7710294, finalFF:1147825 Max relations in full relation-set: 28 Initial matrix: 1004635 x 1147825 with sparse part having weight 130586785. Pruned matrix : 900542 x 905629 with weight 111974876. Total sieving time: 1328.99 hours. Total relation processing time: 0.88 hours. Matrix solve time: 18.70 hours. Time per square root: 0.44 hours. Prototype def-par.txt line would be: snfs,191,5,0,0,0,0,0,0,0,0,7400000,7400000,27,27,48,48,2.6,2.6,100000 total time: 1349.01 hours. --------- CPU info (if available) ----------``` `Core 2 Duo E6300 1.86GHz,Windows Vista and Cygwin45`
14×10191-419
c172
name 名前 apo October 17, 2012 21:31:42 UTC 2012 年 10 月 18 日 (木) 6 時 31 分 42 秒 (日本時間) `9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563<172>` `5354165852117185252135914375321320153508853934084562766957505865511187<70>` `1759089849578231622051771480901303190911856747169695121364422606428678272553353568622080274069992468249<103>` ```-> makeJobFile(): Adjusted to q0=4100000, q1=4200000. -> client 1 q0: 4100000 -> makeJobFile(): Adjusted to q0=4100000, q1=4160000. -> client 1 q0: 4100000 LatSieveTime: 1046 LatSieveTime: 1055 LatSieveTime: 1092 LatSieveTime: 1103 LatSieveTime: 1107 LatSieveTime: 1109 -> makeJobFile(): Adjusted to q0=4160001, q1=4340000. -> client 1 q0: 4160001 LatSieveTime: 3010 LatSieveTime: 3022 LatSieveTime: 3025 LatSieveTime: 3044 LatSieveTime: 3049 LatSieveTime: 3062 -> makeJobFile(): Adjusted to q0=4340001, q1=4520000. -> client 1 q0: 4340001 LatSieveTime: 2985 LatSieveTime: 3007 LatSieveTime: 3024 LatSieveTime: 3046 LatSieveTime: 3069 LatSieveTime: 3074 -> makeJobFile(): Adjusted to q0=4520001, q1=4700000. -> client 1 q0: 4520001 [10/12 22:25:05] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 22:26:26] GGNFS-0.77.1-VC8(UTE) : makefb [10/12 22:26:35] name: [10/12 22:26:35] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/12 22:26:35] c0: -41 [10/12 22:26:35] c1: 0 [10/12 22:26:35] c2: 0 [10/12 22:26:35] c3: 0 [10/12 22:26:35] c4: 0 [10/12 22:26:35] c5: 140 [10/12 22:26:35] RFBsize: 738873 (upto 11199989) [10/12 22:26:35] AFBsize: 551794 (upto 8199979) [10/12 22:26:35] maxNumLargeRatPrimes: 3 [10/12 22:26:35] maxLargeRatPrime: 1073741824 [10/12 22:26:35] maxNumLargeAlgPrimes: 3 [10/12 22:26:35] maxLargeAlgPrime: 268435456 [10/12 22:27:56] GGNFS-0.77.1-VC8(UTE) : makefb [10/12 22:28:02] name: [10/12 22:28:02] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/12 22:28:02] c0: -41 [10/12 22:28:02] c1: 0 [10/12 22:28:02] c2: 0 [10/12 22:28:02] c3: 0 [10/12 22:28:02] c4: 0 [10/12 22:28:02] c5: 140 [10/12 22:28:02] RFBsize: 738873 (upto 11199989) [10/12 22:28:02] AFBsize: 551794 (upto 8199979) [10/12 22:28:02] maxNumLargeRatPrimes: 3 [10/12 22:28:02] maxLargeRatPrime: 1073741824 [10/12 22:28:02] maxNumLargeAlgPrimes: 3 [10/12 22:28:02] maxLargeAlgPrime: 268435456 [10/12 22:28:21] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 22:30:38] There were 12410/1399003 duplicates. [10/12 22:30:38] RelProcTime: 135.0 [10/12 22:30:41] largePrimes: 3512179 , relations: 1386593 [10/12 22:32:20] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 22:32:25] largePrimes: 3512179 , relations: 1386593 [10/12 22:32:26] Heap stats for matbuild run. [10/12 22:32:26] Max heap usage: 0 MB [10/12 22:32:26] malloc/realloc errors: 0 [10/12 22:32:26] total malloc's : 54 [10/12 22:32:26] total realloc's: 30 [10/12 22:32:26] rels:1386593, initialFF:0, finalFF:13093 LatSieveTime: 3022 LatSieveTime: 3039 LatSieveTime: 3041 LatSieveTime: 3075 LatSieveTime: 3090 LatSieveTime: 3091 -> makeJobFile(): Adjusted to q0=4700001, q1=4880000. -> client 1 q0: 4700001 [10/12 23:24:14] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 23:24:53] There were 12431/601308 duplicates. [10/12 23:24:53] RelProcTime: 31.9 [10/12 23:24:59] largePrimes: 4832956 , relations: 1975470 [10/12 23:25:10] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 23:25:18] largePrimes: 4832956 , relations: 1975470 [10/12 23:25:19] Heap stats for matbuild run. [10/12 23:25:19] Max heap usage: 0 MB [10/12 23:25:19] malloc/realloc errors: 0 [10/12 23:25:19] total malloc's : 72 [10/12 23:25:19] total realloc's: 37 [10/12 23:25:19] rels:1975470, initialFF:0, finalFF:18444 LatSieveTime: 3049 LatSieveTime: 3057 LatSieveTime: 3059 LatSieveTime: 3066 LatSieveTime: 3083 LatSieveTime: 3114 -> makeJobFile(): Adjusted to q0=4880001, q1=5060000. -> client 1 q0: 4880001 [10/13 00:29:15] GGNFS-0.77.1-VC8(UTE) : procrels [10/13 00:29:54] There were 16046/599275 duplicates. [10/13 00:29:54] RelProcTime: 38.3 [10/13 00:30:00] largePrimes: 6059475 , relations: 2558699 [10/13 00:30:09] GGNFS-0.77.1-VC8(UTE) : matbuild [10/13 00:30:19] largePrimes: 6059475 , relations: 2558699 [10/13 00:30:20] Heap stats for matbuild run. [10/13 00:30:20] Max heap usage: 0 MB [10/13 00:30:20] malloc/realloc errors: 0 [10/13 00:30:20] total malloc's : 84 [10/13 00:30:20] total realloc's: 45 [10/13 00:30:20] rels:2558699, initialFF:0, finalFF:23667 LatSieveTime: 3146 LatSieveTime: 3147 LatSieveTime: 3161 LatSieveTime: 3199 LatSieveTime: 3231 LatSieveTime: 3245 -> makeJobFile(): Adjusted to q0=5060001, q1=5240000. -> client 1 q0: 5060001 LatSieveTime: 3248 LatSieveTime: 3250 LatSieveTime: 3277 LatSieveTime: 3276 LatSieveTime: 3283 LatSieveTime: 3287 -> makeJobFile(): Adjusted to q0=5240001, q1=5420000. -> client 1 q0: 5240001 LatSieveTime: 3311 LatSieveTime: 3340 LatSieveTime: 3344 LatSieveTime: 3354 LatSieveTime: 3361 LatSieveTime: 3364 -> makeJobFile(): Adjusted to q0=5420001, q1=5600000. -> client 1 q0: 5420001 LatSieveTime: 3308 LatSieveTime: 3334 LatSieveTime: 3337 LatSieveTime: 3343 LatSieveTime: 3356 LatSieveTime: 3358 -> makeJobFile(): Adjusted to q0=5600001, q1=5780000. -> client 1 q0: 5600001 LatSieveTime: 3412 LatSieveTime: 3448 LatSieveTime: 3448 LatSieveTime: 3455 LatSieveTime: 3463 LatSieveTime: 3536 -> makeJobFile(): Adjusted to q0=5780001, q1=5960000. -> client 1 q0: 5780001 LatSieveTime: 3416 LatSieveTime: 3446 LatSieveTime: 3447 LatSieveTime: 3463 LatSieveTime: 3496 LatSieveTime: 3504 -> makeJobFile(): Adjusted to q0=5960001, q1=6140000. -> client 1 q0: 5960001 LatSieveTime: 3322 LatSieveTime: 3360 LatSieveTime: 3368 LatSieveTime: 3374 LatSieveTime: 3375 LatSieveTime: 3391 -> makeJobFile(): Adjusted to q0=6140001, q1=6320000. -> client 1 q0: 6140001 [10/13 06:56:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/13 07:01:15] There were 205772/4196107 duplicates. [10/13 07:01:15] RelProcTime: 265.7 [10/13 07:01:35] largePrimes: 12928705 , relations: 6549034 [10/13 07:07:25] GGNFS-0.77.1-VC8(UTE) : procrels [10/13 07:08:00] There were 599275/599275 duplicates. [10/13 07:08:00] RelProcTime: 7.0 [10/13 07:08:19] largePrimes: 12928705 , relations: 6549034 [10/13 07:08:24] GGNFS-0.77.1-VC8(UTE) : matbuild [10/13 07:09:06] largePrimes: 12928705 , relations: 6549034 [10/13 07:09:09] Heap stats for matbuild run. [10/13 07:09:09] Max heap usage: 0 MB [10/13 07:09:09] malloc/realloc errors: 0 [10/13 07:09:09] total malloc's : 127 [10/13 07:09:09] total realloc's: 89 [10/13 07:09:09] rels:6549034, initialFF:0, finalFF:58788 LatSieveTime: 3096 LatSieveTime: 3100 LatSieveTime: 3127 LatSieveTime: 3136 LatSieveTime: 3139 LatSieveTime: 3145 -> makeJobFile(): Adjusted to q0=6320001, q1=6500000. -> client 1 q0: 6320001 LatSieveTime: 2940 LatSieveTime: 2969 LatSieveTime: 2979 LatSieveTime: 2982 LatSieveTime: 2992 LatSieveTime: 3004 -> makeJobFile(): Adjusted to q0=6500001, q1=6680000. -> client 1 q0: 6500001 LatSieveTime: 2949 LatSieveTime: 2979 LatSieveTime: 3001 LatSieveTime: 3033 LatSieveTime: 3035 LatSieveTime: 3037 -> makeJobFile(): Adjusted to q0=6680001, q1=6860000. -> client 1 q0: 6680001 LatSieveTime: 2983 LatSieveTime: 3025 LatSieveTime: 3026 LatSieveTime: 3032 LatSieveTime: 3034 LatSieveTime: 3040 -> makeJobFile(): Adjusted to q0=6860001, q1=7040000. -> client 1 q0: 6860001 LatSieveTime: 3013 LatSieveTime: 3030 LatSieveTime: 3037 LatSieveTime: 3040 LatSieveTime: 3058 LatSieveTime: 3067 -> makeJobFile(): Adjusted to q0=7040001, q1=7220000. -> client 1 q0: 7040001 LatSieveTime: 2988 LatSieveTime: 2997 LatSieveTime: 2997 LatSieveTime: 3017 LatSieveTime: 3043 LatSieveTime: 3065 -> makeJobFile(): Adjusted to q0=7220001, q1=7400000. -> client 1 q0: 7220001 LatSieveTime: 3049 LatSieveTime: 3068 LatSieveTime: 3132 LatSieveTime: 3148 LatSieveTime: 3159 LatSieveTime: 3165 -> makeJobFile(): Adjusted to q0=7400001, q1=7580000. -> client 1 q0: 7400001 LatSieveTime: 3138 LatSieveTime: 3141 LatSieveTime: 3173 LatSieveTime: 3185 LatSieveTime: 3196 LatSieveTime: 3202 -> makeJobFile(): Adjusted to q0=7580001, q1=7760000. -> client 1 q0: 7580001 LatSieveTime: 3124 LatSieveTime: 3154 LatSieveTime: 3177 LatSieveTime: 3199 LatSieveTime: 3214 LatSieveTime: 3242 -> makeJobFile(): Adjusted to q0=7760001, q1=7940000. -> client 1 q0: 7760001 LatSieveTime: 3166 LatSieveTime: 3173 LatSieveTime: 3198 LatSieveTime: 3211 LatSieveTime: 3217 LatSieveTime: 3220 -> makeJobFile(): Adjusted to q0=7940001, q1=8120000. -> client 1 q0: 7940001 LatSieveTime: 3143 LatSieveTime: 3171 LatSieveTime: 3180 LatSieveTime: 3188 LatSieveTime: 3201 LatSieveTime: 3215 -> makeJobFile(): Adjusted to q0=8120001, q1=8300000. -> client 1 q0: 8120001 LatSieveTime: 3187 LatSieveTime: 3201 LatSieveTime: 3211 LatSieveTime: 3229 LatSieveTime: 3239 LatSieveTime: 3244 -> makeJobFile(): Adjusted to q0=8300001, q1=8480000. -> client 1 q0: 8300001 LatSieveTime: 3128 LatSieveTime: 3182 LatSieveTime: 3206 LatSieveTime: 3210 LatSieveTime: 3232 LatSieveTime: 3238 -> makeJobFile(): Adjusted to q0=8480001, q1=8660000. -> client 1 q0: 8480001 LatSieveTime: 3229 LatSieveTime: 3236 LatSieveTime: 3245 LatSieveTime: 3251 LatSieveTime: 3265 LatSieveTime: 3267 -> makeJobFile(): Adjusted to q0=8660001, q1=8840000. -> client 1 q0: 8660001 LatSieveTime: 3194 LatSieveTime: 3208 LatSieveTime: 3224 LatSieveTime: 3239 LatSieveTime: 3253 LatSieveTime: 3256 -> makeJobFile(): Adjusted to q0=8840001, q1=9020000. -> client 1 q0: 8840001 LatSieveTime: 3156 LatSieveTime: 3190 LatSieveTime: 3202 LatSieveTime: 3244 LatSieveTime: 3251 LatSieveTime: 3254 -> makeJobFile(): Adjusted to q0=9020001, q1=9200000. -> client 1 q0: 9020001 LatSieveTime: 3225 LatSieveTime: 3255 LatSieveTime: 3256 LatSieveTime: 3266 LatSieveTime: 3294 LatSieveTime: 3300 -> makeJobFile(): Adjusted to q0=9200001, q1=9380000. -> client 1 q0: 9200001 LatSieveTime: 3228 LatSieveTime: 3250 LatSieveTime: 3270 LatSieveTime: 3281 LatSieveTime: 3288 LatSieveTime: 3296 -> makeJobFile(): Adjusted to q0=9380001, q1=9560000. -> client 1 q0: 9380001 LatSieveTime: 3224 LatSieveTime: 3230 LatSieveTime: 3229 LatSieveTime: 3234 LatSieveTime: 3241 LatSieveTime: 3275 -> makeJobFile(): Adjusted to q0=9560001, q1=9740000. -> client 1 q0: 9560001 LatSieveTime: 3241 LatSieveTime: 3294 LatSieveTime: 3296 LatSieveTime: 3297 LatSieveTime: 3299 LatSieveTime: 3332 -> makeJobFile(): Adjusted to q0=9740001, q1=9920000. -> client 1 q0: 9740001 LatSieveTime: 3248 LatSieveTime: 3272 LatSieveTime: 3275 LatSieveTime: 3280 LatSieveTime: 3315 LatSieveTime: 3317 -> makeJobFile(): Adjusted to q0=9920001, q1=10100000. -> client 1 q0: 9920001 LatSieveTime: 3332 LatSieveTime: 3380 LatSieveTime: 3381 LatSieveTime: 3406 LatSieveTime: 3408 LatSieveTime: 3435 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 [10/14 15:41:45] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 15:58:07] There were 1329163/12935188 duplicates. [10/14 15:58:07] RelProcTime: 924.9 [10/14 15:58:53] largePrimes: 25494353 , relations: 18155059 [10/14 16:00:35] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 16:02:47] There were 1399003/1399003 duplicates. [10/14 16:02:47] RelProcTime: 13.0 [10/14 16:03:40] largePrimes: 25494353 , relations: 18155059 [10/14 16:03:49] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 16:05:37] There were 601308/601308 duplicates. [10/14 16:05:37] RelProcTime: 11.1 [10/14 16:06:37] largePrimes: 25494353 , relations: 18155059 [10/14 16:08:34] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 16:10:16] There were 601308/601308 duplicates. [10/14 16:10:16] RelProcTime: 11.4 [10/14 16:11:24] largePrimes: 25494353 , relations: 18155059 [10/14 16:11:35] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 16:13:28] There were 601308/601308 duplicates. [10/14 16:13:28] RelProcTime: 11.4 [10/14 16:14:43] largePrimes: 25494353 , relations: 18155059 [10/14 16:15:29] GGNFS-0.77.1-VC8(UTE) : matbuild [10/14 16:17:10] largePrimes: 25494353 , relations: 18155059 [10/14 16:17:17] Heap stats for matbuild run. [10/14 16:17:17] Max heap usage: 0 MB [10/14 16:17:17] malloc/realloc errors: 0 [10/14 16:17:17] total malloc's : 323 [10/14 16:17:17] total realloc's: 218 [10/14 16:17:17] rels:18155059, initialFF:0, finalFF:164934 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 -> makeJobFile(): Adjusted to q0=10100001, q1=10280000. -> client 1 q0: 10100001 LatSieveTime: 3595 LatSieveTime: 3610 LatSieveTime: 3649 LatSieveTime: 3650 LatSieveTime: 3655 LatSieveTime: 3682 -> makeJobFile(): Adjusted to q0=10280001, q1=10460000. -> client 1 q0: 10280001 [10/14 23:48:44] GGNFS-0.77.1-VC8(UTE) : procrels [10/14 23:51:20] There were 76320/574898 duplicates. [10/14 23:51:20] RelProcTime: 155.8 [10/14 23:53:24] largePrimes: 25899781 , relations: 18653637 [10/14 23:56:13] GGNFS-0.77.1-VC8(UTE) : matbuild [10/14 23:58:57] largePrimes: 25899781 , relations: 18653637 [10/14 23:59:10] Heap stats for matbuild run. [10/14 23:59:10] Max heap usage: 0 MB [10/14 23:59:10] malloc/realloc errors: 0 [10/14 23:59:10] total malloc's : 325 [10/14 23:59:10] total realloc's: 225 [10/14 23:59:10] rels:18653637, initialFF:0, finalFF:170096 LatSieveTime: 3839 LatSieveTime: 3845 LatSieveTime: 3868 LatSieveTime: 3881 LatSieveTime: 3882 LatSieveTime: 3904 -> makeJobFile(): Adjusted to q0=10460001, q1=10640000. -> client 1 q0: 10460001 LatSieveTime: 3872 LatSieveTime: 3885 LatSieveTime: 3890 LatSieveTime: 3899 LatSieveTime: 3920 LatSieveTime: 3928 -> makeJobFile(): Adjusted to q0=10640001, q1=10820000. -> client 1 q0: 10640001 LatSieveTime: 3803 LatSieveTime: 3865 LatSieveTime: 3901 LatSieveTime: 3921 LatSieveTime: 3927 LatSieveTime: 3966 -> makeJobFile(): Adjusted to q0=10820001, q1=11000000. -> client 1 q0: 10820001 LatSieveTime: 3885 LatSieveTime: 3891 LatSieveTime: 3892 LatSieveTime: 3895 LatSieveTime: 3932 LatSieveTime: 3970 -> makeJobFile(): Adjusted to q0=11000001, q1=11180000. -> client 1 q0: 11000001 LatSieveTime: 3832 LatSieveTime: 3861 LatSieveTime: 3911 LatSieveTime: 3948 LatSieveTime: 3971 LatSieveTime: 3979 -> makeJobFile(): Adjusted to q0=11180001, q1=11360000. -> client 1 q0: 11180001 LatSieveTime: 3850 LatSieveTime: 3890 LatSieveTime: 3912 LatSieveTime: 3922 LatSieveTime: 3937 LatSieveTime: 3975 -> makeJobFile(): Adjusted to q0=11360001, q1=11540000. -> client 1 q0: 11360001 LatSieveTime: 3874 LatSieveTime: 3878 LatSieveTime: 3888 LatSieveTime: 3954 LatSieveTime: 3968 LatSieveTime: 3975 -> makeJobFile(): Adjusted to q0=11540001, q1=11720000. -> client 1 q0: 11540001 LatSieveTime: 3878 LatSieveTime: 3881 LatSieveTime: 3888 LatSieveTime: 3905 LatSieveTime: 3947 LatSieveTime: 4042 -> makeJobFile(): Adjusted to q0=11720001, q1=11900000. -> client 1 q0: 11720001 LatSieveTime: 3843 LatSieveTime: 3844 LatSieveTime: 3890 LatSieveTime: 3891 LatSieveTime: 3901 LatSieveTime: 4038 -> makeJobFile(): Adjusted to q0=11900001, q1=12080000. -> client 1 q0: 11900001 LatSieveTime: 3853 LatSieveTime: 3876 LatSieveTime: 3884 LatSieveTime: 3891 LatSieveTime: 3910 LatSieveTime: 3974 -> makeJobFile(): Adjusted to q0=12080001, q1=12260000. -> client 1 q0: 12080001 LatSieveTime: 3837 LatSieveTime: 3854 LatSieveTime: 3866 LatSieveTime: 3878 LatSieveTime: 3907 LatSieveTime: 3929 -> makeJobFile(): Adjusted to q0=12260001, q1=12440000. -> client 1 q0: 12260001 LatSieveTime: 3795 LatSieveTime: 3881 LatSieveTime: 3881 LatSieveTime: 3893 LatSieveTime: 3950 LatSieveTime: 3976 -> makeJobFile(): Adjusted to q0=12440001, q1=12620000. -> client 1 q0: 12440001 LatSieveTime: 3791 LatSieveTime: 3831 LatSieveTime: 3859 LatSieveTime: 3862 LatSieveTime: 3897 LatSieveTime: 3909 -> makeJobFile(): Adjusted to q0=12620001, q1=12800000. -> client 1 q0: 12620001 LatSieveTime: 3794 LatSieveTime: 3863 LatSieveTime: 3888 LatSieveTime: 3902 LatSieveTime: 3907 LatSieveTime: 4035 -> makeJobFile(): Adjusted to q0=12800001, q1=12980000. -> client 1 q0: 12800001 LatSieveTime: 3807 LatSieveTime: 3826 LatSieveTime: 3830 LatSieveTime: 3838 LatSieveTime: 3847 LatSieveTime: 3888 -> makeJobFile(): Adjusted to q0=12980001, q1=13160000. -> client 1 q0: 12980001 LatSieveTime: 3790 LatSieveTime: 3797 LatSieveTime: 3833 LatSieveTime: 3854 LatSieveTime: 3870 LatSieveTime: 3870 -> makeJobFile(): Adjusted to q0=13160001, q1=13340000. -> client 1 q0: 13160001 [10/15 17:50:06] GGNFS-0.77.1-VC8(UTE) : makefb [10/15 17:50:17] name: [10/15 17:50:17] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/15 17:50:17] c0: -41 [10/15 17:50:17] c1: 0 [10/15 17:50:17] c2: 0 [10/15 17:50:17] c3: 0 [10/15 17:50:17] c4: 0 [10/15 17:50:17] c5: 140 [10/15 17:50:17] RFBsize: 922193 (upto 14199989) [10/15 17:50:17] AFBsize: 737893 (upto 11199989) [10/15 17:50:17] maxNumLargeRatPrimes: 3 [10/15 17:50:17] maxLargeRatPrime: 1073741824 [10/15 17:50:17] maxNumLargeAlgPrimes: 3 [10/15 17:50:17] maxLargeAlgPrime: 1073741824 [10/15 17:52:37] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 17:54:04] There were 12410/1399003 duplicates. [10/15 17:54:04] RelProcTime: 85.5 [10/15 17:54:06] largePrimes: 3336400 , relations: 1386593 [10/15 17:55:16] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 17:56:02] There were 12431/601308 duplicates. [10/15 17:56:02] RelProcTime: 38.7 [10/15 17:56:06] largePrimes: 4611573 , relations: 1975470 [10/15 17:56:34] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 17:56:40] There were 601308/601308 duplicates. [10/15 17:56:40] RelProcTime: 4.8 [10/15 17:56:44] largePrimes: 4611573 , relations: 1975470 [10/15 17:56:55] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 17:57:35] There were 16046/599275 duplicates. [10/15 17:57:35] RelProcTime: 38.9 [10/15 17:57:40] largePrimes: 5803976 , relations: 2558699 [10/15 17:58:10] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:03:07] There were 205772/4196107 duplicates. [10/15 18:03:07] RelProcTime: 290.7 [10/15 18:03:22] largePrimes: 12579494 , relations: 6549034 [10/15 18:03:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:04:08] There were 599275/599275 duplicates. [10/15 18:04:08] RelProcTime: 8.4 [10/15 18:04:29] largePrimes: 12579494 , relations: 6549034 [10/15 18:04:31] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:05:00] There were 599275/599275 duplicates. [10/15 18:05:00] RelProcTime: 8.1 [10/15 18:05:26] largePrimes: 12579494 , relations: 6549034 [10/15 18:07:00] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:26:31] There were 1329163/12935188 duplicates. [10/15 18:26:31] RelProcTime: 1135.1 LatSieveTime: 4017 LatSieveTime: 4019 LatSieveTime: 4031 LatSieveTime: 4040 LatSieveTime: 4041 LatSieveTime: 4049 -> makeJobFile(): Adjusted to q0=13340001, q1=13520000. -> client 1 q0: 13340001 [10/15 18:27:34] largePrimes: 25125094 , relations: 18155059 [10/15 18:31:36] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:33:03] There were 599275/599275 duplicates. [10/15 18:33:03] RelProcTime: 16.2 [10/15 18:34:27] largePrimes: 25125094 , relations: 18155059 [10/15 18:34:30] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:36:17] There were 599275/599275 duplicates. [10/15 18:36:17] RelProcTime: 12.9 [10/15 18:37:51] largePrimes: 25125094 , relations: 18155059 [10/15 18:37:58] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:39:31] There were 599275/599275 duplicates. [10/15 18:39:31] RelProcTime: 15.9 [10/15 18:41:19] largePrimes: 25125094 , relations: 18155059 [10/15 18:42:47] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 18:45:56] There were 76320/574898 duplicates. [10/15 18:45:56] RelProcTime: 69.3 [10/15 18:47:57] largePrimes: 25530486 , relations: 18653637 [10/15 18:48:08] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 19:05:53] There were 1355506/8967184 duplicates. [10/15 19:05:53] RelProcTime: 1051.7 [10/15 19:08:51] largePrimes: 30890408 , relations: 26265315 [10/15 19:08:58] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 19:12:39] There were 599275/599275 duplicates. [10/15 19:12:39] RelProcTime: 23.8 [10/15 19:15:37] largePrimes: 30890408 , relations: 26265315 [10/15 19:17:03] GGNFS-0.77.1-VC8(UTE) : matbuild [10/15 19:21:13] largePrimes: 30890408 , relations: 26265315 [10/15 19:22:17] Heap stats for matbuild run. [10/15 19:22:17] Max heap usage: 0 MB [10/15 19:22:17] malloc/realloc errors: 0 [10/15 19:22:17] total malloc's : 1543 [10/15 19:22:17] total realloc's: 863 [10/15 19:22:17] rels:26265315, initialFF:0, finalFF:547402 [10/15 19:23:45] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 19:28:10] There were 90934/538992 duplicates. [10/15 19:28:10] RelProcTime: 82.4 [10/15 19:31:27] largePrimes: 31165213 , relations: 26713373 [10/15 19:34:07] GGNFS-0.77.1-VC8(UTE) : matbuild LatSieveTime: 4046 LatSieveTime: 4091 LatSieveTime: 4110 LatSieveTime: 4121 LatSieveTime: 4133 LatSieveTime: 4147 -> makeJobFile(): Adjusted to q0=13520001, q1=13700000. -> client 1 q0: 13520001 [10/15 19:37:45] largePrimes: 31165213 , relations: 26713373 [10/15 19:39:04] Heap stats for matbuild run. [10/15 19:39:04] Max heap usage: 0 MB [10/15 19:39:04] malloc/realloc errors: 0 [10/15 19:39:04] total malloc's : 1786 [10/15 19:39:04] total realloc's: 973 [10/15 19:39:04] rels:26713373, initialFF:0, finalFF:576245 [10/15 19:52:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 19:58:39] There were 91615/533925 duplicates. [10/15 19:58:39] RelProcTime: 80.2 [10/15 20:02:15] largePrimes: 31432748 , relations: 27155683 [10/15 20:03:12] GGNFS-0.77.1-VC8(UTE) : matbuild [10/15 20:07:44] largePrimes: 31432748 , relations: 27155683 [10/15 20:09:18] Heap stats for matbuild run. [10/15 20:09:18] Max heap usage: 0 MB [10/15 20:09:18] malloc/realloc errors: 0 [10/15 20:09:18] total malloc's : 1912 [10/15 20:09:18] total realloc's: 1026 [10/15 20:09:18] rels:27155683, initialFF:0, finalFF:607736 LatSieveTime: 3948 LatSieveTime: 3950 LatSieveTime: 3983 LatSieveTime: 3999 LatSieveTime: 4006 LatSieveTime: 4042 -> makeJobFile(): Adjusted to q0=13700001, q1=13880000. -> client 1 q0: 13700001 LatSieveTime: 3822 LatSieveTime: 3862 LatSieveTime: 3873 LatSieveTime: 3879 LatSieveTime: 3881 LatSieveTime: 3921 -> makeJobFile(): Adjusted to q0=13880001, q1=14060000. -> client 1 q0: 13880001 [10/15 21:50:28] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 21:55:25] There were 185186/1067393 duplicates. [10/15 21:55:25] RelProcTime: 296.3 [10/15 21:59:05] largePrimes: 31955260 , relations: 28037890 [10/15 21:59:35] GGNFS-0.77.1-VC8(UTE) : matbuild [10/15 22:04:15] largePrimes: 31955260 , relations: 28037890 [10/15 22:06:25] Heap stats for matbuild run. [10/15 22:06:25] Max heap usage: 0 MB [10/15 22:06:25] malloc/realloc errors: 0 [10/15 22:06:25] total malloc's : 1877 [10/15 22:06:25] total realloc's: 1017 [10/15 22:06:25] rels:28037890, initialFF:0, finalFF:681817 LatSieveTime: 3762 LatSieveTime: 3773 LatSieveTime: 3781 LatSieveTime: 3790 LatSieveTime: 3804 LatSieveTime: 3942 -> makeJobFile(): Adjusted to q0=14060001, q1=14240000. -> client 1 q0: 14060001 [10/15 22:56:54] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 22:58:12] There were 92555/524389 duplicates. [10/15 22:58:12] RelProcTime: 77.1 [10/15 23:01:49] largePrimes: 32205123 , relations: 28469724 [10/15 23:04:25] GGNFS-0.77.1-VC8(UTE) : matbuild [10/15 23:09:04] largePrimes: 32205123 , relations: 28469724 [10/15 23:11:30] Heap stats for matbuild run. [10/15 23:11:30] Max heap usage: 0 MB [10/15 23:11:30] malloc/realloc errors: 0 [10/15 23:11:30] total malloc's : 1858 [10/15 23:11:30] total realloc's: 1011 [10/15 23:11:30] rels:28469724, initialFF:0, finalFF:723582 [10/15 23:16:22] GGNFS-0.77.1-VC8(UTE) : makefb [10/15 23:16:34] name: [10/15 23:16:34] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/15 23:16:34] c0: -41 [10/15 23:16:34] c1: 0 [10/15 23:16:34] c2: 0 [10/15 23:16:34] c3: 0 [10/15 23:16:34] c4: 0 [10/15 23:16:34] c5: 140 [10/15 23:16:34] RFBsize: 1270607 (upto 19999999) [10/15 23:16:34] AFBsize: 737893 (upto 11199989) [10/15 23:16:34] maxNumLargeRatPrimes: 3 [10/15 23:16:34] maxLargeRatPrime: 4 [10/15 23:16:34] maxNumLargeAlgPrimes: 3 [10/15 23:16:34] maxLargeAlgPrime: 4 [10/15 23:17:28] GGNFS-0.77.1-VC8(UTE) : makefb [10/15 23:17:40] name: [10/15 23:17:40] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/15 23:17:40] c0: -41 [10/15 23:17:40] c1: 0 [10/15 23:17:40] c2: 0 [10/15 23:17:40] c3: 0 [10/15 23:17:40] c4: 0 [10/15 23:17:40] c5: 140 [10/15 23:17:40] RFBsize: 1270607 (upto 19999999) [10/15 23:17:40] AFBsize: 737893 (upto 11199989) [10/15 23:17:40] maxNumLargeRatPrimes: 3 [10/15 23:17:40] maxLargeRatPrime: 1 [10/15 23:17:40] maxNumLargeAlgPrimes: 3 [10/15 23:17:40] maxLargeAlgPrime: 1 [10/15 23:18:31] GGNFS-0.77.1-VC8(UTE) : makefb [10/15 23:18:43] name: [10/15 23:18:43] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/15 23:18:43] c0: -41 [10/15 23:18:43] c1: 0 [10/15 23:18:43] c2: 0 [10/15 23:18:43] c3: 0 [10/15 23:18:43] c4: 0 [10/15 23:18:43] c5: 140 [10/15 23:18:43] RFBsize: 1270607 (upto 19999999) [10/15 23:18:43] AFBsize: 737893 (upto 11199989) [10/15 23:18:43] maxNumLargeRatPrimes: 3 [10/15 23:18:43] maxLargeRatPrime: 1073741824 [10/15 23:18:43] maxNumLargeAlgPrimes: 3 [10/15 23:18:43] maxLargeAlgPrime: 1073741824 [10/15 23:19:06] GGNFS-0.77.1-VC8(UTE) : makefb [10/15 23:19:18] name: [10/15 23:19:18] n=9418458803417723740888717803533233710133479438269932663459345303689160281670363146397952544429060881101532530146048621903681308124945425051323363886343174649261020251801563 (172 digits) [10/15 23:19:18] c0: -41 [10/15 23:19:18] c1: 0 [10/15 23:19:18] c2: 0 [10/15 23:19:18] c3: 0 [10/15 23:19:18] c4: 0 [10/15 23:19:18] c5: 140 [10/15 23:19:18] RFBsize: 1270607 (upto 19999999) [10/15 23:19:18] AFBsize: 737893 (upto 11199989) [10/15 23:19:18] maxNumLargeRatPrimes: 3 [10/15 23:19:18] maxLargeRatPrime: -2147483648 [10/15 23:19:18] maxNumLargeAlgPrimes: 3 [10/15 23:19:18] maxLargeAlgPrime: -2147483648 [10/15 23:19:51] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 23:45:13] There were 1471579/19131606 duplicates. [10/15 23:45:13] RelProcTime: 1482.4 [10/15 23:45:13] Memory allocation error (3108362544 bytes requested). [10/15 23:52:05] GGNFS-0.77.1-VC8(UTE) : procrels [10/15 23:52:41] There were 2007/599275 duplicates. [10/15 23:52:41] RelProcTime: 35.7 [10/15 23:52:41] Memory allocation error (3213765480 bytes requested). [10/15 23:56:06] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 3903 LatSieveTime: 3958 LatSieveTime: 3969 LatSieveTime: 3996 LatSieveTime: 4011 LatSieveTime: 4013 -> makeJobFile(): Adjusted to q0=14240001, q1=14420000. -> client 1 q0: 14240001 [10/16 00:06:44] There were 447123/9583958 duplicates. [10/16 00:06:44] RelProcTime: 628.4 [10/16 00:07:20] largePrimes: 15750178 , relations: 9136835 [10/16 00:07:34] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:08:55] There were 61154/599275 duplicates. [10/16 00:08:55] RelProcTime: 50.2 [10/16 00:09:17] largePrimes: 16398497 , relations: 9674956 [10/16 00:09:28] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:10:02] There were 599275/599275 duplicates. [10/16 00:10:02] RelProcTime: 10.6 [10/16 00:10:31] largePrimes: 16398497 , relations: 9674956 [10/16 00:12:15] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:12:53] There were 599275/599275 duplicates. [10/16 00:12:53] RelProcTime: 10.4 [10/16 00:13:30] largePrimes: 16398497 , relations: 9674956 [10/16 00:14:18] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:15:09] There were 599275/599275 duplicates. [10/16 00:15:09] RelProcTime: 10.3 [10/16 00:15:53] largePrimes: 16398497 , relations: 9674956 [10/16 00:16:08] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:30:47] There were 1067545/9547648 duplicates. [10/16 00:30:47] RelProcTime: 843.5 [10/16 00:32:13] largePrimes: 24778173 , relations: 18155059 [10/16 00:32:58] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:34:33] There were 599275/599275 duplicates. [10/16 00:34:33] RelProcTime: 16.9 [10/16 00:36:12] largePrimes: 24778173 , relations: 18155059 [10/16 00:36:23] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 00:38:27] There were 599275/599275 duplicates. [10/16 00:38:27] RelProcTime: 17.6 [10/16 00:40:17] largePrimes: 24778173 , relations: 18155059 [10/16 00:40:36] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 01:04:41] There were 1614373/10614999 duplicates. [10/16 01:04:41] RelProcTime: 1252.6 [10/16 01:08:55] largePrimes: 31084412 , relations: 27155685 [10/16 01:09:10] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 4205 LatSieveTime: 4206 LatSieveTime: 4236 LatSieveTime: 4236 LatSieveTime: 4244 LatSieveTime: 4297 -> makeJobFile(): Adjusted to q0=14420001, q1=14600000. -> client 1 q0: 14420001 [10/16 01:14:26] There were 185186/1067393 duplicates. [10/16 01:14:26] RelProcTime: 123.1 [10/16 01:17:34] largePrimes: 31606906 , relations: 28037892 [10/16 01:17:48] GGNFS-0.77.1-VC8(UTE) : procrels [10/16 01:23:22] There were 92555/524389 duplicates. [10/16 01:23:22] RelProcTime: 93.5 [10/16 01:26:54] largePrimes: 31856761 , relations: 28469726 [10/16 01:27:18] GGNFS-0.77.1-VC8(UTE) : matbuild [10/16 01:31:50] largePrimes: 31856761 , relations: 28469726 [10/16 01:33:55] Heap stats for matbuild run. [10/16 01:33:55] Max heap usage: 0 MB [10/16 01:33:55] malloc/realloc errors: 0 [10/16 01:33:55] total malloc's : 1301 [10/16 01:33:55] total realloc's: 740 [10/16 01:33:55] rels:28469726, initialFF:0, finalFF:1081727 LatSieveTime: 4070 LatSieveTime: 4093 LatSieveTime: 4120 LatSieveTime: 4146 LatSieveTime: 4155 LatSieveTime: 4174 -> makeJobFile(): Adjusted to q0=14600001, q1=14780000. -> client 1 q0: 14600001 LatSieveTime: 3973 LatSieveTime: 4010 LatSieveTime: 4048 LatSieveTime: 4083 LatSieveTime: 4117 LatSieveTime: 4118 -> makeJobFile(): Adjusted to q0=14780001, q1=14960000. -> client 1 q0: 14780001 LatSieveTime: 4015 LatSieveTime: 4023 LatSieveTime: 4068 LatSieveTime: 4074 LatSieveTime: 4075 LatSieveTime: 4101 -> makeJobFile(): Adjusted to q0=14960001, q1=15140000. -> client 1 q0: 14960001 LatSieveTime: 3972 LatSieveTime: 4029 LatSieveTime: 4032 LatSieveTime: 4054 LatSieveTime: 4067 LatSieveTime: 4078 -> makeJobFile(): Adjusted to q0=15140001, q1=15320000. -> client 1 q0: 15140001 LatSieveTime: 3992 LatSieveTime: 4013 LatSieveTime: 4069 LatSieveTime: 4092 LatSieveTime: 4100 LatSieveTime: 4101 -> makeJobFile(): Adjusted to q0=15320001, q1=15500000. -> client 1 q0: 15320001 LatSieveTime: 3957 LatSieveTime: 3972 LatSieveTime: 4016 LatSieveTime: 4023 LatSieveTime: 4027 LatSieveTime: 4051 -> makeJobFile(): Adjusted to q0=15500001, q1=15680000. -> client 1 q0: 15500001 LatSieveTime: 4039 LatSieveTime: 4052 LatSieveTime: 4061 LatSieveTime: 4078 LatSieveTime: 4099 LatSieveTime: 4126 -> makeJobFile(): Adjusted to q0=15680001, q1=15860000. -> client 1 q0: 15680001 LatSieveTime: 3986 LatSieveTime: 3995 LatSieveTime: 4035 LatSieveTime: 4054 LatSieveTime: 4067 LatSieveTime: 4093 -> makeJobFile(): Adjusted to q0=15860001, q1=16040000. -> client 1 q0: 15860001 LatSieveTime: 3960 LatSieveTime: 4011 LatSieveTime: 4016 LatSieveTime: 4029 LatSieveTime: 4062 LatSieveTime: 4068 -> makeJobFile(): Adjusted to q0=16040001, q1=16220000. -> client 1 q0: 16040001 LatSieveTime: 3985 LatSieveTime: 3994 LatSieveTime: 4003 LatSieveTime: 4012 LatSieveTime: 4021 LatSieveTime: 4072 -> makeJobFile(): Adjusted to q0=16220001, q1=16400000. -> client 1 q0: 16220001 LatSieveTime: 3882 LatSieveTime: 3995 LatSieveTime: 3998 LatSieveTime: 4026 LatSieveTime: 4042 LatSieveTime: 4048 -> makeJobFile(): Adjusted to q0=16400001, q1=16580000. -> client 1 q0: 16400001 LatSieveTime: 3940 LatSieveTime: 3949 LatSieveTime: 3962 LatSieveTime: 4007 LatSieveTime: 4031 LatSieveTime: 4037 -> makeJobFile(): Adjusted to q0=16580001, q1=16760000. -> client 1 q0: 16580001 LatSieveTime: 3978 LatSieveTime: 4001 LatSieveTime: 4015 LatSieveTime: 4033 LatSieveTime: 4035 LatSieveTime: 4088 -> makeJobFile(): Adjusted to q0=16760001, q1=16940000. -> client 1 q0: 16760001 LatSieveTime: 3946 LatSieveTime: 3979 LatSieveTime: 3986 LatSieveTime: 3997 LatSieveTime: 4006 LatSieveTime: 4041 -> makeJobFile(): Adjusted to q0=16940001, q1=17120000. -> client 1 q0: 16940001 [10/16 17:59:26] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 4012 LatSieveTime: 4065 LatSieveTime: 4073 LatSieveTime: 4073 LatSieveTime: 4114 LatSieveTime: 4138 -> makeJobFile(): Adjusted to q0=17120001, q1=17300000. -> client 1 q0: 17120001 [10/16 18:26:06] There were 1550796/8152824 duplicates. [10/16 18:26:06] RelProcTime: 1115.8 [10/16 18:33:08] largePrimes: 35325945 , relations: 35071754 [10/16 18:33:35] GGNFS-0.77.1-VC8(UTE) : matbuild [10/16 18:45:30] largePrimes: 35325945 , relations: 35071754 [10/16 18:56:55] reduceRelSets dropped relation-set weight from 45211101 to 37151772. [10/16 19:00:10] reduceRelSets dropped relation-set weight from 37151772 to 37102828. [10/16 19:00:10] After removing heavy rel-sets, weight is 28934538. [10/16 19:13:01] Heap stats for matbuild run. [10/16 19:13:01] Max heap usage: 0 MB [10/16 19:13:01] malloc/realloc errors: 0 [10/16 19:13:01] total malloc's : 1568 [10/16 19:13:01] total realloc's: 820 [10/16 19:13:01] rels:35071754, initialFF:0, finalFF:2455021 [10/16 19:13:01] depinf file written. Run matprune. [10/16 19:13:35] GGNFS-0.77.1-VC8(UTE) : matprune [10/16 19:13:50] Pruning matrix with wt=0.700 [10/16 19:13:50] Initial matrix is 2008567 x 2455018 with sparse part having weight 303753471. [10/16 19:13:50] (total weight is 459972825) [10/16 19:53:34] Matrix pruned to 1823579 x 1833685 with weight 173762309. [10/16 19:53:44] Matrix is pruned. Run matsolve. [10/16 19:53:44] Heap stats for matprune run: [10/16 19:53:44] Max heap usage: 0 MB [10/16 19:53:44] malloc/realloc errors: 0 [10/16 19:53:44] total malloc's : 13 [10/16 19:53:44] total realloc's: 0 [10/16 19:57:01] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=3226487518) [10/17 21:44:51] BLanczosTime: 92867.6 [10/17 21:44:51] Heap stats for matsolve run: [10/17 21:44:51] Max heap usage: 0 MB [10/17 21:44:51] malloc/realloc errors: 0 [10/17 21:44:51] total malloc's : 11 [10/17 21:44:51] total realloc's: 0 [10/17 21:45:44] GGNFS-0.77.1-VC8(UTE) : sqrt [10/17 21:45:46] bmultiplier=2 [10/17 22:13:28] From dependence 0, sqrt obtained: [10/17 22:13:28] r1=5354165852117185252135914375321320153508853934084562766957505865511187 (pp70) [10/17 22:13:28] r2=1759089849578231622051771480901303190911856747169695121364422606428678272553353568622080274069992468249 (c103) [10/17 22:13:28] (pp=probable prime, c=composite) [10/17 22:13:28] sqrtTime: 1664.1```
c103
name 名前 apo October 18, 2012 15:04:59 UTC 2012 年 10 月 19 日 (金) 0 時 4 分 59 秒 (日本時間) `1759089849578231622051771480901303190911856747169695121364422606428678272553353568622080274069992468249<103>` ```692659106326086370144243932616122488217213337<45> 2539618455185799085486796485203664599969880932717451374977<58>``` ```-> makeJobFile(): Adjusted to q0=1150000, q1=1250000. -> client 1 q0: 1150000 LatSieveTime: 676 LatSieveTime: 702 LatSieveTime: 709 LatSieveTime: 718 LatSieveTime: 720 LatSieveTime: 745 -> makeJobFile(): Adjusted to q0=1250001, q1=1350000. -> client 1 q0: 1250001 [10/17 23:44:30] GGNFS-0.77.1-VC8(UTE) : makefb [10/17 23:44:34] name: n103 [10/17 23:44:34] n=1759089849578231622051771480901303190911856747169695121364422606428678272553353568622080274069992468249 (103 digits) [10/17 23:44:34] c0: -2754664553643739114455504 [10/17 23:44:34] c1: 748229372051421576924 [10/17 23:44:34] c2: -238552518335959716 [10/17 23:44:34] c3: -11233985120407 [10/17 23:44:34] c4: 540277310 [10/17 23:44:34] c5: 3780 [10/17 23:44:34] RFBsize: 169511 (upto 2299963) [10/17 23:44:34] AFBsize: 169633 (upto 2299963) [10/17 23:44:34] maxNumLargeRatPrimes: 3 [10/17 23:44:34] maxLargeRatPrime: 67108864 [10/17 23:44:34] maxNumLargeAlgPrimes: 3 [10/17 23:44:34] maxLargeAlgPrime: 67108864 [10/17 23:45:28] GGNFS-0.77.1-VC8(UTE) : procrels [10/17 23:45:29] DISC_P_DIV=4124378227 [10/17 23:46:50] There were 8361/749192 duplicates. [10/17 23:46:50] RelProcTime: 63.6 [10/17 23:46:51] largePrimes: 1360659 , relations: 740831 [10/17 23:47:16] GGNFS-0.77.1-VC8(UTE) : matbuild [10/17 23:47:18] largePrimes: 1360659 , relations: 740831 [10/17 23:47:18] Heap stats for matbuild run. [10/17 23:47:18] Max heap usage: 0 MB [10/17 23:47:18] malloc/realloc errors: 0 [10/17 23:47:18] total malloc's : 71 [10/17 23:47:18] total realloc's: 29 [10/17 23:47:18] rels:740831, initialFF:0, finalFF:21968 LatSieveTime: 668 LatSieveTime: 687 LatSieveTime: 703 LatSieveTime: 712 LatSieveTime: 732 LatSieveTime: 748 -> makeJobFile(): Adjusted to q0=1350001, q1=1450000. -> client 1 q0: 1350001 [10/17 23:55:47] GGNFS-0.77.1-VC8(UTE) : procrels [10/17 23:56:42] There were 23384/744485 duplicates. [10/17 23:56:42] RelProcTime: 53.4 [10/17 23:56:44] largePrimes: 2264625 , relations: 1461932 [10/17 23:56:51] GGNFS-0.77.1-VC8(UTE) : matbuild [10/17 23:56:56] largePrimes: 2264625 , relations: 1461932 [10/17 23:56:56] Heap stats for matbuild run. [10/17 23:56:56] Max heap usage: 0 MB [10/17 23:56:56] malloc/realloc errors: 0 [10/17 23:56:56] total malloc's : 119 [10/17 23:56:56] total realloc's: 46 [10/17 23:56:56] rels:1461932, initialFF:0, finalFF:48176 LatSieveTime: 716 LatSieveTime: 719 LatSieveTime: 722 LatSieveTime: 731 LatSieveTime: 752 LatSieveTime: 781 -> makeJobFile(): Adjusted to q0=1450001, q1=1550000. -> client 1 q0: 1450001 LatSieveTime: 734 LatSieveTime: 737 LatSieveTime: 758 LatSieveTime: 759 LatSieveTime: 762 LatSieveTime: 774 -> makeJobFile(): Adjusted to q0=1550001, q1=1650000. -> client 1 q0: 1550001 LatSieveTime: 684 LatSieveTime: 713 LatSieveTime: 721 LatSieveTime: 741 LatSieveTime: 758 LatSieveTime: 759 -> makeJobFile(): Adjusted to q0=1650001, q1=1750000. -> client 1 q0: 1650001 [10/18 00:41:16] GGNFS-0.77.1-VC8(UTE) : procrels [10/18 00:44:25] There were 145677/2242452 duplicates. [10/18 00:44:25] RelProcTime: 185.3 [10/18 00:44:31] largePrimes: 3902284 , relations: 3558707 [10/18 00:44:37] GGNFS-0.77.1-VC8(UTE) : matbuild [10/18 00:44:48] largePrimes: 3902284 , relations: 3558707 [10/18 00:44:56] Heap stats for matbuild run. [10/18 00:44:56] Max heap usage: 0 MB [10/18 00:44:56] malloc/realloc errors: 0 [10/18 00:44:56] total malloc's : 388 [10/18 00:44:56] total realloc's: 169 [10/18 00:44:56] rels:3558707, initialFF:0, finalFF:236092 LatSieveTime: 725 LatSieveTime: 744 LatSieveTime: 744 LatSieveTime: 745 LatSieveTime: 748 LatSieveTime: 768 -> makeJobFile(): Adjusted to q0=1750001, q1=1850000. -> client 1 q0: 1750001 [10/18 00:47:10] GGNFS-0.77.1-VC8(UTE) : procrels [10/18 00:48:21] There were 69033/734670 duplicates. [10/18 00:48:21] RelProcTime: 58.5 [10/18 00:48:30] largePrimes: 4260643 , relations: 4224344 [10/18 00:48:36] GGNFS-0.77.1-VC8(UTE) : matbuild [10/18 00:48:52] largePrimes: 4260643 , relations: 4224344 [10/18 00:49:11] reduceRelSets dropped relation-set weight from 3530467 to 3467612. [10/18 00:49:11] After removing heavy rel-sets, weight is 3086086. [10/18 00:49:41] Heap stats for matbuild run. [10/18 00:49:41] Max heap usage: 0 MB [10/18 00:49:41] malloc/realloc errors: 0 [10/18 00:49:41] total malloc's : 395 [10/18 00:49:41] total realloc's: 155 [10/18 00:49:41] rels:4224344, initialFF:0, finalFF:413533 [10/18 00:49:41] depinf file written. Run matprune. [10/18 00:49:51] GGNFS-0.77.1-VC8(UTE) : matprune [10/18 00:49:52] Pruning matrix with wt=0.700 [10/18 00:49:52] Initial matrix is 339224 x 413533 with sparse part having weight 30024240. [10/18 00:49:52] (total weight is 53404810) [10/18 00:50:19] Matrix pruned to 302655 x 304415 with weight 14434218. [10/18 00:50:20] Matrix is pruned. Run matsolve. [10/18 00:50:20] Heap stats for matprune run: [10/18 00:50:20] Max heap usage: 0 MB [10/18 00:50:20] malloc/realloc errors: 0 [10/18 00:50:20] total malloc's : 12 [10/18 00:50:20] total realloc's: 0 [10/18 00:50:27] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=4070449336) [10/18 01:03:02] BLanczosTime: 755.4 [10/18 01:03:02] Heap stats for matsolve run: [10/18 01:03:02] Max heap usage: 0 MB [10/18 01:03:02] malloc/realloc errors: 0 [10/18 01:03:02] total malloc's : 10 [10/18 01:03:02] total realloc's: 0 [10/18 01:03:29] GGNFS-0.77.1-VC8(UTE) : sqrt [10/18 01:03:30] bmultiplier=2 [10/18 01:08:20] From dependence 0, sqrt obtained: [10/18 01:08:20] r1=692659106326086370144243932616122488217213337 (pp45) [10/18 01:08:20] r2=2539618455185799085486796485203664599969880932717451374977 (pp58) [10/18 01:08:20] (pp=probable prime, c=composite) [10/18 01:08:20] sqrtTime: 290.8```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:05:14 UTC 2009 年 3 月 7 日 (土) 8 時 5 分 14 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 08:33:52 UTC 2010 年 9 月 26 日 (日) 17 時 33 分 52 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 08:34:06 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 6 秒 (日本時間)
403e62150110Ignacio SantosSeptember 26, 2010 08:34:06 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 6 秒 (日本時間)
2040apoOctober 14, 2012 20:45:51 UTC 2012 年 10 月 15 日 (月) 5 時 45 分 51 秒 (日本時間)
4511e632 / 3478Ignacio SantosSeptember 26, 2010 08:34:06 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 6 秒 (日本時間)
5043e60 / 7294--
5511e770 / 17750apoOctober 15, 2012 17:37:21 UTC 2012 年 10 月 16 日 (火) 2 時 37 分 21 秒 (日本時間)
14×10192-419
c183
name 名前 Robert Backstrom March 10, 2012 06:27:17 UTC 2012 年 3 月 10 日 (土) 15 時 27 分 17 秒 (日本時間) `432608828906239735752006257349769652301089218581949976550742800349041299376583481901085656381970744537046678729059708264747657063639459787132402509230250357505487333138601946953803467<183>` ```1423659297182882545031785117937981624522356875754080374233<58> 303871038360287564571164152498014867302528978346549869257661231794260296209831899355347630315789010531288693036209286587096899<126>``` ```Number: n N=432608828906239735752006257349769652301089218581949976550742800349041299376583481901085656381970744537046678729059708264747657063639459787132402509230250357505487333138601946953803467 ( 183 digits) SNFS difficulty: 193 digits. Divisors found: Sat Mar 10 17:23:10 2012 prp58 factor: 1423659297182882545031785117937981624522356875754080374233 Sat Mar 10 17:23:10 2012 prp126 factor: 303871038360287564571164152498014867302528978346549869257661231794260296209831899355347630315789010531288693036209286587096899 Sat Mar 10 17:23:10 2012 elapsed time 03:39:51 (Msieve 1.44 - dependency 2) Version: Total time: 0.00 hours. Scaled time: 0.00 units (timescale=1.628). Factorization parameters were as follows: name: KA_15551_192 n: 432608828906239735752006257349769652301089218581949976550742800349041299376583481901085656381970744537046678729059708264747657063639459787132402509230250357505487333138601946953803467 m: 100000000000000000000000000000000000000 # c183, diff: 193.15 skew: 0.49 deg: 5 c5: 1400 c0: -41 type: snfs lss: 1 rlim: 15000000 alim: 15000000 lpbr: 29 lpba: 29 mfbr: 57 mfba: 57 rlambda: 2.5 alambda: 2.5 qintsize: 100000 Factor base limits: 15000000/15000000 Large primes per side: 3 Large prime bits: 29/29 Max factor residue bits: 57/57 Sieved special-q in [100000, 13100000) Primes: , , Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 7919159 hash collisions in 77880516 relations (72425061 unique) Msieve: matrix is 1260729 x 1260960 (355.5 MB) Total sieving time: 0.00 hours. Total relation processing time: 0.00 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,193,5,0,0,0,0,0,0,0,0,15000000,15000000,29,29,57,57,2.5,2.5,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- CPU0: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU1: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU2: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU3: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU4: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU5: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU6: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 CPU7: Intel(R) Core(TM) i7 CPU 930 @ 2.80GHz stepping 05 Memory: 6059348k/6815744k available (3972k kernel code, 525828k absent, 230568k reserved, 2498k data, 1292k init) Calibrating delay loop (skipped), value calculated using timer frequency.. 5596.93 BogoMIPS (lpj=2798468) Calibrating delay using timer specific routine.. 5595.11 BogoMIPS (lpj=2797555) Calibrating delay using timer specific routine.. 5595.10 BogoMIPS (lpj=2797554) Calibrating delay using timer specific routine.. 5595.12 BogoMIPS (lpj=2797560) Calibrating delay using timer specific routine.. 5595.10 BogoMIPS (lpj=2797554) Calibrating delay using timer specific routine.. 5595.11 BogoMIPS (lpj=2797555) Calibrating delay using timer specific routine.. 5595.10 BogoMIPS (lpj=2797554) Calibrating delay using timer specific routine.. 5595.11 BogoMIPS (lpj=2797556) Total of 8 processors activated (44762.71 BogoMIPS).```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:07:07 UTC 2009 年 3 月 7 日 (土) 8 時 7 分 7 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 08:34:33 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 33 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 08:34:48 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 48 秒 (日本時間)
403e6110 / 2144Ignacio SantosSeptember 26, 2010 08:34:48 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 48 秒 (日本時間)
4511e632 / 4441Ignacio SantosSeptember 26, 2010 08:34:48 UTC 2010 年 9 月 26 日 (日) 17 時 34 分 48 秒 (日本時間)
14×10193-419
c190
name 名前 matsui January 15, 2008 08:56:46 UTC 2008 年 1 月 15 日 (火) 17 時 56 分 46 秒 (日本時間) `4272330556318471726326711220971039702157526930940828221794989166590375049589551100125118252006469529128139400042723305563184717263267112209710397021575269309408282217949891665903750495895511<190>` `62526939757042506698769902098831<32>` `68327837135788697792225797137899512369957580856225051975788498022627748463669114687246453722274284444945237019649575567790089868646790503352013257536766804281<158>` ```GMP-ECM 6.0 [powered by GMP 4.1.4] [ECM] Input number is 4272330556318471726326711220971039702157526930940828221794989166590375049589551100125118252006469529128139400042723305563184717263267112209710397021575269309408282217949891665903750495895511 (190 digits) Using B1=34782088, B2=192388936756, polynomial Dickson(12), sigma=3772883920 Step 1 took 1056665ms Step 2 took 269522ms ********** Factor found in step 2: 62526939757042506698769902098831 Found probable prime factor of 32 digits: 62526939757042506698769902098831 Composite cofactor 68327837135788697792225797137899512369957580856225051975788498022627748463669114687246453722274284444945237019649575567790089868646790503352013257536766804281 has 158 digits```
c158
name 名前 apo October 18, 2012 06:58:41 UTC 2012 年 10 月 18 日 (木) 15 時 58 分 41 秒 (日本時間) `68327837135788697792225797137899512369957580856225051975788498022627748463669114687246453722274284444945237019649575567790089868646790503352013257536766804281<158>` ```5488210472850165760816428615594143417399<40> 12449930168276569958522089984705967423870346402837080988629159287297661389498138542354436477159569871368343980563746319<119>``` ```Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=85931928 Step 1 took 63960ms Step 2 took 31153ms ********** Factor found in step 2: 5488210472850165760816428615594143417399 Found probable prime factor of 40 digits: 5488210472850165760816428615594143417399 Probable prime cofactor 12449930168276569958522089984705967423870346402837080988629159287297661389498138542354436477159569871368343980563746319 has 119 digits``` `GMP-ECM 6.4.2 [configured with MPIR 2.5.1]`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:07:49 UTC 2009 年 3 月 7 日 (土) 8 時 7 分 49 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 08:35:14 UTC 2010 年 9 月 26 日 (日) 17 時 35 分 14 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 08:35:29 UTC 2010 年 9 月 26 日 (日) 17 時 35 分 29 秒 (日本時間)
403e62144110Ignacio SantosSeptember 26, 2010 08:35:29 UTC 2010 年 9 月 26 日 (日) 17 時 35 分 29 秒 (日本時間)
2034Pawel ApostolOctober 5, 2012 14:47:39 UTC 2012 年 10 月 5 日 (金) 23 時 47 分 39 秒 (日本時間)
4511e632 / 3991Ignacio SantosSeptember 26, 2010 08:35:29 UTC 2010 年 9 月 26 日 (日) 17 時 35 分 29 秒 (日本時間)
14×10194-419
c168
name 名前 Pawel Apostol October 3, 2012 13:14:49 UTC 2012 年 10 月 3 日 (水) 22 時 14 分 49 秒 (日本時間) `143425899295298493547667776972810307359568695526747206375172911744562804904622830704851532009679085160647535787468722279278620890129250378743839728699888544866528719657<168>` `265226968457979428288586070906929307<36>` `540766650273808100154187099866892723436596010820382251905045683083448000449468302727044122305591370349001752000657983358062043975051<132>` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=2108643891 Step 1 took 74282ms Step 2 took 27256ms ********** Factor found in step 2: 265226968457979428288586070906929307 Found probable prime factor of 36 digits: 265226968457979428288586070906929307 Composite cofactor 540766650273808100154187099866892723436596010820382251905045683083448000449468302727044122305591370349001752000657983358062043975051 has 132 digits``` `GMP-ECM 6.4.2`
c132
name 名前 Erik Branger October 13, 2012 16:37:55 UTC 2012 年 10 月 14 日 (日) 1 時 37 分 55 秒 (日本時間) `540766650273808100154187099866892723436596010820382251905045683083448000449468302727044122305591370349001752000657983358062043975051<132>` ```124483856359675894502614351089777451500106499101<48> 4344070517154856217773971182492603248742872623377195897054574166612610989370060330951<85>``` ```Number: 15551_194 N = 540766650273808100154187099866892723436596010820382251905045683083448000449468302727044122305591370349001752000657983358062043975051 (132 digits) Divisors found: r1=124483856359675894502614351089777451500106499101 (pp48) r2=4344070517154856217773971182492603248742872623377195897054574166612610989370060330951 (pp85) Version: Msieve v. 1.49 (SVN unknown) Total time: 127.91 hours. Factorization parameters were as follows: # Murphy_E = 6.683e-11, selected by Erik Branger n: 540766650273808100154187099866892723436596010820382251905045683083448000449468302727044122305591370349001752000657983358062043975051 Y0: -41084186607019687020817167 Y1: 94620224996867 c0: -1086422341937417329145413702029440 c1: -6020889710238913945142789632 c2: 7539324880541915694810 c3: 5585757945384571 c4: -25593677836 c5: 4620 skew: 978212.7 type: gnfs # selected mechanically rlim: 10900000 alim: 10900000 lpbr: 28 lpba: 28 mfbr: 54 mfba: 54 rlambda: 2.6 alambda: 2.6 Factor base limits: 10900000/10900000 Large primes per side: 3 Large prime bits: 28/28 Sieved algebraic special-q in [5450000, 11650000) Relations: 20238641 Relations in full relation-set: 2496542 relations Pruned matrix : 1513244 x 1513474 Polynomial selection time: 0.00 hours. Total sieving time: 125.64 hours. Total relation processing time: 0.12 hours. Matrix solve time: 2.03 hours. time per square root: 0.12 hours. Prototype def-par.txt line would be: gnfs,131,5,65,2000,1e-05,0.28,250,20,50000,3600,10900000,10900000,28,28,54,54,2.6,2.6,100000 total time: 127.91 hours. --------- CPU info (if available) ----------``` `GGNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:08:28 UTC 2009 年 3 月 7 日 (土) 8 時 8 分 28 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 08:35:56 UTC 2010 年 9 月 26 日 (日) 17 時 35 分 56 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 08:36:10 UTC 2010 年 9 月 26 日 (日) 17 時 36 分 10 秒 (日本時間)
403e6110 / 2144Ignacio SantosSeptember 26, 2010 08:36:10 UTC 2010 年 9 月 26 日 (日) 17 時 36 分 10 秒 (日本時間)
4511e632 / 4441Ignacio SantosSeptember 26, 2010 08:36:10 UTC 2010 年 9 月 26 日 (日) 17 時 36 分 10 秒 (日本時間)
14×10195-419
c189
name 名前 Ignacio Santos August 19, 2010 08:50:40 UTC 2010 年 8 月 19 日 (木) 17 時 50 分 40 秒 (日本時間) `103556177573133018405682967747850116427840601982119975586483199737357863691089377776123837684538310426669869807176400203973123591834775881982231816258262183397236001296257055901865905534679<189>` ```2497423954786642973860786832895298814702978735865761670081297693784718741907640129361669<88> 41465197518688776105915699806998452805026785191311770592220943494392203591081015105473183899056981291<101>``` ```Number: 6 N=103556177573133018405682967747850116427840601982119975586483199737357863691089377776123837684538310426669869807176400203973123591834775881982231816258262183397236001296257055901865905534679 ( 189 digits) SNFS difficulty: 196 digits. Divisors found: r1=2497423954786642973860786832895298814702978735865761670081297693784718741907640129361669 (pp88) r2=41465197518688776105915699806998452805026785191311770592220943494392203591081015105473183899056981291 (pp101) Version: Msieve-1.40 Total time: 353.21 hours. Scaled time: 614.24 units (timescale=1.739). Factorization parameters were as follows: n: 103556177573133018405682967747850116427840601982119975586483199737357863691089377776123837684538310426669869807176400203973123591834775881982231816258262183397236001296257055901865905534679 m: 1000000000000000000000000000000000000000 deg: 5 c5: 14 c0: -41 skew: 1.24 type: snfs lss: 1 rlim: 13000000 alim: 13000000 lpbr: 28 lpba: 28 mfbr: 55 mfba: 55 rlambda: 2.5 alambda: 2.5 Factor base limits: 13000000/13000000 Large primes per side: 3 Large prime bits: 28/28 Max factor residue bits: 55/55 Sieved rational special-q in [6500000, 11900001) Primes: , , Relations: relations Max relations in full relation-set: Initial matrix: Pruned matrix : 2291021 x 2291247 Total sieving time: 343.25 hours. Total relation processing time: 0.28 hours. Matrix solve time: 8.84 hours. Time per square root: 0.85 hours. Prototype def-par.txt line would be: snfs,196.000,5,0,0,0,0,0,0,0,0,13000000,13000000,28,28,55,55,2.5,2.5,100000 total time: 353.21 hours. --------- CPU info (if available) ----------``` `GGNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:09:09 UTC 2009 年 3 月 7 日 (土) 8 時 9 分 9 秒 (日本時間)
255e40--
3025e40--
351e61024120Serge BatalovJune 18, 2010 07:34:13 UTC 2010 年 6 月 18 日 (金) 16 時 34 分 13 秒 (日本時間)
904Wataru SakaiJune 18, 2010 08:19:48 UTC 2010 年 6 月 18 日 (金) 17 時 19 分 48 秒 (日本時間)
403e62071Wataru SakaiJuly 2, 2010 08:30:07 UTC 2010 年 7 月 2 日 (金) 17 時 30 分 7 秒 (日本時間)
14×10196-419
c195
name 名前 matsuix November 11, 2007 10:28:49 UTC 2007 年 11 月 11 日 (日) 19 時 28 分 49 秒 (日本時間) `361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757105943152454780361757<195>` `12991941439670998826484083573<29>` `27844730337109139843652566781414443973010019917901912124608905800374293093913322511112990355973246182909621326436655829024147474787381586660026718061075677175452115209<167>` ```GMP-ECM 6.0 [powered by GMP 4.1.4] [ECM] Input number is 36175710594315245478036175710594315245478036175710594315245478036175710594315245478036175710594315245478036175710594315245478036175710594315245478036175 7105943152454780361757105943152454780361757 (195 digits) Using B1=6700417, B2=14271659230, polynomial Dickson(12), sigma=1028308536 Step 1 took 218775ms Step 2 took 72716ms ********** Factor found in step 2: 12991941439670998826484083573 Found probable prime factor of 29 digits: 12991941439670998826484083573 Composite cofactor 27844730337109139843652566781414443973010019917901912124608905800374293093913322511112990355973246182909621326436655829024147474787381586660026718061075 677175452115209 has 167 digits```
c167
name 名前 Serge Batalov July 11, 2008 20:19:55 UTC 2008 年 7 月 12 日 (土) 5 時 19 分 55 秒 (日本時間) `27844730337109139843652566781414443973010019917901912124608905800374293093913322511112990355973246182909621326436655829024147474787381586660026718061075677175452115209<167>` `31549870079323557671928299889097<32>` `882562440577446030345916906651940795515837407245481246517202735480841275954656517617252892526279545364367486770393761420751543094322497<135>` ```Input number is 27844730337109139843652566781414443973010019917901912124608905800374293093913322511112990355973246182909621326436655829024147474787381586660026718061075677175452115209 (167 digits) Using B1=1000000, B2=2139490090, polynomial Dickson(6), sigma=1933614541 Step 1 took 6404ms Step 2 took 4024ms ********** Factor found in step 2: 31549870079323557671928299889097 Found probable prime factor of 32 digits: 31549870079323557671928299889097 Composite cofactor 882562440577446030345916906651940795515837407245481246517202735480841275954656517617252892526279545364367486770393761420751543094322497 has 135 digits``` `GMP-ECM 6.2.1`
c135
name 名前 Erik Branger December 6, 2011 06:15:30 UTC 2011 年 12 月 6 日 (火) 15 時 15 分 30 秒 (日本時間) `882562440577446030345916906651940795515837407245481246517202735480841275954656517617252892526279545364367486770393761420751543094322497<135>` ```9066926448394133632569682408595490287284175174922766475773<58> 97338656666147204198058713229311734186191430359290834526675518894586306371989<77>``` ```Number: 15551_196 N = 882562440577446030345916906651940795515837407245481246517202735480841275954656517617252892526279545364367486770393761420751543094322497 (135 digits) Divisors found: r1=9066926448394133632569682408595490287284175174922766475773 (pp58) r2=97338656666147204198058713229311734186191430359290834526675518894586306371989 (pp77) Version: Msieve v. 1.49 (SVN unknown) Total time: 223.10 hours. Factorization parameters were as follows: # Murphy_E = 4.162e-11, selected by Erik Branger n: 882562440577446030345916906651940795515837407245481246517202735480841275954656517617252892526279545364367486770393761420751543094322497 Y0: -289769798912249698571380770 Y1: 367615944399599 c0: -8725966989815133058488209266795177 c1: 57278632827855469348364563366 c2: 35899892825060996410765 c3: -9478411026394714 c4: -3059998272 c5: 432 skew: 3240176.75 type: gnfs # selected mechanically rlim: 13300000 alim: 13300000 lpbr: 28 lpba: 28 mfbr: 54 mfba: 54 rlambda: 2.6 alambda: 2.6 Factor base limits: 13300000/13300000 Large primes per side: 3 Large prime bits: 28/28 Sieved algebraic special-q in [6650000, 15750000) Relations: 21127498 Relations in full relation-set: 3260358 relations Pruned matrix : 1972354 x 1972581 Polynomial selection time: 0.00 hours. Total sieving time: 218.21 hours. Total relation processing time: 0.12 hours. Matrix solve time: 4.42 hours. time per square root: 0.34 hours. Prototype def-par.txt line would be: gnfs,134,5,65,2000,1e-05,0.28,250,20,50000,3600,13300000,13300000,28,28,54,54,2.6,2.6,100000 total time: 223.10 hours. --------- CPU info (if available) ----------``` `GGNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
403e62350500Erik BrangerJanuary 31, 2009 09:47:38 UTC 2009 年 1 月 31 日 (土) 18 時 47 分 38 秒 (日本時間)
1850Wataru SakaiMay 3, 2009 04:49:03 UTC 2009 年 5 月 3 日 (日) 13 時 49 分 3 秒 (日本時間)
4511e64017817Dmitry DomanovApril 1, 2010 17:51:37 UTC 2010 年 4 月 2 日 (金) 2 時 51 分 37 秒 (日本時間)
3200Wataru SakaiAugust 27, 2010 08:35:45 UTC 2010 年 8 月 27 日 (金) 17 時 35 分 45 秒 (日本時間)
14×10197-419
c152
name 名前 apo October 31, 2012 17:18:49 UTC 2012 年 11 月 1 日 (木) 2 時 18 分 49 秒 (日本時間) `55100796409547873505726910732839026411783896951704656274308614672145830600178494357451253742962116332186925549223018022410337811652503655098378132362717<152>` ```750457255821884523130598577814932184632912548356261079022859<60> 73422964442128919638964497419330013227626433313870429164198871373341032307608541285834881463<92>``` ```-> makeJobFile(): Adjusted to q0=7000000, q1=7180000. -> client 1 q0: 7000000 LatSieveTime: 7199 LatSieveTime: 7255 LatSieveTime: 7343 LatSieveTime: 7507 LatSieveTime: 9047 LatSieveTime: 9208 LatSieveTime: 9587 LatSieveTime: 9712 -> makeJobFile(): Adjusted to q0=7180001, q1=7360000. -> client 1 q0: 7180001 [10/25 00:06:37] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 00:07:27] GGNFS-0.77.1-VC8(UTE) : makefb [10/25 00:07:39] name: [10/25 00:07:39] n=55100796409547873505726910732839026411783896951704656274308614672145830600178494357451253742962116332186925549223018022410337811652503655098378132362717 (152 digits) [10/25 00:07:39] c0: -41 [10/25 00:07:39] c1: 0 [10/25 00:07:39] c2: 0 [10/25 00:07:39] c3: 0 [10/25 00:07:39] c4: 0 [10/25 00:07:39] c5: 1400 [10/25 00:07:39] RFBsize: 910077 (upto 13999981) [10/25 00:07:39] AFBsize: 908519 (upto 13999969) [10/25 00:07:39] maxNumLargeRatPrimes: 3 [10/25 00:07:39] maxLargeRatPrime: 268435456 [10/25 00:07:39] maxNumLargeAlgPrimes: 3 [10/25 00:07:39] maxLargeAlgPrime: 268435456 [10/25 00:07:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 00:08:42] There were 2067/627650 duplicates. [10/25 00:08:42] RelProcTime: 47.8 [10/25 00:08:43] largePrimes: 1520100 , relations: 625583 [10/25 00:09:20] GGNFS-0.77.1-VC8(UTE) : matbuild [10/25 00:09:22] largePrimes: 1520100 , relations: 625583 [10/25 00:09:22] Heap stats for matbuild run. [10/25 00:09:22] Max heap usage: 0 MB [10/25 00:09:22] malloc/realloc errors: 0 [10/25 00:09:22] total malloc's : 52 [10/25 00:09:22] total realloc's: 21 [10/25 00:09:22] rels:625583, initialFF:0, finalFF:13053 LatSieveTime: 6370 LatSieveTime: 6462 LatSieveTime: 6467 LatSieveTime: 6479 LatSieveTime: 6487 LatSieveTime: 6491 LatSieveTime: 6497 LatSieveTime: 6561 -> makeJobFile(): Adjusted to q0=7360001, q1=7540000. -> client 1 q0: 7360001 LatSieveTime: 6373 LatSieveTime: 6506 LatSieveTime: 6517 LatSieveTime: 6578 LatSieveTime: 6591 LatSieveTime: 6597 LatSieveTime: 6641 LatSieveTime: 6655 -> makeJobFile(): Adjusted to q0=7540001, q1=7720000. -> client 1 q0: 7540001 LatSieveTime: 6668 LatSieveTime: 6741 LatSieveTime: 6779 LatSieveTime: 6808 LatSieveTime: 6821 LatSieveTime: 6828 LatSieveTime: 6879 LatSieveTime: 6940 -> makeJobFile(): Adjusted to q0=7720001, q1=7900000. -> client 1 q0: 7720001 LatSieveTime: 6921 LatSieveTime: 6941 LatSieveTime: 6974 LatSieveTime: 6975 LatSieveTime: 6977 LatSieveTime: 7020 LatSieveTime: 7051 LatSieveTime: 7138 -> makeJobFile(): Adjusted to q0=7900001, q1=8080000. -> client 1 q0: 7900001 LatSieveTime: 6967 LatSieveTime: 7000 LatSieveTime: 7003 LatSieveTime: 7008 LatSieveTime: 7020 LatSieveTime: 7042 LatSieveTime: 7081 LatSieveTime: 7086 -> makeJobFile(): Adjusted to q0=8080001, q1=8260000. -> client 1 q0: 8080001 LatSieveTime: 7186 LatSieveTime: 7192 LatSieveTime: 7282 LatSieveTime: 7323 LatSieveTime: 7331 LatSieveTime: 7391 LatSieveTime: 7393 LatSieveTime: 7412 -> makeJobFile(): Adjusted to q0=8260001, q1=8440000. -> client 1 q0: 8260001 LatSieveTime: 7186 LatSieveTime: 7261 LatSieveTime: 7295 LatSieveTime: 7316 LatSieveTime: 7328 LatSieveTime: 7345 LatSieveTime: 7370 LatSieveTime: 7393 -> makeJobFile(): Adjusted to q0=8440001, q1=8620000. -> client 1 q0: 8440001 LatSieveTime: 7228 LatSieveTime: 7241 LatSieveTime: 7260 LatSieveTime: 7270 LatSieveTime: 7272 LatSieveTime: 7272 LatSieveTime: 7294 LatSieveTime: 7330 -> makeJobFile(): Adjusted to q0=8620001, q1=8800000. -> client 1 q0: 8620001 LatSieveTime: 7205 LatSieveTime: 7316 LatSieveTime: 7327 LatSieveTime: 7346 LatSieveTime: 7357 LatSieveTime: 7362 LatSieveTime: 7423 LatSieveTime: 7441 -> makeJobFile(): Adjusted to q0=8800001, q1=8980000. -> client 1 q0: 8800001 [10/25 18:36:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 18:45:21] There were 179084/5630834 duplicates. [10/25 18:45:21] RelProcTime: 498.4 [10/25 18:45:33] largePrimes: 10659590 , relations: 6077330 [10/25 18:48:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 18:49:33] There were 627650/627650 duplicates. [10/25 18:49:33] RelProcTime: 11.8 [10/25 18:49:48] largePrimes: 10659590 , relations: 6077330 [10/25 18:50:10] GGNFS-0.77.1-VC8(UTE) : matbuild [10/25 18:50:46] largePrimes: 10659590 , relations: 6077330 [10/25 18:50:50] Heap stats for matbuild run. [10/25 18:50:50] Max heap usage: 0 MB [10/25 18:50:50] malloc/realloc errors: 0 [10/25 18:50:50] total malloc's : 130 [10/25 18:50:50] total realloc's: 93 [10/25 18:50:50] rels:6077330, initialFF:0, finalFF:137419 LatSieveTime: 7296 LatSieveTime: 7380 LatSieveTime: 7428 LatSieveTime: 7446 LatSieveTime: 7458 LatSieveTime: 7462 LatSieveTime: 7541 LatSieveTime: 7586 -> makeJobFile(): Adjusted to q0=8980001, q1=9160000. -> client 1 q0: 8980001 [10/25 19:58:56] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 20:00:42] There were 35249/620844 duplicates. [10/25 20:00:42] RelProcTime: 76.4 [10/25 20:01:09] largePrimes: 11350799 , relations: 6662925 [10/25 20:01:16] GGNFS-0.77.1-VC8(UTE) : matbuild [10/25 20:01:52] largePrimes: 11350799 , relations: 6662925 [10/25 20:01:56] Heap stats for matbuild run. [10/25 20:01:56] Max heap usage: 0 MB [10/25 20:01:56] malloc/realloc errors: 0 [10/25 20:01:56] total malloc's : 156 [10/25 20:01:56] total realloc's: 99 [10/25 20:01:56] rels:6662925, initialFF:0, finalFF:152100 [10/25 20:03:14] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 20:03:55] There were 620844/620844 duplicates. [10/25 20:03:55] RelProcTime: 10.5 [10/25 20:04:22] largePrimes: 11350799 , relations: 6662925 LatSieveTime: 7481 LatSieveTime: 7494 LatSieveTime: 7500 LatSieveTime: 7560 LatSieveTime: 7606 LatSieveTime: 7665 LatSieveTime: 7683 LatSieveTime: 7815 -> makeJobFile(): Adjusted to q0=9160001, q1=9340000. -> client 1 q0: 9160001 [10/25 22:09:19] GGNFS-0.77.1-VC8(UTE) : procrels [10/25 22:10:16] There were 37803/619049 duplicates. [10/25 22:10:16] RelProcTime: 55.7 [10/25 22:10:45] largePrimes: 11999695 , relations: 7244171 [10/25 22:10:55] GGNFS-0.77.1-VC8(UTE) : matbuild [10/25 22:11:40] largePrimes: 11999695 , relations: 7244171 [10/25 22:11:44] Heap stats for matbuild run. [10/25 22:12:55] Max heap usage: 0 MB [10/25 22:12:55] malloc/realloc errors: 0 [10/25 22:12:55] total malloc's : 197 [10/25 22:12:55] total realloc's: 109 [10/25 22:12:55] rels:7244171, initialFF:0, finalFF:167291 LatSieveTime: 8004 LatSieveTime: 8045 LatSieveTime: 8061 LatSieveTime: 8085 LatSieveTime: 8090 LatSieveTime: 8105 LatSieveTime: 8221 LatSieveTime: 8280 -> makeJobFile(): Adjusted to q0=9340001, q1=9520000. -> client 1 q0: 9340001 LatSieveTime: 7839 LatSieveTime: 7906 LatSieveTime: 7916 LatSieveTime: 7926 LatSieveTime: 7947 LatSieveTime: 7970 LatSieveTime: 7975 LatSieveTime: 7990 -> makeJobFile(): Adjusted to q0=9520001, q1=9700000. -> client 1 q0: 9520001 LatSieveTime: 7885 LatSieveTime: 7902 LatSieveTime: 7912 LatSieveTime: 7961 LatSieveTime: 7986 LatSieveTime: 8002 LatSieveTime: 8014 LatSieveTime: 8180 -> makeJobFile(): Adjusted to q0=9700001, q1=9880000. -> client 1 q0: 9700001 LatSieveTime: 7783 LatSieveTime: 7837 LatSieveTime: 7892 LatSieveTime: 7907 LatSieveTime: 7930 LatSieveTime: 8009 LatSieveTime: 8015 LatSieveTime: 8073 -> makeJobFile(): Adjusted to q0=9880001, q1=10060000. -> client 1 q0: 9880001 [10/26 08:25:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/26 08:29:23] There were 177338/2474429 duplicates. [10/26 08:29:23] RelProcTime: 213.8 [10/26 08:29:59] largePrimes: 14246969 , relations: 9541262 [10/26 08:31:45] GGNFS-0.77.1-VC8(UTE) : matbuild [10/26 08:32:35] largePrimes: 14246969 , relations: 9541262 [10/26 08:32:41] Heap stats for matbuild run. [10/26 08:32:41] Max heap usage: 0 MB [10/26 08:32:41] malloc/realloc errors: 0 [10/26 08:32:41] total malloc's : 199 [10/26 08:32:41] total realloc's: 131 [10/26 08:32:41] rels:9541262, initialFF:0, finalFF:232486 LatSieveTime: 7730 LatSieveTime: 7915 LatSieveTime: 7998 LatSieveTime: 7999 LatSieveTime: 8010 LatSieveTime: 8032 LatSieveTime: 8045 LatSieveTime: 8075 -> makeJobFile(): Adjusted to q0=10060001, q1=10240000. -> client 1 q0: 10060001 LatSieveTime: 8049 LatSieveTime: 8078 LatSieveTime: 8150 LatSieveTime: 8156 LatSieveTime: 8211 LatSieveTime: 8216 LatSieveTime: 8219 LatSieveTime: 8273 -> makeJobFile(): Adjusted to q0=10240001, q1=10420000. -> client 1 q0: 10240001 LatSieveTime: 8166 LatSieveTime: 8206 LatSieveTime: 8253 LatSieveTime: 8255 LatSieveTime: 8281 LatSieveTime: 8282 LatSieveTime: 8332 LatSieveTime: 8345 -> makeJobFile(): Adjusted to q0=10420001, q1=10600000. -> client 1 q0: 10420001 LatSieveTime: 8105 LatSieveTime: 8169 LatSieveTime: 8285 LatSieveTime: 8289 LatSieveTime: 8295 LatSieveTime: 8305 LatSieveTime: 8318 LatSieveTime: 8529 -> makeJobFile(): Adjusted to q0=10600001, q1=10780000. -> client 1 q0: 10600001 LatSieveTime: 8111 LatSieveTime: 8145 LatSieveTime: 8220 LatSieveTime: 8229 LatSieveTime: 8256 LatSieveTime: 8257 LatSieveTime: 8268 LatSieveTime: 8448 -> makeJobFile(): Adjusted to q0=10780001, q1=10960000. -> client 1 q0: 10780001 [10/26 19:27:03] GGNFS-0.77.1-VC8(UTE) : procrels [10/26 19:33:47] There were 272322/3065880 duplicates. [10/26 19:33:47] RelProcTime: 302.3 [10/26 19:34:44] largePrimes: 16446702 , relations: 12334819 [10/26 19:35:41] GGNFS-0.77.1-VC8(UTE) : matbuild [10/26 19:37:04] largePrimes: 16446702 , relations: 12334819 [10/26 19:37:14] Heap stats for matbuild run. [10/26 19:37:14] Max heap usage: 0 MB [10/26 19:37:14] malloc/realloc errors: 0 [10/26 19:37:14] total malloc's : 298 [10/26 19:37:14] total realloc's: 178 [10/26 19:37:14] rels:12334819, initialFF:0, finalFF:328130 LatSieveTime: 7933 LatSieveTime: 7986 LatSieveTime: 8020 LatSieveTime: 8070 LatSieveTime: 8105 LatSieveTime: 8107 LatSieveTime: 8259 LatSieveTime: 8387 -> makeJobFile(): Adjusted to q0=10960001, q1=11140000. -> client 1 q0: 10960001 -> makeJobFile(): Adjusted to q0=10960001, q1=11140000. -> client 1 q0: 10960001 LatSieveTime: 40947 LatSieveTime: 40988 LatSieveTime: 41201 LatSieveTime: 41225 LatSieveTime: 41259 LatSieveTime: 41284 -> makeJobFile(): Adjusted to q0=11140001, q1=11320000. -> client 1 q0: 11140001 LatSieveTime: 7805 LatSieveTime: 7867 LatSieveTime: 7869 LatSieveTime: 7887 LatSieveTime: 7914 LatSieveTime: 7962 -> makeJobFile(): Adjusted to q0=11320001, q1=11500000. -> client 1 q0: 11320001 -> makeJobFile(): Adjusted to q0=11320001, q1=11500000. -> client 1 q0: 11320001 [10/27 14:30:59] GGNFS-0.77.1-VC8(UTE) : procrels [10/27 14:36:15] There were 188710/1829554 duplicates. [10/27 14:36:15] RelProcTime: 197.4 [10/27 14:37:34] largePrimes: 17528360 , relations: 13975663 [10/27 15:56:25] GGNFS-0.77.1-VC8(UTE) : matbuild [10/27 15:58:02] largePrimes: 17528360 , relations: 13975663 [10/27 15:58:17] Heap stats for matbuild run. [10/27 15:58:17] Max heap usage: 0 MB [10/27 15:58:17] malloc/realloc errors: 0 [10/27 15:58:17] total malloc's : 382 [10/27 15:58:17] total realloc's: 215 [10/27 15:58:17] rels:13975663, initialFF:0, finalFF:400968 -> makeJobFile(): Adjusted to q0=11320001, q1=11500000. -> client 1 q0: 11320001 LatSieveTime: 11492 LatSieveTime: 11529 LatSieveTime: 11568 LatSieveTime: 11585 LatSieveTime: 11600 LatSieveTime: 11676 -> makeJobFile(): Adjusted to q0=11500001, q1=11680000. -> client 1 q0: 11500001 [10/27 22:34:15] GGNFS-0.77.1-VC8(UTE) : procrels [10/27 22:35:43] There were 1829554/1829554 duplicates. [10/27 22:35:43] RelProcTime: 81.2 [10/27 22:37:56] largePrimes: 17528360 , relations: 13975663 [10/27 22:41:44] GGNFS-0.77.1-VC8(UTE) : procrels [10/27 22:42:58] There were 66252/606012 duplicates. [10/27 22:42:58] RelProcTime: 73.0 [10/27 22:44:13] largePrimes: 17856924 , relations: 14515423 [10/27 22:46:10] GGNFS-0.77.1-VC8(UTE) : matbuild [10/27 22:47:56] largePrimes: 17856924 , relations: 14515423 [10/27 22:48:16] Heap stats for matbuild run. [10/27 22:48:16] Max heap usage: 0 MB [10/27 22:48:16] malloc/realloc errors: 0 [10/27 22:48:16] total malloc's : 411 [10/27 22:48:16] total realloc's: 235 [10/27 22:48:16] rels:14515423, initialFF:0, finalFF:429683 LatSieveTime: 10708 LatSieveTime: 10741 LatSieveTime: 10774 LatSieveTime: 10789 LatSieveTime: 10894 LatSieveTime: 10978 -> makeJobFile(): Adjusted to q0=11680001, q1=11860000. -> client 1 q0: 11680001 [10/28 01:48:01] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 01:52:53] There were 68545/608751 duplicates. [10/28 01:52:53] RelProcTime: 136.9 [10/28 01:55:24] largePrimes: 18172610 , relations: 15055628 [10/28 01:57:10] GGNFS-0.77.1-VC8(UTE) : matbuild [10/28 01:59:09] largePrimes: 18172610 , relations: 15055628 [10/28 01:59:31] Heap stats for matbuild run. [10/28 01:59:31] Max heap usage: 0 MB [10/28 01:59:31] malloc/realloc errors: 0 [10/28 01:59:31] total malloc's : 487 [10/28 01:59:31] total realloc's: 260 [10/28 01:59:31] rels:15055628, initialFF:0, finalFF:461804 LatSieveTime: 10545 LatSieveTime: 10729 LatSieveTime: 10732 LatSieveTime: 10770 LatSieveTime: 10805 LatSieveTime: 10841 -> makeJobFile(): Adjusted to q0=11860001, q1=12040000. -> client 1 q0: 11860001 LatSieveTime: 9729 LatSieveTime: 9787 LatSieveTime: 9850 LatSieveTime: 9854 LatSieveTime: 9873 LatSieveTime: 9887 -> makeJobFile(): Adjusted to q0=12040001, q1=12220000. -> client 1 q0: 12040001 LatSieveTime: 9015 LatSieveTime: 9089 LatSieveTime: 9322 LatSieveTime: 9355 LatSieveTime: 9413 LatSieveTime: 9419 -> makeJobFile(): Adjusted to q0=12220001, q1=12400000. -> client 1 q0: 12220001 [10/28 11:59:32] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 11:59:32] RelProcTime: 0.0 [10/28 12:01:01] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 12:04:34] There were 215714/1816455 duplicates. [10/28 12:04:34] RelProcTime: 210.3 [10/28 12:06:10] largePrimes: 19037517 , relations: 16656369 [10/28 12:06:53] GGNFS-0.77.1-VC8(UTE) : matbuild [10/28 12:09:03] largePrimes: 19037517 , relations: 16656369 [10/28 12:09:56] Heap stats for matbuild run. [10/28 12:09:56] Max heap usage: 0 MB [10/28 12:09:56] malloc/realloc errors: 0 [10/28 12:09:56] total malloc's : 1290 [10/28 12:09:56] total realloc's: 672 [10/28 12:09:56] rels:16656369, initialFF:0, finalFF:593088 LatSieveTime: 9163 LatSieveTime: 9664 LatSieveTime: 9744 LatSieveTime: 9798 LatSieveTime: 9870 LatSieveTime: 10090 -> makeJobFile(): Adjusted to q0=12400001, q1=12580000. -> client 1 q0: 12400001 [10/28 12:40:01] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 12:41:16] There were 75158/603774 duplicates. [10/28 12:41:16] RelProcTime: 73.3 [10/28 12:42:56] largePrimes: 19302131 , relations: 17184985 [10/28 12:49:15] GGNFS-0.77.1-VC8(UTE) : matbuild [10/28 12:51:27] largePrimes: 19302131 , relations: 17184985 [10/28 12:52:40] Heap stats for matbuild run. [10/28 12:52:40] Max heap usage: 0 MB [10/28 12:52:40] malloc/realloc errors: 0 [10/28 12:52:40] total malloc's : 1599 [10/28 12:52:40] total realloc's: 833 [10/28 12:52:40] rels:17184985, initialFF:0, finalFF:654941 LatSieveTime: 12351 LatSieveTime: 12689 LatSieveTime: 12716 LatSieveTime: 12782 LatSieveTime: 12799 LatSieveTime: 13169 -> makeJobFile(): Adjusted to q0=12580001, q1=12760000. -> client 1 q0: 12580001 [10/28 16:24:00] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 16:28:43] There were 76287/600738 duplicates. [10/28 16:28:43] RelProcTime: 81.2 [10/28 16:30:34] largePrimes: 19555249 , relations: 17709436 [10/28 16:30:44] GGNFS-0.77.1-VC8(UTE) : matbuild [10/28 16:33:15] largePrimes: 19555249 , relations: 17709436 [10/28 16:34:54] Heap stats for matbuild run. [10/28 16:34:54] Max heap usage: 0 MB [10/28 16:34:54] malloc/realloc errors: 0 [10/28 16:34:54] total malloc's : 1489 [10/28 16:34:54] total realloc's: 772 [10/28 16:34:54] rels:17709436, initialFF:0, finalFF:729159 LatSieveTime: 10717 LatSieveTime: 10859 LatSieveTime: 10936 LatSieveTime: 10937 LatSieveTime: 10973 LatSieveTime: 11043 -> makeJobFile(): Adjusted to q0=12760001, q1=12940000. -> client 1 q0: 12760001 LatSieveTime: 10463 LatSieveTime: 10563 LatSieveTime: 10576 LatSieveTime: 10577 LatSieveTime: 10653 LatSieveTime: 10734 -> makeJobFile(): Adjusted to q0=12940001, q1=13120000. -> client 1 q0: 12940001 [10/28 22:28:28] GGNFS-0.77.1-VC8(UTE) : procrels [10/28 22:32:37] There were 157916/1202909 duplicates. [10/28 22:32:37] RelProcTime: 247.1 [10/28 22:34:35] largePrimes: 20033122 , relations: 18754428 [10/28 22:34:58] GGNFS-0.77.1-VC8(UTE) : matbuild [10/28 22:37:44] largePrimes: 20033122 , relations: 18754428 [10/28 22:41:50] Heap stats for matbuild run. [10/28 22:41:50] Max heap usage: 0 MB [10/28 22:41:50] malloc/realloc errors: 0 [10/28 22:41:50] total malloc's : 1455 [10/28 22:41:50] total realloc's: 755 [10/28 22:41:50] rels:18754428, initialFF:0, finalFF:921630 [10/29 00:35:00] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 00:35:27] There were 658/658 duplicates. [10/29 00:35:27] RelProcTime: 26.3 [10/29 00:37:21] largePrimes: 20033122 , relations: 18754428 [10/29 00:37:38] GGNFS-0.77.1-VC8(UTE) : matbuild [10/29 00:40:36] largePrimes: 20033122 , relations: 18754428 [10/29 00:42:55] Heap stats for matbuild run. [10/29 00:42:55] Max heap usage: 0 MB [10/29 00:42:55] malloc/realloc errors: 0 [10/29 00:42:55] total malloc's : 1455 [10/29 00:42:55] total realloc's: 755 [10/29 00:42:55] rels:18754428, initialFF:0, finalFF:921630 LatSieveTime: 10641 LatSieveTime: 10683 LatSieveTime: 10688 LatSieveTime: 10726 LatSieveTime: 10799 LatSieveTime: 11022 -> makeJobFile(): Adjusted to q0=13120001, q1=13300000. -> client 1 q0: 13120001 LatSieveTime: 10667 LatSieveTime: 10719 LatSieveTime: 10725 LatSieveTime: 10800 LatSieveTime: 10865 LatSieveTime: 10869 -> makeJobFile(): Adjusted to q0=13300001, q1=13480000. -> client 1 q0: 13300001 LatSieveTime: 10653 LatSieveTime: 10673 LatSieveTime: 10718 LatSieveTime: 10736 LatSieveTime: 10777 LatSieveTime: 10814 -> makeJobFile(): Adjusted to q0=13480001, q1=13660000. -> client 1 q0: 13480001 [10/29 08:08:01] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 08:11:44] There were 246430/1786442 duplicates. [10/29 08:11:44] RelProcTime: 219.0 [10/29 08:13:45] largePrimes: 20679114 , relations: 20294440 [10/29 08:13:52] GGNFS-0.77.1-VC8(UTE) : matbuild [10/29 08:16:41] largePrimes: 20679114 , relations: 20294440 [10/29 08:20:27] Heap stats for matbuild run. [10/29 08:20:27] Max heap usage: 0 MB [10/29 08:20:27] malloc/realloc errors: 0 [10/29 08:20:27] total malloc's : 1365 [10/29 08:20:27] total realloc's: 709 [10/29 08:20:27] rels:20294440, initialFF:0, finalFF:1313267 LatSieveTime: 10795 LatSieveTime: 10828 LatSieveTime: 10869 LatSieveTime: 10936 LatSieveTime: 10983 LatSieveTime: 11048 -> makeJobFile(): Adjusted to q0=13660001, q1=13840000. -> client 1 q0: 13660001 LatSieveTime: 9531 LatSieveTime: 9632 LatSieveTime: 9690 LatSieveTime: 9757 LatSieveTime: 9832 LatSieveTime: 9832 -> makeJobFile(): Adjusted to q0=13840001, q1=14020000. -> client 1 q0: 13840001 LatSieveTime: 9204 LatSieveTime: 9220 LatSieveTime: 9301 LatSieveTime: 9390 LatSieveTime: 9417 LatSieveTime: 9445 -> makeJobFile(): Adjusted to q0=14020001, q1=14200000. -> client 1 q0: 14020001 [10/29 17:44:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 17:52:17] There were 258001/1782164 duplicates. [10/29 17:52:17] RelProcTime: 208.4 [10/29 17:54:27] largePrimes: 21255891 , relations: 21818601 [10/29 17:54:37] GGNFS-0.77.1-VC8(UTE) : matbuild [10/29 17:57:40] largePrimes: 21255891 , relations: 21818601 [10/29 18:06:57] reduceRelSets dropped relation-set weight from 32698936 to 26030795. [10/29 18:09:53] reduceRelSets dropped relation-set weight from 26030795 to 25992742. [10/29 18:09:53] After removing heavy rel-sets, weight is 19813286. [10/29 18:09:53] Heap stats for matbuild run. [10/29 18:09:53] Max heap usage: 0 MB [10/29 18:09:53] malloc/realloc errors: 0 [10/29 18:09:53] total malloc's : 1395 [10/29 18:09:53] total realloc's: 714 [10/29 18:09:53] rels:21818601, initialFF:0, finalFF:1696604 [10/29 18:11:30] GGNFS-0.77.1-VC8(UTE) : makefb [10/29 18:11:46] name: [10/29 18:11:46] n=55100796409547873505726910732839026411783896951704656274308614672145830600178494357451253742962116332186925549223018022410337811652503655098378132362717 (152 digits) [10/29 18:11:46] c0: -41 [10/29 18:11:46] c1: 0 [10/29 18:11:46] c2: 0 [10/29 18:11:46] c3: 0 [10/29 18:11:46] c4: 0 [10/29 18:11:46] c5: 1400 [10/29 18:11:46] RFBsize: 922193 (upto 14199989) [10/29 18:11:46] AFBsize: 908519 (upto 13999969) [10/29 18:11:46] maxNumLargeRatPrimes: 3 [10/29 18:11:46] maxLargeRatPrime: 268435456 [10/29 18:11:46] maxNumLargeAlgPrimes: 3 [10/29 18:11:46] maxLargeAlgPrime: 268435456 [10/29 18:20:55] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:21:38] There were 2067/627650 duplicates. [10/29 18:21:38] RelProcTime: 40.3 [10/29 18:21:39] largePrimes: 1517192 , relations: 625583 [10/29 18:21:44] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 9315 [10/29 18:29:03] There were 179084/5630834 duplicates. [10/29 18:29:03] RelProcTime: 421.0 [10/29 18:29:13] largePrimes: 10648347 , relations: 6077330 [10/29 18:30:01] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 9524 [10/29 18:32:14] There were 35249/620844 duplicates. [10/29 18:32:14] RelProcTime: 47.6 LatSieveTime: 9541 [10/29 18:32:26] largePrimes: 11339375 , relations: 6662925 LatSieveTime: 9556 [10/29 18:33:15] GGNFS-0.77.1-VC8(UTE) : procrels LatSieveTime: 9635 LatSieveTime: 9641 -> makeJobFile(): Adjusted to q0=14200001, q1=14380000. -> client 1 q0: 14200001 [10/29 18:34:47] There were 37803/619049 duplicates. [10/29 18:34:47] RelProcTime: 50.8 [10/29 18:35:02] largePrimes: 11988109 , relations: 7244171 [10/29 18:35:10] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:40:19] There were 177338/2474429 duplicates. [10/29 18:40:19] RelProcTime: 255.6 [10/29 18:41:05] largePrimes: 14235040 , relations: 9541262 [10/29 18:41:10] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:46:34] There were 272322/3065880 duplicates. [10/29 18:46:34] RelProcTime: 255.7 [10/29 18:47:27] largePrimes: 16434634 , relations: 12334819 [10/29 18:47:35] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:51:48] There were 188710/1829554 duplicates. [10/29 18:51:48] RelProcTime: 173.2 [10/29 18:52:57] largePrimes: 17516261 , relations: 13975663 [10/29 18:53:06] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:54:14] There were 66252/606012 duplicates. [10/29 18:54:14] RelProcTime: 64.9 [10/29 18:55:25] largePrimes: 17844821 , relations: 14515423 [10/29 18:55:32] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 18:58:50] There were 68545/608751 duplicates. [10/29 18:58:50] RelProcTime: 67.5 [10/29 19:00:12] largePrimes: 18160504 , relations: 15055628 [10/29 19:00:23] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 19:03:45] There were 215714/1816455 duplicates. [10/29 19:03:45] RelProcTime: 195.8 [10/29 19:05:14] largePrimes: 19025406 , relations: 16656369 [10/29 19:07:22] GGNFS-0.77.1-VC8(UTE) : procrels [10/29 19:19:34] There were 903331/6568572 duplicates. [10/29 19:19:34] RelProcTime: 706.2 [10/29 19:21:26] largePrimes: 21421678 , relations: 22321607 [10/29 19:23:37] GGNFS-0.77.1-VC8(UTE) : matbuild [10/29 19:26:14] largePrimes: 21421678 , relations: 22321607 [10/29 19:36:25] reduceRelSets dropped relation-set weight from 38516408 to 30437777. [10/29 19:39:52] reduceRelSets dropped relation-set weight from 30437777 to 30389382. [10/29 19:39:52] After removing heavy rel-sets, weight is 23486401. [10/29 19:48:04] Heap stats for matbuild run. [10/29 19:48:04] Max heap usage: 0 MB [10/29 19:48:04] malloc/realloc errors: 0 [10/29 19:48:04] total malloc's : 1337 [10/29 19:48:04] total realloc's: 679 [10/29 19:48:04] rels:22321607, initialFF:0, finalFF:1919842 [10/29 19:48:04] depinf file written. Run matprune. [10/29 19:48:27] GGNFS-0.77.1-VC8(UTE) : matprune [10/29 19:48:40] Pruning matrix with wt=0.700 [10/29 19:48:40] Initial matrix is 1830779 x 1919841 with sparse part having weight 253861759. [10/29 19:48:40] (total weight is 385477299) [10/29 19:59:47] Matrix pruned to 1803382 x 1812599 with weight 222611206. [10/29 19:59:58] Matrix is pruned. Run matsolve. [10/29 19:59:58] Heap stats for matprune run: [10/29 19:59:58] Max heap usage: 0 MB [10/29 19:59:58] malloc/realloc errors: 0 [10/29 19:59:58] total malloc's : 14 [10/29 19:59:58] total realloc's: 0 [10/29 20:00:24] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=1807880816) [10/29 23:46:15] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=1053353268) [10/31 05:48:30] BLanczosTime: 108121.5 [10/31 05:48:30] Heap stats for matsolve run: [10/31 05:48:30] Max heap usage: 0 MB [10/31 05:48:30] malloc/realloc errors: 0 [10/31 05:48:30] total malloc's : 12 [10/31 05:48:30] total realloc's: 0 [10/31 08:51:34] GGNFS-0.77.1-VC8(UTE) : sqrt [10/31 08:51:36] bmultiplier=20 [10/31 09:19:14] From dependence 0, sqrt obtained: [10/31 09:19:14] r1=750457255821884523130598577814932184632912548356261079022859 (pp60) [10/31 09:19:14] r2=73422964442128919638964497419330013227626433313870429164198871373341032307608541285834881463 (pp92) [10/31 09:19:14] (pp=probable prime, c=composite) [10/31 09:19:14] sqrtTime: 1659.1```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:09:51 UTC 2009 年 3 月 7 日 (土) 8 時 9 分 51 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 09:19:05 UTC 2010 年 9 月 26 日 (日) 18 時 19 分 5 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 09:19:20 UTC 2010 年 9 月 26 日 (日) 18 時 19 分 20 秒 (日本時間)
403e62144110Ignacio SantosSeptember 26, 2010 09:19:20 UTC 2010 年 9 月 26 日 (日) 18 時 19 分 20 秒 (日本時間)
2034Pawel ApostolOctober 4, 2012 16:43:41 UTC 2012 年 10 月 5 日 (金) 1 時 43 分 41 秒 (日本時間)
4511e632Ignacio SantosSeptember 26, 2010 09:19:20 UTC 2010 年 9 月 26 日 (日) 18 時 19 分 20 秒 (日本時間)
5043e62578 / 7463778apoOctober 14, 2012 14:37:15 UTC 2012 年 10 月 14 日 (日) 23 時 37 分 15 秒 (日本時間)
1800apoOctober 17, 2012 22:12:16 UTC 2012 年 10 月 18 日 (木) 7 時 12 分 16 秒 (日本時間)
14×10199-419
c175
name 名前 Serge Batalov October 21, 2008 07:30:23 UTC 2008 年 10 月 21 日 (火) 16 時 30 分 23 秒 (日本時間) `2888593773388598163263708212746244530413575744205044102462119203482209689452341197946715597603602687304850906910991431365199840709380312375827360928063823410216618927214589497<175>` `93342682385042014936376105033<29>` `30946119177002457256257679064752877429329252807104046662651498568475129246685538584195558365545659808057878297832367685917174350969470564322807409<146>` ```Using B1=3000000, B2=11414255590, polynomial Dickson(12), sigma=3253425714 Step 1 took 28821ms Step 2 took 18723ms ********** Factor found in step 2: 93342682385042014936376105033 Found probable prime factor of 29 digits: 93342682385042014936376105033 Composite cofactor 30946119177002457256257679064752877429329252807104046662651498568475129246685538584195558365545659808057878297832367685917174350969470564322807409 has 146 digits``` `GMP-ECM 6.2.1`
c146
name 名前 Pawel Apostol October 2, 2012 07:16:58 UTC 2012 年 10 月 2 日 (火) 16 時 16 分 58 秒 (日本時間) `30946119177002457256257679064752877429329252807104046662651498568475129246685538584195558365545659808057878297832367685917174350969470564322807409<146>` `824789929518275606354968148456179<33>` `37520001238468996708555838724021582916031204217399954248169904353694100250544879496795878786149349124520726735371<113>` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=2606282689 Step 1 took 60388ms Step 2 took 17176ms ********** Factor found in step 2: 824789929518275606354968148456179 Found probable prime factor of 33 digits: 824789929518275606354968148456179 Composite cofactor 37520001238468996708555838724021582916031204217399954248169904353694100250544879496795878786149349124520726735371 has 113 digits``` `GMP-ECM 6.4.3`
c113
name 名前 Pawel Apostol October 12, 2012 19:00:10 UTC 2012 年 10 月 13 日 (土) 4 時 0 分 10 秒 (日本時間) `37520001238468996708555838724021582916031204217399954248169904353694100250544879496795878786149349124520726735371<113>` ```35935157536142967989605044657709419838606709311522126889<56> 1044102873369402095385581358668033655912741249962371480339<58>``` ```-> makeJobFile(): Adjusted to q0=1750000, q1=1850000. -> client 1 q0: 1750000 -> makeJobFile(): Adjusted to q0=1750183, q1=1850000. -> makeJobFile(): Adjusted to q0=1750183, q1=1850000. -> client 1 q0: 1750183 LatSieveTime: 2265 LatSieveTime: 2290 LatSieveTime: 2302 LatSieveTime: 2359 -> makeJobFile(): Adjusted to q0=1850001, q1=1950000. -> client 1 q0: 1850001 LatSieveTime: 2209 LatSieveTime: 2236 LatSieveTime: 2270 LatSieveTime: 2417 -> makeJobFile(): Adjusted to q0=1950001, q1=2050000. -> client 1 q0: 1950001 LatSieveTime: 2735 LatSieveTime: 2794 LatSieveTime: 2869 LatSieveTime: 2986 -> makeJobFile(): Adjusted to q0=2050001, q1=2150000. -> client 1 q0: 2050001 LatSieveTime: 2754 LatSieveTime: 2963 LatSieveTime: 3118 LatSieveTime: 3148 -> makeJobFile(): Adjusted to q0=2150001, q1=2250000. -> client 1 q0: 2150001 [10/11 22:46:41] GGNFS-0.77.1-VC8(UTE) : makefb [10/11 22:46:58] name: 15551_199 [10/11 22:46:58] n=37520001238468996708555838724021582916031204217399954248169904353694100250544879496795878786149349124520726735371 (113 digits) [10/11 22:46:58] c0: 5611183967166046261917576750 [10/11 22:46:58] c1: -140321958894425839176945 [10/11 22:46:58] c2: -4367426202921211278 [10/11 22:46:58] c3: 24632793011390 [10/11 22:46:58] c4: 1324640242 [10/11 22:46:58] c5: 2220 [10/11 22:46:58] RFBsize: 250150 (upto 3499999) [10/11 22:46:58] AFBsize: 250157 (upto 3499973) [10/11 22:46:58] maxNumLargeRatPrimes: 3 [10/11 22:46:58] maxLargeRatPrime: 134217728 [10/11 22:46:58] maxNumLargeAlgPrimes: 3 [10/11 22:46:58] maxLargeAlgPrime: 134217728 [10/11 22:47:52] GGNFS-0.77.1-VC8(UTE) : procrels [10/11 22:47:55] DISC_P_DIV=9684297179 [10/11 22:57:56] There were 96969/3408184 duplicates. [10/11 22:57:56] RelProcTime: 577.7 [10/11 22:58:02] largePrimes: 4693906 , relations: 3311215 [10/11 22:58:10] GGNFS-0.77.1-VC8(UTE) : procrels [10/11 22:58:39] There were 3408184/3408184 duplicates. [10/11 22:58:39] RelProcTime: 19.3 [10/11 22:58:44] largePrimes: 4693906 , relations: 3311215 [10/11 22:58:46] GGNFS-0.77.1-VC8(UTE) : procrels [10/11 22:59:56] There were 3408184/3408184 duplicates. [10/11 22:59:56] RelProcTime: 51.1 [10/11 23:00:08] largePrimes: 4693906 , relations: 3311215 [10/11 23:00:53] GGNFS-0.77.1-VC8(UTE) : matbuild [10/11 23:01:11] largePrimes: 4693906 , relations: 3311215 [10/11 23:01:15] Heap stats for matbuild run. [10/11 23:01:15] Max heap usage: 0 MB [10/11 23:01:15] malloc/realloc errors: 0 [10/11 23:01:15] total malloc's : 175 [10/11 23:01:15] total realloc's: 73 [10/11 23:01:15] rels:3311215, initialFF:0, finalFF:88589 LatSieveTime: 2602 LatSieveTime: 2702 LatSieveTime: 2804 LatSieveTime: 3205 -> makeJobFile(): Adjusted to q0=2250001, q1=2350000. -> client 1 q0: 2250001 [10/11 23:39:35] GGNFS-0.77.1-VC8(UTE) : procrels [10/11 23:40:52] There were 50401/859197 duplicates. [10/11 23:40:52] RelProcTime: 74.1 [10/11 23:41:01] largePrimes: 5386437 , relations: 4120011 [10/11 23:41:10] GGNFS-0.77.1-VC8(UTE) : matbuild [10/11 23:41:30] largePrimes: 5386437 , relations: 4120011 [10/11 23:41:33] Heap stats for matbuild run. [10/11 23:41:33] Max heap usage: 0 MB [10/11 23:41:33] malloc/realloc errors: 0 [10/11 23:41:33] total malloc's : 231 [10/11 23:41:33] total realloc's: 101 [10/11 23:41:33] rels:4120011, initialFF:0, finalFF:123614 LatSieveTime: 2091 LatSieveTime: 2167 LatSieveTime: 2264 LatSieveTime: 2275 -> makeJobFile(): Adjusted to q0=2350001, q1=2450000. -> client 1 q0: 2350001 [10/12 00:16:53] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 00:18:16] There were 57904/836328 duplicates. [10/12 00:18:16] RelProcTime: 77.5 [10/12 00:18:37] largePrimes: 5974183 , relations: 4898435 [10/12 00:18:43] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 00:19:00] largePrimes: 5974183 , relations: 4898435 [10/12 00:19:04] Heap stats for matbuild run. [10/12 00:19:04] Max heap usage: 0 MB [10/12 00:19:04] malloc/realloc errors: 0 [10/12 00:19:04] total malloc's : 306 [10/12 00:19:04] total realloc's: 133 [10/12 00:19:04] rels:4898435, initialFF:0, finalFF:172413 LatSieveTime: 2155 LatSieveTime: 2192 LatSieveTime: 2200 LatSieveTime: 2295 -> makeJobFile(): Adjusted to q0=2450001, q1=2550000. -> client 1 q0: 2450001 [10/12 00:55:49] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 00:57:07] There were 67964/858287 duplicates. [10/12 00:57:07] RelProcTime: 76.3 [10/12 00:57:18] largePrimes: 6508764 , relations: 5688758 [10/12 00:57:25] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 00:57:44] largePrimes: 6508764 , relations: 5688758 [10/12 00:57:53] Heap stats for matbuild run. [10/12 00:57:53] Max heap usage: 0 MB [10/12 00:57:53] malloc/realloc errors: 0 [10/12 00:57:53] total malloc's : 539 [10/12 00:57:53] total realloc's: 253 [10/12 00:57:53] rels:5688758, initialFF:0, finalFF:257218 LatSieveTime: 2136 LatSieveTime: 2161 LatSieveTime: 2187 LatSieveTime: 2273 -> makeJobFile(): Adjusted to q0=2550001, q1=2650000. -> client 1 q0: 2550001 [10/12 01:32:47] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:35:24] There were 76766/863079 duplicates. [10/12 01:35:24] RelProcTime: 90.6 [10/12 01:35:39] largePrimes: 6988453 , relations: 6475071 [10/12 01:35:47] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 01:36:28] largePrimes: 6988453 , relations: 6475071 [10/12 01:36:46] Heap stats for matbuild run. [10/12 01:36:46] Max heap usage: 0 MB [10/12 01:36:46] malloc/realloc errors: 0 [10/12 01:36:46] total malloc's : 482 [10/12 01:36:46] total realloc's: 221 [10/12 01:36:46] rels:6475071, initialFF:0, finalFF:410027 [10/12 01:43:41] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:44:13] There were 21727/230941 duplicates. [10/12 01:44:13] RelProcTime: 30.2 [10/12 01:44:27] largePrimes: 7108708 , relations: 6684282 [10/12 01:44:39] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 01:45:01] largePrimes: 7108708 , relations: 6684282 [10/12 01:45:14] Heap stats for matbuild run. [10/12 01:45:14] Max heap usage: 0 MB [10/12 01:45:14] malloc/realloc errors: 0 [10/12 01:45:14] total malloc's : 496 [10/12 01:45:14] total realloc's: 228 [10/12 01:45:14] rels:6684282, initialFF:0, finalFF:464034 [10/12 01:46:22] GGNFS-0.77.1-VC8(UTE) : makefb [10/12 01:46:26] name: 15551_199 [10/12 01:46:26] n=37520001238468996708555838724021582916031204217399954248169904353694100250544879496795878786149349124520726735371 (113 digits) [10/12 01:46:26] c0: 5611183967166046261917576750 [10/12 01:46:26] c1: -140321958894425839176945 [10/12 01:46:26] c2: -4367426202921211278 [10/12 01:46:26] c3: 24632793011390 [10/12 01:46:26] c4: 1324640242 [10/12 01:46:26] c5: 2220 [10/12 01:46:26] RFBsize: 250150 (upto 3499999) [10/12 01:46:26] AFBsize: 196337 (upto 2699981) [10/12 01:46:26] maxNumLargeRatPrimes: 3 [10/12 01:46:26] maxLargeRatPrime: 134217728 [10/12 01:46:26] maxNumLargeAlgPrimes: 3 [10/12 01:46:26] maxLargeAlgPrime: 134217728 [10/12 01:47:03] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:50:19] There were 96969/3408184 duplicates. [10/12 01:50:19] RelProcTime: 182.6 [10/12 01:50:22] largePrimes: 4747624 , relations: 3311215 [10/12 01:50:29] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:51:22] There were 50401/859197 duplicates. [10/12 01:51:22] RelProcTime: 48.8 [10/12 01:51:27] largePrimes: 5440239 , relations: 4120011 [10/12 01:51:32] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:52:32] There were 57904/836328 duplicates. [10/12 01:52:32] RelProcTime: 48.7 [10/12 01:52:41] largePrimes: 6028000 , relations: 4898435 [10/12 01:52:48] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 01:53:02] largePrimes: 6028000 , relations: 4898435 [10/12 01:53:06] Heap stats for matbuild run. [10/12 01:53:06] Max heap usage: 0 MB [10/12 01:53:06] malloc/realloc errors: 0 [10/12 01:53:06] total malloc's : 341 [10/12 01:53:06] total realloc's: 148 [10/12 01:53:06] rels:4898435, initialFF:0, finalFF:133316 [10/12 01:53:23] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:54:14] There were 67964/858287 duplicates. [10/12 01:54:14] RelProcTime: 49.7 [10/12 01:54:23] largePrimes: 6562584 , relations: 5688758 [10/12 01:54:29] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:55:41] There were 76766/863079 duplicates. [10/12 01:55:41] RelProcTime: 52.4 [10/12 01:55:54] largePrimes: 7042273 , relations: 6475071 [10/12 01:55:57] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 01:56:14] There were 21727/230941 duplicates. [10/12 01:56:14] RelProcTime: 15.5 [10/12 01:56:27] largePrimes: 7162528 , relations: 6684282 [10/12 01:56:31] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 01:56:53] largePrimes: 7162528 , relations: 6684282 [10/12 01:57:09] Heap stats for matbuild run. [10/12 01:57:09] Max heap usage: 0 MB [10/12 01:57:09] malloc/realloc errors: 0 [10/12 01:57:09] total malloc's : 557 [10/12 01:57:09] total realloc's: 261 [10/12 01:57:09] rels:6684282, initialFF:0, finalFF:404161 -> makeJobFile(): Adjusted to q0=2550001, q1=2650000. -> client 1 q0: 2550001 [10/12 02:08:32] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 02:08:42] There were 232955/252075 duplicates. [10/12 02:08:42] RelProcTime: 8.1 [10/12 02:09:04] largePrimes: 7173353 , relations: 6703398 [10/12 02:09:12] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 02:09:41] largePrimes: 7173353 , relations: 6703398 [10/12 02:10:01] Heap stats for matbuild run. [10/12 02:10:01] Max heap usage: 0 MB [10/12 02:10:01] malloc/realloc errors: 0 [10/12 02:10:01] total malloc's : 533 [10/12 02:10:01] total realloc's: 249 [10/12 02:10:01] rels:6703398, initialFF:0, finalFF:409491 [10/12 02:16:06] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 02:16:32] There were 268549/423053 duplicates. [10/12 02:16:32] RelProcTime: 22.8 [10/12 02:17:04] largePrimes: 7260026 , relations: 6857898 [10/12 02:17:11] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 02:17:41] largePrimes: 7260026 , relations: 6857898 [10/12 02:19:07] reduceRelSets dropped relation-set weight from 5801704 to 5568478. [10/12 02:19:07] After removing heavy rel-sets, weight is 4158860. [10/12 02:19:07] Heap stats for matbuild run. [10/12 02:19:07] Max heap usage: 0 MB [10/12 02:19:07] malloc/realloc errors: 0 [10/12 02:19:07] total malloc's : 468 [10/12 02:19:07] total realloc's: 219 [10/12 02:19:07] rels:6857898, initialFF:0, finalFF:425223 [10/12 02:22:21] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 02:22:49] There were 429876/492898 duplicates. [10/12 02:22:49] RelProcTime: 26.6 [10/12 02:23:04] largePrimes: 7295138 , relations: 6920916 [10/12 02:23:08] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 02:23:33] largePrimes: 7295138 , relations: 6920916 [10/12 02:24:10] reduceRelSets dropped relation-set weight from 6151303 to 5887586. [10/12 02:24:10] After removing heavy rel-sets, weight is 4414834. [10/12 02:24:10] Heap stats for matbuild run. [10/12 02:24:10] Max heap usage: 0 MB [10/12 02:24:10] malloc/realloc errors: 0 [10/12 02:24:10] total malloc's : 521 [10/12 02:24:10] total realloc's: 243 [10/12 02:24:10] rels:6920916, initialFF:0, finalFF:442430 Parameters change detected. [10/12 02:26:55] GGNFS-0.77.1-VC8(UTE) : procrels -> makeJobFile(): Adjusted to q0=2550001, q1=2600000. -> client 1 q0: 2550001 -> makeJobFile(): Adjusted to q0=2550089, q1=2600000. -> makeJobFile(): Adjusted to q0=2650089, q1=2700000. -> client 1 q0: 2650089 [10/12 02:32:57] GGNFS-0.77.1-VC8(UTE) : procrels [10/12 02:33:13] There were 2186/21793 duplicates. [10/12 02:33:13] RelProcTime: 14.7 [10/12 02:33:30] largePrimes: 7305840 , relations: 6940522 [10/12 02:33:33] GGNFS-0.77.1-VC8(UTE) : matbuild [10/12 02:33:59] largePrimes: 7305840 , relations: 6940522 [10/12 02:34:41] reduceRelSets dropped relation-set weight from 6263447 to 5988676. [10/12 02:34:41] After removing heavy rel-sets, weight is 4495317. -> makeJobFile(): Adjusted to q0=2654543, q1=2700000. [10/12 02:35:31] Heap stats for matbuild run. [10/12 02:35:31] Max heap usage: 0 MB [10/12 02:35:31] malloc/realloc errors: 0 [10/12 02:35:31] total malloc's : 527 [10/12 02:35:31] total realloc's: 227 [10/12 02:35:31] rels:6940522, initialFF:0, finalFF:447871 [10/12 02:35:31] depinf file written. Run matprune. [10/12 02:36:01] GGNFS-0.77.1-VC8(UTE) : matprune [10/12 02:36:03] Pruning matrix with wt=0.700 [10/12 02:36:03] Initial matrix is 446571 x 447871 with sparse part having weight 45542400. [10/12 02:36:03] (total weight is 73890044) [10/12 02:36:04] Matrix pruned to 443499 x 444799 with weight 45741276. [10/12 02:36:10] Matrix is pruned. Run matsolve. [10/12 02:36:10] Heap stats for matprune run: [10/12 02:36:10] Max heap usage: 0 MB [10/12 02:36:10] malloc/realloc errors: 0 [10/12 02:36:10] total malloc's : 13 [10/12 02:36:10] total realloc's: 0 [10/12 02:36:38] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=3993005452) [10/12 08:04:24] GGNFS-0.77.1-VC8(UTE) : matsolve (seed=257957576) [10/12 08:34:58] BLanczosTime: 1831.3 [10/12 08:34:58] Heap stats for matsolve run: [10/12 08:34:58] Max heap usage: 0 MB [10/12 08:34:58] malloc/realloc errors: 0 [10/12 08:34:58] total malloc's : 11 [10/12 08:34:58] total realloc's: 0 [10/12 19:45:46] GGNFS-0.77.1-VC8(UTE) : sqrt [10/12 19:45:48] bmultiplier=2 [10/12 19:55:28] From dependence 0, sqrt obtained: [10/12 19:55:28] r1=35935157536142967989605044657709419838606709311522126889 (pp56) [10/12 19:55:28] r2=1044102873369402095385581358668033655912741249962371480339 (pp58) [10/12 19:55:28] (pp=probable prime, c=composite) [10/12 19:55:28] sqrtTime: 582.0```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:10:30 UTC 2009 年 3 月 7 日 (土) 8 時 10 分 30 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 09:20:06 UTC 2010 年 9 月 26 日 (日) 18 時 20 分 6 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 09:20:20 UTC 2010 年 9 月 26 日 (日) 18 時 20 分 20 秒 (日本時間)
403e6110 / 2144Ignacio SantosSeptember 26, 2010 09:20:20 UTC 2010 年 9 月 26 日 (日) 18 時 20 分 20 秒 (日本時間)
4511e632 / 4441Ignacio SantosSeptember 26, 2010 09:20:20 UTC 2010 年 9 月 26 日 (日) 18 時 20 分 20 秒 (日本時間)
14×10200-419
c180
name 名前 Robert Backstrom September 23, 2012 13:25:27 UTC 2012 年 9 月 23 日 (日) 22 時 25 分 27 秒 (日本時間) `480214762492167420793787400491431084313934014485359873574914042071447247239418688432759258453760822240814313777809628357286069747561190094419577627058883782684992011645476546026049<180>` ```4877097562737991821538593892177796992063947325757979053800763353284653<70> 98463226604508586379055660475243392593618405451073303028847639624158258692336070291158185670651352528292002533<110>``` ```Number: n N=480214762492167420793787400491431084313934014485359873574914042071447247239418688432759258453760822240814313777809628357286069747561190094419577627058883782684992011645476546026049 ( 180 digits) SNFS difficulty: 201 digits. Divisors found: Sun Sep 23 23:20:31 2012 prp70 factor: 4877097562737991821538593892177796992063947325757979053800763353284653 Sun Sep 23 23:20:31 2012 prp110 factor: 98463226604508586379055660475243392593618405451073303028847639624158258692336070291158185670651352528292002533 Sun Sep 23 23:20:31 2012 elapsed time 06:04:02 (Msieve 1.44 - dependency 9) Version: GGNFS-0.77.1-20060513-nocona Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.086). Factorization parameters were as follows: name: KA_15551_200 n: 480214762492167420793787400491431084313934014485359873574914042071447247239418688432759258453760822240814313777809628357286069747561190094419577627058883782684992011645476546026049 m: 10000000000000000000000000000000000000000 # c180, diff: 201.15 skew: 1.24 deg: 5 c5: 14 c0: -41 # Murphy_E = 1.765e-11 type: snfs lss: 1 rlim: 15800000 alim: 15800000 lpbr: 29 lpba: 29 mfbr: 56 mfba: 56 rlambda: 2.6 alambda: 2.6 qintsize: 100000 Factor base limits: 15800000/15800000 Large primes per side: 3 Large prime bits: 29/29 Max factor residue bits: 56/56 Sieved special-q in [100000, 7900000) Primes: RFBsize:1019012, AFBsize:1019168, Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 8216761 hash collisions in 66076967 relations (59479609 unique) Msieve: matrix is 1702798 x 1703025 (480.4 MB) Total sieving time: 0.00 hours. Total relation processing time: 0.00 hours. Matrix solve time: 0.00 hours. Total square root time: 0.00 hours, sqrts: 0. Prototype def-par.txt line would be: snfs,201,5,0,0,0,0,0,0,0,0,15800000,15800000,29,29,56,56,2.6,2.6,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- CPU0: Intel(R) Core(TM)2 Quad CPU Q9550 @ 2.83GHz stepping 07 CPU1: Intel(R) Core(TM)2 Quad CPU Q9550 @ 2.83GHz stepping 07 CPU2: Intel(R) Core(TM)2 Quad CPU Q9550 @ 2.83GHz stepping 07 CPU3: Intel(R) Core(TM)2 Quad CPU Q9550 @ 2.83GHz stepping 07 Memory: 8109188k/9175040k available (3972k kernel code, 787464k absent, 278388k reserved, 2498k data, 1292k init) Calibrating delay loop (skipped), value calculated using timer frequency.. 5661.42 BogoMIPS (lpj=2830713) Calibrating delay using timer specific routine.. 5660.89 BogoMIPS (lpj=2830448) Calibrating delay using timer specific routine.. 5660.91 BogoMIPS (lpj=2830455) Calibrating delay using timer specific routine.. 5660.92 BogoMIPS (lpj=2830461) Total of 4 processors activated (22644.15 BogoMIPS).```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
2011e374Max DettweilerMarch 6, 2009 23:11:11 UTC 2009 年 3 月 7 日 (土) 8 時 11 分 11 秒 (日本時間)
255e40--
3025e4430Ignacio SantosSeptember 26, 2010 10:03:42 UTC 2010 年 9 月 26 日 (日) 19 時 3 分 42 秒 (日本時間)
351e6300Ignacio SantosSeptember 26, 2010 10:03:56 UTC 2010 年 9 月 26 日 (日) 19 時 3 分 56 秒 (日本時間)
403e6110Ignacio SantosSeptember 26, 2010 10:03:56 UTC 2010 年 9 月 26 日 (日) 19 時 3 分 56 秒 (日本時間)
4511e6632 / 444132Ignacio SantosSeptember 26, 2010 10:03:56 UTC 2010 年 9 月 26 日 (日) 19 時 3 分 56 秒 (日本時間)
600Dmitry DomanovFebruary 23, 2011 10:40:10 UTC 2011 年 2 月 23 日 (水) 19 時 40 分 10 秒 (日本時間)
14×10201-419
c200
name 名前 Dmitry Domanov October 30, 2012 06:44:26 UTC 2012 年 10 月 30 日 (火) 15 時 44 分 26 秒 (日本時間) `47138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047138047<200>` ```4487595935769585276123410306176624361461<40> 10504075637095735112689582825029595511637956613355484445782226998159322056607575172227934330401636195206933515257550099369321483206005339916052803948831310782627<161>``` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=487747711 Step 1 took 37336ms Step 2 took 12445ms ********** Factor found in step 2: 4487595935769585276123410306176624361461 Found probable prime factor of 40 digits: 4487595935769585276123410306176624361461```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:35:59 UTC 2012 年 10 月 28 日 (日) 22 時 35 分 59 秒 (日本時間)
14×10202-419
c187
name 名前 Warut Roonguthai October 27, 2012 03:18:43 UTC 2012 年 10 月 27 日 (土) 12 時 18 分 43 秒 (日本時間) `5233116555127333000186156104747360490245783688366975370763317063372682199635471794360645484345799819398950935858746412856709791562881420913061349011136309337724431134913497806611408469177<187>` `2473242123634150426508661466391<31>` `2115893347084780478520704579444794725489172820416404142979788609676390712523803315231680632768685112118529858948795206523930676439865988697232738011531996847<157>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=900751127 Step 1 took 10546ms Step 2 took 6536ms ********** Factor found in step 2: 2473242123634150426508661466391 Found probable prime factor of 31 digits: 2473242123634150426508661466391 Composite cofactor 2115893347084780478520704579444794725489172820416404142979788609676390712523803315231680632768685112118529858948795206523930676439865988697232738011531996847 has 157 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:36:13 UTC 2012 年 10 月 28 日 (日) 22 時 36 分 13 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:03:02 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 2 秒 (日本時間)
4511e630001000Dmitry DomanovNovember 2, 2012 07:03:02 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 2 秒 (日本時間)
1000Ignacio SantosOctober 17, 2021 17:51:12 UTC 2021 年 10 月 18 日 (月) 2 時 51 分 12 秒 (日本時間)
1000Ignacio SantosOctober 18, 2021 13:20:07 UTC 2021 年 10 月 18 日 (月) 22 時 20 分 7 秒 (日本時間)
5043e6500 / 6836Dmitry DomanovNovember 8, 2012 09:02:44 UTC 2012 年 11 月 8 日 (木) 18 時 2 分 44 秒 (日本時間)
14×10203-419
c187
name 名前 Warut Roonguthai October 27, 2012 02:57:43 UTC 2012 年 10 月 27 日 (土) 11 時 57 分 43 秒 (日本時間) `7887050187339830637918088010494985273344905501027843191364849906846077889626080015023091916961404138339374452278078696714791733477536939088438851630696571726955893181416283089659210540757<187>` `11055045869657519809722384960461<32>` `713434415409093892095989425128467439060502542067377499874346865286594617775655879454999908680815170848397420422895649137868643306672712157586988377825021737<156>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=1719024382 Step 1 took 10468ms Step 2 took 6521ms ********** Factor found in step 2: 11055045869657519809722384960461 Found probable prime factor of 32 digits: 11055045869657519809722384960461 Composite cofactor 713434415409093892095989425128467439060502542067377499874346865286594617775655879454999908680815170848397420422895649137868643306672712157586988377825021737 has 156 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:36:21 UTC 2012 年 10 月 28 日 (日) 22 時 36 分 21 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:03:12 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 12 秒 (日本時間)
4511e654801000Dmitry DomanovNovember 2, 2012 07:03:12 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 12 秒 (日本時間)
4480Ignacio SantosDecember 13, 2021 19:57:59 UTC 2021 年 12 月 14 日 (火) 4 時 57 分 59 秒 (日本時間)
5043e6500 / 6279Dmitry DomanovNovember 8, 2012 09:02:31 UTC 2012 年 11 月 8 日 (木) 18 時 2 分 31 秒 (日本時間)
14×10206-419
c184
name 名前 Bob Backstrom December 8, 2021 06:48:01 UTC 2021 年 12 月 8 日 (水) 15 時 48 分 1 秒 (日本時間) `2591252327449328928078162018919917291587631397255314171084358812607798397357341696658854713327983473186273073969605836349661099021858576636153361739720898500662478224168655660776588923<184>` ```36258103132509359445638611376513789762912778073271286817214413631<65> 71466847506591912527264118948897906389941024569512794529589324704073763509418982277183461571797877937403136852321096133<119>``` ```Number: n N=2591252327449328928078162018919917291587631397255314171084358812607798397357341696658854713327983473186273073969605836349661099021858576636153361739720898500662478224168655660776588923 ( 184 digits) SNFS difficulty: 207 digits. Divisors found: Wed Dec 8 16:46:13 2021 p65 factor: 36258103132509359445638611376513789762912778073271286817214413631 Wed Dec 8 16:46:13 2021 p119 factor: 71466847506591912527264118948897906389941024569512794529589324704073763509418982277183461571797877937403136852321096133 Wed Dec 8 16:46:13 2021 elapsed time 02:21:23 (Msieve 1.54 - dependency 1) Version: Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.342). Factorization parameters were as follows: # # N = 14x10^206-41 = 15(205)1 # n: 2591252327449328928078162018919917291587631397255314171084358812607798397357341696658854713327983473186273073969605836349661099021858576636153361739720898500662478224168655660776588923 m: 100000000000000000000000000000000000000000 deg: 5 c5: 140 c0: -41 skew: 0.78 # Murphy_E = 8.254e-12 type: snfs lss: 1 rlim: 19800000 alim: 19800000 lpbr: 29 lpba: 29 mfbr: 56 mfba: 56 rlambda: 2.6 alambda: 2.6 Factor base limits: 19800000/19800000 Large primes per side: 3 Large prime bits: 29/29 Max factor residue bits: 56/56 Sieved special-q in [100000, 58700000) Primes: , , Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 11195299 hash collisions in 73816175 relations (64016636 unique) Msieve: matrix is 2468568 x 2468795 (841.8 MB) Sieving start time : 2021/12/07 15:23:44 Sieving end time : 2021/12/08 14:23:13 Total sieving time: 22hrs 59min 29secs. Total relation processing time: 1hrs 59min 51sec. Matrix solve time: 0.00 hours. Total square root time: 0hrs 2min 52sec. Prototype def-par.txt line would be: snfs,207,5,0,0,0,0,0,0,0,0,19800000,19800000,29,29,56,56,2.6,2.6,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- [ 0.118965] smpboot: CPU0: AMD Ryzen 7 2700 Eight-Core Processor (family: 0x17, model: 0x8, stepping: 0x2) [ 0.000000] Memory: 16239956K/16727236K available (14339K kernel code, 2400K rwdata, 5020K rodata, 2736K init, 4964K bss, 487280K reserved, 0K cma-reserved) [ 0.154164] x86/mm: Memory block size: 128MB [ 0.000004] Calibrating delay loop (skipped), value calculated using timer frequency.. 6399.06 BogoMIPS (lpj=12798120) [ 0.152035] smpboot: Total of 16 processors activated (102384.96 BogoMIPS)```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:36:38 UTC 2012 年 10 月 28 日 (日) 22 時 36 分 38 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:03:24 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 24 秒 (日本時間)
4511e61000 / 4208Dmitry DomanovNovember 2, 2012 07:03:24 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 24 秒 (日本時間)
14×10207-419
c156
composite cofactor 合成数の残り `114795728096964659542139088162053816936429947671799954407727681892758503597593245862875515949865683900295760673045217597136989460096026595365218498929280013<156>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:36:49 UTC 2012 年 10 月 28 日 (日) 22 時 36 分 49 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:03:34 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 34 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:03:34 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 34 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:02:14 UTC 2012 年 11 月 8 日 (木) 18 時 2 分 14 秒 (日本時間)
14×10208-419
c192
name 名前 Dmitry Domanov October 30, 2012 06:43:54 UTC 2012 年 10 月 30 日 (火) 15 時 43 分 54 秒 (日本時間) `660305634769214745897646946140428702232333631604381665470748484340186207555282999468654485025631976835996819548713099007694587001192256747094035843143546127627283170126241581451617504694804037<192>` `50314735166899655641663203240997573<35>` `13123504130130197958251635508777340457005039080776157069462739413754519935064582894541192899070840869809381504587741473795286359345192196640156406258666845569<158>` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=3234712473 Step 1 took 32188ms Step 2 took 6100ms ********** Factor found in step 2: 50314735166899655641663203240997573 Found probable prime factor of 35 digits: 50314735166899655641663203240997573```
c158
name 名前 Bob Backstrom September 30, 2021 03:11:20 UTC 2021 年 9 月 30 日 (木) 12 時 11 分 20 秒 (日本時間) `13123504130130197958251635508777340457005039080776157069462739413754519935064582894541192899070840869809381504587741473795286359345192196640156406258666845569<158>` ```101091449131743825350501899041481556941326667<45> 129818142314168029087082155217506792061859214676064380645135511365876926572305452096071093562884481634812966676707<114>``` ```# # N = 14x10^208-41 = 15(207)1 # n: 13123504130130197958251635508777340457005039080776157069462739413754519935064582894541192899070840869809381504587741473795286359345192196640156406258666845569 m: 100000000000000000000000000000000000000000 deg: 5 c5: 14000 c0: -41 skew: 0.31 # Murphy_E = 6.978e-12 type: snfs lss: 1 rlim: 21000000 alim: 21000000 lpbr: 29 lpba: 29 mfbr: 57 mfba: 57 rlambda: 2.6 alambda: 2.6 GMP-ECM 7.0.4 [configured with GMP 6.2.0, --enable-asm-redc] [ECM] Input number is 13123504130130197958251635508777340457005039080776157069462739413754519935064582894541192899070840869809381504587741473795286359345192196640156406258666845569 (158 digits) Using B1=48910000, B2=240496185946, polynomial Dickson(12), sigma=1:3418354186 Step 1 took 116396ms ********** Factor found in step 1: 101091449131743825350501899041481556941326667 Found prime factor of 45 digits: 101091449131743825350501899041481556941326667 Prime cofactor 129818142314168029087082155217506792061859214676064380645135511365876926572305452096071093562884481634812966676707 has 114 digits```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:37:01 UTC 2012 年 10 月 28 日 (日) 22 時 37 分 1 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:03:53 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 53 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:03:53 UTC 2012 年 11 月 2 日 (金) 16 時 3 分 53 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:02:01 UTC 2012 年 11 月 8 日 (木) 18 時 2 分 1 秒 (日本時間)
14×10209-419
c168
composite cofactor 合成数の残り `229320208945872852578380762895851593076529879422642464668362153513918765022076593378856393164802000083731530053534858395109212207060484034550289803299465533382954418801<168>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:37:08 UTC 2012 年 10 月 28 日 (日) 22 時 37 分 8 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:04:08 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 8 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:04:08 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 8 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:01:46 UTC 2012 年 11 月 8 日 (木) 18 時 1 分 46 秒 (日本時間)
14×10211-419
c95
name 名前 Warut Roonguthai October 27, 2012 15:40:44 UTC 2012 年 10 月 28 日 (日) 0 時 40 分 44 秒 (日本時間) `26659789044586448554888764778713605597347673043493620258411765655818398284247686963613880977027<95>` ```6670726681136226336895976223083512566929<40> 3996534458528509347769645510551792943821797033715524563<55>``` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=2382856706 Step 1 took 9001ms Step 2 took 6677ms ********** Factor found in step 2: 6670726681136226336895976223083512566929 Found probable prime factor of 40 digits: 6670726681136226336895976223083512566929 Probable prime cofactor 3996534458528509347769645510551792943821797033715524563has 55 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10212-419
c213
name 名前 Robert Backstrom December 15, 2014 15:00:40 UTC 2014 年 12 月 16 日 (火) 0 時 0 分 40 秒 (日本時間) `155555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551<213>` ```292980227851544247686627839893207364843593338181149782003<57> 530942161852563558071454855473582374879404560297280976602900684569545608810130064867052447195504909836540244344965450420488823095517847238792740951559420517<156>``` ```Number: n N=155555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551 ( 213 digits) SNFS difficulty: 213 digits. Divisors found: Tue Dec 16 01:45:52 2014 prp57 factor: 292980227851544247686627839893207364843593338181149782003 Tue Dec 16 01:45:52 2014 prp156 factor: 530942161852563558071454855473582374879404560297280976602900684569545608810130064867052447195504909836540244344965450420488823095517847238792740951559420517 Tue Dec 16 01:45:52 2014 elapsed time 11:34:11 (Msieve 1.44 - dependency 2) Version: GGNFS-0.77.1-20060513-nocona Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.266). Factorization parameters were as follows: # # N = 14*10^212-41 15(211)1 # n: 155555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551 m: 100000000000000000000000000000000000 deg: 6 c6: 1400 c0: -41 skew: 0.56 # Murphy_E = 4.073e-12 type: snfs lss: 1 rlim: 25000000 alim: 25000000 lpbr: 29 lpba: 29 mfbr: 57 mfba: 57 rlambda: 2.6 alambda: 2.6 Factor base limits: 25000000/25000000 Large primes per side: 3 Large prime bits: 29/29 Max factor residue bits: 57/57 Sieved special-q in [100000, 74100000) Primes: RFBsize:1565927, AFBsize:1563784, Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 15019563 hash collisions in 80340096 relations (66470018 unique) Msieve: matrix is 3058461 x 3058687 (869.2 MB) Total sieving time: 0.00 hours. Total relation processing time: 10hrs 47min 39sec. Matrix solve time: 0.00 hours. Total square root time: 0hrs 19min 57sec. Prototype def-par.txt line would be: snfs,213,6,0,0,0,0,0,0,0,0,25000000,25000000,29,29,57,57,2.6,2.6,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- [ 0.038797] smpboot: CPU0: Intel(R) Core(TM) i7-4790 CPU @ 3.60GHz (fam: 06, model: 3c, stepping: 03) [ 0.000000] Memory: 16059644K/16661464K available (7373K kernel code, 1159K rwdata, 3228K rodata, 1468K init, 1504K bss, 601820K reserved) [ 1.132245] [drm] Memory usable by graphics device = 2048M [ 0.000019] Calibrating delay loop (skipped), value calculated using timer frequency.. 7200.64 BogoMIPS (lpj=3600320) [ 0.136681] smpboot: Total of 8 processors activated (57605.12 BogoMIPS)```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e6918118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
800Serge BatalovOctober 27, 2012 05:13:25 UTC 2012 年 10 月 27 日 (土) 14 時 13 分 25 秒 (日本時間)
403e61600600Serge BatalovOctober 27, 2012 06:37:34 UTC 2012 年 10 月 27 日 (土) 15 時 37 分 34 秒 (日本時間)
1000Dmitry DomanovNovember 2, 2012 07:04:21 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 21 秒 (日本時間)
4511e649501000Dmitry DomanovNovember 2, 2012 07:04:21 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 21 秒 (日本時間)
850Serge BatalovNovember 8, 2013 17:14:17 UTC 2013 年 11 月 9 日 (土) 2 時 14 分 17 秒 (日本時間)
400Serge BatalovJanuary 6, 2014 02:27:31 UTC 2014 年 1 月 6 日 (月) 11 時 27 分 31 秒 (日本時間)
1800Serge BatalovMay 24, 2014 09:17:10 UTC 2014 年 5 月 24 日 (土) 18 時 17 分 10 秒 (日本時間)
900Serge BatalovMay 24, 2014 19:03:25 UTC 2014 年 5 月 25 日 (日) 4 時 3 分 25 秒 (日本時間)
14×10213-419
c204
name 名前 Ignacio Santos June 7, 2021 07:07:18 UTC 2021 年 6 月 7 日 (月) 16 時 7 分 18 秒 (日本時間) `122123715817281211972021253946000127416889139459398033606864863994769225197246319804019892334748690647652076620510687194153951101607503910218391165876049013824617361306064340726831781138641645385230277907<204>` ```430519064126437536780486386747940391620169200689559031913<57> 283666220600663467417190383935003371226106981516901666130271721137829838159949438517984472720615676962147450641155710659170115130257052919284634139<147>``` ```N=122123715817281211972021253946000127416889139459398033606864863994769225197246319804019892334748690647652076620510687194153951101607503910218391165876049013824617361306064340726831781138641645385230277907 ( 204 digits) SNFS difficulty: 214 digits. Divisors found: r1=430519064126437536780486386747940391620169200689559031913 (pp57) r2=283666220600663467417190383935003371226106981516901666130271721137829838159949438517984472720615676962147450641155710659170115130257052919284634139 (pp147) Version: Msieve v. 1.52 (SVN 927) Total time: 1097.09 hours. Scaled time: 8611.05 units (timescale=7.849). Factorization parameters were as follows: n: 122123715817281211972021253946000127416889139459398033606864863994769225197246319804019892334748690647652076620510687194153951101607503910218391165876049013824617361306064340726831781138641645385230277907 m: 1000000000000000000000000000000000000000000 deg: 5 c5: 14000 c0: -41 skew: 0.31 # Murphy_E = 3.789e-12 type: snfs lss: 1 rlim: 26000000 alim: 26000000 lpbr: 29 lpba: 29 mfbr: 57 mfba: 57 rlambda: 2.6 alambda: 2.6Factor base limits: 26000000/26000000 Large primes per side: 3 Large prime bits: 29/29 Max factor residue bits: 57/57 Sieved rational special-q in [13000000, 34100001) Primes: , , Relations: relations Max relations in full relation-set: Initial matrix: Pruned matrix : 4950434 x 4950665 Total sieving time: 1090.36 hours. Total relation processing time: 0.16 hours. Matrix solve time: 6.13 hours. Time per square root: 0.44 hours. Prototype def-par.txt line would be: snfs,214.000,5,0,0,0,0,0,0,0,0,26000000,26000000,29,29,57,57,2.6,2.6,100000 total time: 1097.09 hours. ``` `GNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:37:40 UTC 2012 年 10 月 28 日 (日) 22 時 37 分 40 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:04:32 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 32 秒 (日本時間)
4511e654801000Dmitry DomanovNovember 2, 2012 07:04:32 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 32 秒 (日本時間)
4480Ignacio SantosJune 1, 2021 16:44:55 UTC 2021 年 6 月 2 日 (水) 1 時 44 分 55 秒 (日本時間)
14×10215-419
c191
composite cofactor 合成数の残り `72318986422833039654480870695389692390009560597898969905138619224333422740995255134985289851231538360784402049750285027879260725244251956977390109853201725548059610923943457642571609570179297<191>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:37:52 UTC 2012 年 10 月 28 日 (日) 22 時 37 分 52 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:04:45 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 45 秒 (日本時間)
4511e61000 / 4208Dmitry DomanovNovember 2, 2012 07:04:45 UTC 2012 年 11 月 2 日 (金) 16 時 4 分 45 秒 (日本時間)
14×10217-419
c161
composite cofactor 合成数の残り `76690139204036875949571915416047648605183985788539694848173230349134198988750926218673006430156509096259470986808884034173866750388875122492301527504523497115559<161>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:38:20 UTC 2012 年 10 月 28 日 (日) 22 時 38 分 20 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:05:02 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 2 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:05:02 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 2 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:01:29 UTC 2012 年 11 月 8 日 (木) 18 時 1 分 29 秒 (日本時間)
14×10218-419
c212
name 名前 Bob Backstrom August 23, 2019 05:39:07 UTC 2019 年 8 月 23 日 (金) 14 時 39 分 7 秒 (日本時間) `68563011832065872599247775607758453292892910011585679792222731057717175911091012353928339801029160191747236775945823075204857708977733819092477447544031288632502403500866563009187492559076706975351058536993638283<212>` ```110152888481431951059319647177043039295304817417<48> 622434988108580338506871644753488851266375781507581031578468481655442138816268474093558454729402461331070250684217116870000542936773250369872770894220766778355642099<165>``` ```GMP-ECM 7.0-dev [configured with MPIR 2.6.0, --enable-openmp] [ECM] Input number is 68563011832065872599247775607758453292892910011585679792222731057717175911091012353928339801029160191747236775945823075204857708977733819092477447544031288632502403500866563009187492559076706975351058536993638283 (212 digits) Using B1=30970000, B2=144289975846, polynomial Dickson(12), sigma=1:1045857993 Step 1 took 405469ms Step 2 took 105375ms ********** Factor found in step 2: 110152888481431951059319647177043039295304817417 Found probable prime factor of 48 digits: 110152888481431951059319647177043039295304817417 Probable prime cofactor 622434988108580338506871644753488851266375781507581031578468481655442138816268474093558454729402461331070250684217116870000542936773250369872770894220766778355642099 has 165 digits```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:38:30 UTC 2012 年 10 月 28 日 (日) 22 時 38 分 30 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:05:19 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 19 秒 (日本時間)
4511e61000 / 4208Dmitry DomanovNovember 2, 2012 07:05:19 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 19 秒 (日本時間)
14×10221-419
c202
name 名前 Warut Roonguthai October 27, 2012 11:09:46 UTC 2012 年 10 月 27 日 (土) 20 時 9 分 46 秒 (日本時間) `8703960895154370925000763346821587185877872800828090254166446847747687942949220645339753995224278038652586096723340896101179325046251178990416031497079527570997178335618008595828292922828578182312731853<202>` `125000453996958415251650726013413<33>` `69631434261560048681554464451895882318787128702942130989346336631226967891676972955840601563401694145911571462361824645605073574993632678519266908786664016625662384455881<170>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2788910512 Step 1 took 11341ms Step 2 took 7239ms ********** Factor found in step 2: 125000453996958415251650726013413 Found probable prime factor of 33 digits: 125000453996958415251650726013413 Composite cofactor 69631434261560048681554464451895882318787128702942130989346336631226967891676972955840601563401694145911571462361824645605073574993632678519266908786664016625662384455881 has 170 digits``` `GMP-ECM 6.3`
c170
name 名前 Erik Branger April 21, 2020 21:09:22 UTC 2020 年 4 月 22 日 (水) 6 時 9 分 22 秒 (日本時間) `69631434261560048681554464451895882318787128702942130989346336631226967891676972955840601563401694145911571462361824645605073574993632678519266908786664016625662384455881<170>` ```935037508782958494937770030487758845836409678777543208634052309<63> 74469134775344014326463964560298530267814712532990539463758817489368045105656001730572936525853455574223909<107>``` ```Number: 15551_221 N = 69631434261560048681554464451895882318787128702942130989346336631226967891676972955840601563401694145911571462361824645605073574993632678519266908786664016625662384455881 (170 digits) SNFS difficulty: 223 digits. Divisors found: r1=935037508782958494937770030487758845836409678777543208634052309 (pp63) r2=74469134775344014326463964560298530267814712532990539463758817489368045105656001730572936525853455574223909 (pp107) Version: Msieve v. 1.52 (SVN unknown) Total time: 102.04 hours. Factorization parameters were as follows: n: 69631434261560048681554464451895882318787128702942130989346336631226967891676972955840601563401694145911571462361824645605073574993632678519266908786664016625662384455881 m: 10000000000000000000000000000000000000000000000000000000 deg: 4 c4: 140 c0: -41 skew: 1.00 type: snfs lss: 1 rlim: 536870912 alim: 44739242 lpbr: 29 lpba: 28 mfbr: 58 mfba: 56 rlambda: 2.8 alambda: 2.8 side: 1 Factor base limits: 536870912/44739242 Large primes per side: 3 Large prime bits: 29/28 Relations: 8996710 relations Pruned matrix : 7741008 x 7741233 Total pre-computation time approximately 1000 CPU-days. Pre-computation saved approximately 18 G relations. Total batch smoothness checking time: 52.10 hours. Total relation processing time: 0.38 hours. Matrix solve time: 47.37 hours. time per square root: 2.19 hours. Prototype def-par.txt line would be: snfs,223,4,0,0,0,0,0,0,0,0,536870912,44739242,29,28,58,56,2.8,2.8,100000 total time: 102.04 hours. Intel64 Family 6 Model 60 Stepping 3, GenuineIntel Windows-10-10.0.17763-SP0 processors: 8, speed: 3.50GHz``` `GGNFS, NFS_factory, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:38:37 UTC 2012 年 10 月 28 日 (日) 22 時 38 分 37 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:05:36 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 36 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:05:36 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 36 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:01:14 UTC 2012 年 11 月 8 日 (木) 18 時 1 分 14 秒 (日本時間)
14×10222-419
c211
name 名前 Dmitry Domanov November 16, 2012 06:29:45 UTC 2012 年 11 月 16 日 (金) 15 時 29 分 45 秒 (日本時間) `3533542549511350702903509546545271054463614850650930972274336227457458890779713738637924843918167758798632890848132795721589777242596809327383688038150762980389375693002486277406773204320879768187246178941643209<211>` ```786511820042365525329403256869651208867994286035479<51> 4492675709973457406292805970211749164013838766719041814311895500864734815638344867118106159834722583727598557376487197179061906425763834272764864480927026179871<160>``` ```Using B1=43000000, B2=240490660426, polynomial Dickson(12), sigma=1181201120 Step 1 took 389506ms Step 2 took 106786ms ********** Factor found in step 2: 786511820042365525329403256869651208867994286035479 Found probable prime factor of 51 digits: 786511820042365525329403256869651208867994286035479```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:38:59 UTC 2012 年 10 月 28 日 (日) 22 時 38 分 59 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:05:53 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 53 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:05:53 UTC 2012 年 11 月 2 日 (金) 16 時 5 分 53 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:10:06 UTC 2012 年 11 月 12 日 (月) 22 時 10 分 6 秒 (日本時間)
14×10223-419
c179
name 名前 Warut Roonguthai October 27, 2012 09:07:04 UTC 2012 年 10 月 27 日 (土) 18 時 7 分 4 秒 (日本時間) `32975153268970917249677803819930399286426907021352315252941091590887704088425027183928512578106503615832634322751850114993391052478189585755332421537199580440501196596025441262483<179>` `649182107271598133243376362936623886681<39>` `50794920099631014279282289958857999560680133593252582845311816951203444385306005236830015269530460146299587522725320057366461431973869079243<140>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3841120287 Step 1 took 10514ms Step 2 took 6381ms ********** Factor found in step 2: 649182107271598133243376362936623886681 Found probable prime factor of 39 digits: 649182107271598133243376362936623886681 Composite cofactor 50794920099631014279282289958857999560680133593252582845311816951203444385306005236830015269530460146299587522725320057366461431973869079243 has 140 digits``` `GMP-ECM 6.3`
c140
name 名前 Erik Branger September 11, 2016 19:24:00 UTC 2016 年 9 月 12 日 (月) 4 時 24 分 0 秒 (日本時間) `50794920099631014279282289958857999560680133593252582845311816951203444385306005236830015269530460146299587522725320057366461431973869079243<140>` ```40921120452795828193602217949313952715920673<44> 1241288594681199268820991359830705124023182641650804313261408513555814443603458562487221323626091<97>``` ```Number: 15551_223 N = 50794920099631014279282289958857999560680133593252582845311816951203444385306005236830015269530460146299587522725320057366461431973869079243 (140 digits) Divisors found: r1=40921120452795828193602217949313952715920673 (pp44) r2=1241288594681199268820991359830705124023182641650804313261408513555814443603458562487221323626091 (pp97) Version: Msieve v. 1.51 (SVN 845) Total time: 325.23 hours. Factorization parameters were as follows: # Murphy_E = 2.184e-11, selected by Erik Branger # expecting poly E from 2.14e-011 to > 2.46e-011 n: 50794920099631014279282289958857999560680133593252582845311816951203444385306005236830015269530460146299587522725320057366461431973869079243 Y0: -1825149955007619806634197570 Y1: 578626842734011 c0: 351684523396806048007231323605718573 c1: 2370709651331576866044890945556 c2: 76914155340586990665807 c3: -168029242146640504 c4: -198962420 c5: 2508 skew: 5278400.3 type: gnfs # selected mechanically rlim: 17900000 alim: 17900000 lpbr: 28 lpba: 28 mfbr: 55 mfba: 55 rlambda: 2.6 alambda: 2.6 Factor base limits: 17900000/17900000 Large primes per side: 3 Large prime bits: 28/28 Sieved algebraic special-q in [0, 0) Total raw relations: 23081931 Relations: 3358046 relations Pruned matrix : 2096688 x 2096914 Polynomial selection time: 0.00 hours. Total sieving time: 320.61 hours. Total relation processing time: 0.20 hours. Matrix solve time: 4.16 hours. time per square root: 0.27 hours. Prototype def-par.txt line would be: gnfs,139,5,65,2000,1e-05,0.28,250,20,50000,3600,17900000,17900000,28,28,55,55,2.6,2.6,100000 total time: 325.23 hours. Intel64 Family 6 Model 58 Stepping 9, GenuineIntel Windows-post2008Server-6.2.9200 processors: 8, speed: 2.29GHz``` `GGNFS, Msieve`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 27, 2012 22:56:31 UTC 2012 年 10 月 28 日 (日) 7 時 56 分 31 秒 (日本時間)
403e66300Warut RoonguthaiOctober 27, 2012 22:56:31 UTC 2012 年 10 月 28 日 (日) 7 時 56 分 31 秒 (日本時間)
4511e630361000Dmitry DomanovNovember 2, 2012 07:06:06 UTC 2012 年 11 月 2 日 (金) 16 時 6 分 6 秒 (日本時間)
850Serge BatalovNovember 8, 2013 01:46:12 UTC 2013 年 11 月 8 日 (金) 10 時 46 分 12 秒 (日本時間)
850Serge BatalovNovember 8, 2013 17:09:20 UTC 2013 年 11 月 9 日 (土) 2 時 9 分 20 秒 (日本時間)
336KTakahashiJanuary 4, 2014 14:30:51 UTC 2014 年 1 月 4 日 (土) 23 時 30 分 51 秒 (日本時間)
5043e60 / 6592--
5511e70 / 17512--
6026e77 / 419601KTakahashiAugust 20, 2014 19:53:06 UTC 2014 年 8 月 21 日 (木) 4 時 53 分 6 秒 (日本時間)
6KTakahashiOctober 27, 2014 21:10:58 UTC 2014 年 10 月 28 日 (火) 6 時 10 分 58 秒 (日本時間)
14×10224-419
c222
name 名前 Bob Backstrom July 22, 2018 19:29:15 UTC 2018 年 7 月 23 日 (月) 4 時 29 分 15 秒 (日本時間) `685266764561918746940773372491434165442976015663240332843857072931962799804209495839451786588350465002447381302006852667645619187469407733724914341654429760156632403328438570729319627998042094958394517865883504650024473813<222>` ```460498946470892421716401930017020677047647964788870280241331535206714527927570983826769591577047863<99> 1488096269955817641836977842387573244415087159843768767925270426982777946592788224754953477969866143506344401270812729005651<124>``` ```Number: n N=685266764561918746940773372491434165442976015663240332843857072931962799804209495839451786588350465002447381302006852667645619187469407733724914341654429760156632403328438570729319627998042094958394517865883504650024473813 ( 222 digits) SNFS difficulty: 225 digits. Divisors found: Mon Jul 23 05:03:55 2018 p99 factor: 460498946470892421716401930017020677047647964788870280241331535206714527927570983826769591577047863 Mon Jul 23 05:03:55 2018 p124 factor: 1488096269955817641836977842387573244415087159843768767925270426982777946592788224754953477969866143506344401270812729005651 Mon Jul 23 05:03:55 2018 elapsed time 13:38:42 (Msieve 1.53 - dependency 1) Version: Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.138). Factorization parameters were as follows: # # 14x10^224-41 = 15(223)1 # n: 685266764561918746940773372491434165442976015663240332843857072931962799804209495839451786588350465002447381302006852667645619187469407733724914341654429760156632403328438570729319627998042094958394517865883504650024473813 m: 10000000000000000000000000000000000000 deg: 6 c6: 1400 c0: -41 skew: 0.56 # Murphy_E = 1.605e-12 type: snfs lss: 1 rlim: 40000000 alim: 40000000 lpbr: 30 lpba: 30 mfbr: 59 mfba: 59 rlambda: 2.7 alambda: 2.7 Factor base limits: 40000000/40000000 Large primes per side: 3 Large prime bits: 30/30 Max factor residue bits: 59/59 Sieved special-q in [100000, 100000) Primes: , , Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 21761018 hash collisions in 127141907 relations (108870784 unique) Msieve: matrix is 5566036 x 5566262 (1589.6 MB) Sieving start time: 2018/07/19 05:21:25 Sieving end time : 2018/07/22 15:21:45 Total sieving time: 82hrs 0min 20secs. Total relation processing time: 12hrs 43min 0sec. Matrix solve time: 0.00 hours. Total square root time: 0hrs 11min 13sec. Prototype def-par.txt line would be: snfs,225,6,0,0,0,0,0,0,0,0,40000000,40000000,30,30,59,59,2.7,2.7,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- [ 0.036000] smpboot: CPU0: AMD Ryzen 7 1700 Eight-Core Processor (family: 0x17, model: 0x1, stepping: 0x1) [ 0.000000] Memory: 16286680K/16703460K available (12300K kernel code, 2470K rwdata, 4240K rodata, 2408K init, 2416K bss, 416780K reserved, 0K cma-reserved) [ 0.068654] x86/mm: Memory block size: 128MB [ 0.024000] Calibrating delay loop (skipped), value calculated using timer frequency.. 5988.68 BogoMIPS (lpj=11977364) [ 0.066212] smpboot: Total of 16 processors activated (95818.91 BogoMIPS)```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:39:19 UTC 2012 年 10 月 28 日 (日) 22 時 39 分 19 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:06:28 UTC 2012 年 11 月 2 日 (金) 16 時 6 分 28 秒 (日本時間)
4511e624541000Dmitry DomanovNovember 2, 2012 07:06:28 UTC 2012 年 11 月 2 日 (金) 16 時 6 分 28 秒 (日本時間)
850Serge BatalovNovember 8, 2013 17:15:28 UTC 2013 年 11 月 9 日 (土) 2 時 15 分 28 秒 (日本時間)
400Serge BatalovJanuary 6, 2014 02:28:28 UTC 2014 年 1 月 6 日 (月) 11 時 28 分 28 秒 (日本時間)
204KTakahashiFebruary 12, 2014 21:20:17 UTC 2014 年 2 月 13 日 (木) 6 時 20 分 17 秒 (日本時間)
5043e6500 / 6958Dmitry DomanovNovember 12, 2012 13:09:50 UTC 2012 年 11 月 12 日 (月) 22 時 9 分 50 秒 (日本時間)
14×10225-419
c209
name 名前 Warut Roonguthai October 27, 2012 11:32:23 UTC 2012 年 10 月 27 日 (土) 20 時 32 分 23 秒 (日本時間) `28699289517843633443366336175456004382650874958922936911889481031393775852457165430157546500721871891219400851066352574941209246512302874326346614581059987873128311285049960292947388878808435482059891406711377<209>` ```11389316008792374900966025113186791<35> 2519843113992818189742370043494612359309090730388817148466950498393808093869094127711488949254573140526971835294724952852690332801025745335715010214055161557177478078529514247<175>``` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=4187493853 Step 1 took 11778ms ********** Factor found in step 1: 11389316008792374900966025113186791 Found probable prime factor of 35 digits: 11389316008792374900966025113186791 Probable prime cofactor 2519843113992818189742370043494612359309090730388817148466950498393808093869094127711488949254573140526971835294724952852690332801025745335715010214055161557177478078529514247 has 175 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10227-419
c220
name 名前 Warut Roonguthai October 27, 2012 11:38:50 UTC 2012 年 10 月 27 日 (土) 20 時 38 分 50 秒 (日本時間) `1435297847949981666171678573482402096557610244912448117917050530668521210915151843834722332060953068465908478985231626844261236465770518156086674606508814721287389666565557109767830208193202769557698850974665812963809259<220>` `297166502357380593400485071188250539<36>` `4829944951951054944785577488885004229379690478109856007075799174138759116132328459572087313679705746663046772009931411818049286340741917540369738263851391635845500608119529050642700481<184>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2216187166 Step 1 took 12434ms Step 2 took 7363ms ********** Factor found in step 2: 297166502357380593400485071188250539 Found probable prime factor of 36 digits: 297166502357380593400485071188250539 Composite cofactor 4829944951951054944785577488885004229379690478109856007075799174138759116132328459572087313679705746663046772009931411818049286340741917540369738263851391635845500608119529050642700481 has 184 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e62318118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
2200Warut RoonguthaiOctober 27, 2012 23:07:36 UTC 2012 年 10 月 28 日 (日) 8 時 7 分 36 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:06:44 UTC 2012 年 11 月 2 日 (金) 16 時 6 分 44 秒 (日本時間)
4511e61000 / 2398Dmitry DomanovNovember 2, 2012 07:06:44 UTC 2012 年 11 月 2 日 (金) 16 時 6 分 44 秒 (日本時間)
5043e6500 / 7277Dmitry DomanovNovember 12, 2012 13:09:34 UTC 2012 年 11 月 12 日 (月) 22 時 9 分 34 秒 (日本時間)
14×10228-419
c188
composite cofactor 合成数の残り `13006795617517976006243969253823956780537266045492535654231954999208789760025095172792362455780810960765620581603647731219943684827437944942991084908436810240185802964344479027072031786611<188>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:39:47 UTC 2012 年 10 月 28 日 (日) 22 時 39 分 47 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:07:02 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 2 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:07:02 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 2 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:09:23 UTC 2012 年 11 月 12 日 (月) 22 時 9 分 23 秒 (日本時間)
14×10231-419
c230
name 名前 Bob Backstrom December 26, 2018 17:13:25 UTC 2018 年 12 月 27 日 (木) 2 時 13 分 25 秒 (日本時間) `15712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349<230>` ```8441940533801698999711626927372646868494009110066068077743246381918212147735676434087018523528417303733275891973<112> 1861264281172693769234533111461897749296802393779646153058079475477346384432620887470626419543453237575160648453296913<118>``` ```Number: n N=15712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349 ( 230 digits) SNFS difficulty: 232 digits. Divisors found: Thu Dec 27 04:09:23 2018 p112 factor: 8441940533801698999711626927372646868494009110066068077743246381918212147735676434087018523528417303733275891973 Thu Dec 27 04:09:23 2018 p118 factor: 1861264281172693769234533111461897749296802393779646153058079475477346384432620887470626419543453237575160648453296913 Thu Dec 27 04:09:23 2018 elapsed time 31:12:18 (Msieve 1.54 - dependency 1) Version: Total time: 0.00 hours. Scaled time: 0.00 units (timescale=2.122). Factorization parameters were as follows: # # N = 14x10^231-41 = 15(230)1 # n: 15712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349046015712682379349 m: 100000000000000000000000000000000000000 deg: 6 c6: 14000 c0: -41 skew: 0.38 # Murphy_E = 7.547e-13 type: snfs lss: 1 rlim: 52000000 alim: 52000000 lpbr: 30 lpba: 30 mfbr: 60 mfba: 60 rlambda: 2.7 alambda: 2.7 Factor base limits: 52000000/52000000 Large primes per side: 3 Large prime bits: 30/30 Max factor residue bits: 60/60 Sieved special-q in [100000, 109200000) Primes: , , Relations: Max relations in full relation-set: 28 Initial matrix: Pruned matrix : Msieve: found 20916446 hash collisions in 112168776 relations (93280104 unique) Msieve: matrix is 7689499 x 7689724 (2711.3 MB) Sieving start time: 2018/12/23 09:28:12 Sieving end time : 2018/12/25 20:52:27 Total sieving time: 59hrs 24min 15secs. Total relation processing time: 30hrs 9min 59sec. Matrix solve time: 0.00 hours. Total square root time: 0hrs 21min 33sec. Prototype def-par.txt line would be: snfs,232,6,0,0,0,0,0,0,0,0,52000000,52000000,30,30,60,60,2.7,2.7,100000 total time: 0.00 hours. --------- CPU info (if available) ---------- [ 0.048000] smpboot: CPU0: AMD Ryzen 7 1700 Eight-Core Processor (family: 0x17, model: 0x1, stepping: 0x1) [ 0.000000] Memory: 16285120K/16703460K available (12300K kernel code, 2472K rwdata, 4252K rodata, 2408K init, 2416K bss, 418340K reserved, 0K cma-reserved) [ 0.080545] x86/mm: Memory block size: 128MB [ 0.028000] Calibrating delay loop (skipped), value calculated using timer frequency.. 5988.82 BogoMIPS (lpj=11977644) [ 0.078213] smpboot: Total of 16 processors activated (95821.15 BogoMIPS)```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:39:56 UTC 2012 年 10 月 28 日 (日) 22 時 39 分 56 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:07:33 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 33 秒 (日本時間)
4511e624541000Dmitry DomanovNovember 2, 2012 07:07:33 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 33 秒 (日本時間)
850Serge BatalovNovember 8, 2013 17:15:56 UTC 2013 年 11 月 9 日 (土) 2 時 15 分 56 秒 (日本時間)
400Serge BatalovJanuary 6, 2014 02:28:51 UTC 2014 年 1 月 6 日 (月) 11 時 28 分 51 秒 (日本時間)
204KTakahashiFebruary 12, 2014 21:20:34 UTC 2014 年 2 月 13 日 (木) 6 時 20 分 34 秒 (日本時間)
5043e6500 / 6958Dmitry DomanovNovember 12, 2012 13:09:06 UTC 2012 年 11 月 12 日 (月) 22 時 9 分 6 秒 (日本時間)
14×10233-419
c207
composite cofactor 合成数の残り `132501795159122220019113560794849168640664061238430890509050372491653798795429830871823216446348777935460754273748693172453945080001473021907141001171831088984960145615229385497666510252546546587559753994589<207>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:40:13 UTC 2012 年 10 月 28 日 (日) 22 時 40 分 13 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:07:54 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 54 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:07:54 UTC 2012 年 11 月 2 日 (金) 16 時 7 分 54 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:08:51 UTC 2012 年 11 月 12 日 (月) 22 時 8 分 51 秒 (日本時間)
14×10234-419
c147
name 名前 Eric Jeancolas November 6, 2021 14:53:07 UTC 2021 年 11 月 6 日 (土) 23 時 53 分 7 秒 (日本時間) `340887745700144956504475017344424883100767674566392189408880951845135365430573666413822458389237538161356596705359315417007409694838571119800676911<147>` ```8057734420595692608988255268415951236165750403862305780821243222099<67> 42305656640758803952386916502522386594673701339711076773807650316499751071879989<80>``` ```340887745700144956504475017344424883100767674566392189408880951845135365430573666413822458389237538161356596705359315417007409694838571119800676911=8057734420595692608988255268415951236165750403862305780821243222099*42305656640758803952386916502522386594673701339711076773807650316499751071879989 cado polynomial n: 340887745700144956504475017344424883100767674566392189408880951845135365430573666413822458389237538161356596705359315417007409694838571119800676911 skew: 4605763.991 c0: -1424829515760229808416960398086593920 c1: -1691767251894870088302417078978 c2: -425691517204005571246746 c3: -268257433565132513 c4: 39498060074 c5: 2040 Y0: -44079891755062048608016310851 Y1: 9356456370481659179 # MurphyE (Bf=5.369e+08,Bg=5.369e+08,area=3.355e+14) = 1.478e-07 # f(x) = 2040*x^5+39498060074*x^4-268257433565132513*x^3-425691517204005571246746*x^2-1691767251894870088302417078978*x-1424829515760229808416960398086593920 # g(x) = 9356456370481659179*x-44079891755062048608016310851 cado parameters (extracts) tasks.lim0 = 10000000 tasks.lim1 = 20000000 tasks.lpb0 = 29 tasks.lpb1 = 29 tasks.sieve.mfb0 = 58 tasks.sieve.mfb1 = 87 tasks.I = 13 tasks.linalg.m = 64 tasks.linalg.n = 64 tasks.linalg.characters.nchar = 50 cado log (extracts) Info:Square Root: Factors: 42305656640758803952386916502522386594673701339711076773807650316499751071879989 8057734420595692608988255268415951236165750403862305780821243222099 Info:Square Root: Total cpu/real time for sqrt: 3274.31/1016 Info:Filtering - Duplicate Removal, splitting pass: Total cpu/real time for dup1: 235.49/285.591 Info:Filtering - Duplicate Removal, splitting pass: Aggregate statistics: Info:Filtering - Duplicate Removal, splitting pass: CPU time for dup1: 284.79999999999995s Info:Generate Free Relations: Total cpu/real time for freerel: 500.23/126.361 Info:Filtering - Duplicate Removal, removal pass: Total cpu/real time for dup2: 1139.72/1136.67 Info:Filtering - Duplicate Removal, removal pass: Aggregate statistics: Info:Filtering - Duplicate Removal, removal pass: CPU time for dup2: 949.5999999999999s Info:Generate Factor Base: Total cpu/real time for makefb: 18.46/4.91517 Info:Quadratic Characters: Total cpu/real time for characters: 148.53/64.6624 Info:Filtering - Singleton removal: Total cpu/real time for purge: 955.13/1068.58 Info:Polynomial Selection (root optimized): Aggregate statistics: Info:Polynomial Selection (root optimized): Total time: 7184.63 Info:Polynomial Selection (root optimized): Rootsieve time: 7182.24 Info:Filtering - Merging: Merged matrix has 3789525 rows and total weight 648142804 (171.0 entries per row on average) Info:Filtering - Merging: Total cpu/real time for merge: 1029.53/289.237 Info:Filtering - Merging: Total cpu/real time for replay: 163.25/141.254 Info:Square Root: Total cpu/real time for sqrt: 3274.31/1016 Info:Polynomial Selection (size optimized): Aggregate statistics: Info:Polynomial Selection (size optimized): potential collisions: 77228.6 Info:Polynomial Selection (size optimized): raw lognorm (nr/min/av/max/std): 78123/44.300/52.829/57.050/0.889 Info:Polynomial Selection (size optimized): optimized lognorm (nr/min/av/max/std): 63430/43.290/47.454/53.740/1.275 Info:Polynomial Selection (size optimized): Total time: 40891.7 Info:Linear Algebra: Total cpu/real time for bwc: 266335/68174.5 Info:Linear Algebra: Aggregate statistics: Info:Linear Algebra: Krylov: CPU time 171784.39, WCT time 43861.01, iteration CPU time 0.35, COMM 0.02, cpu-wait 0.0, comm-wait 0.0 (118784 iterations) Info:Linear Algebra: Lingen CPU time 840.59, WCT time 213.29 Info:Linear Algebra: Mksol: CPU time 92188.74, WCT time 23540.88, iteration CPU time 0.38, COMM 0.02, cpu-wait 0.0, comm-wait 0.0 (59392 iterations) Info:Lattice Sieving: Aggregate statistics: Info:Lattice Sieving: Total number of relations: 51735811 Info:Lattice Sieving: Average J: 4043.27 for 1429401 special-q, max bucket fill -bkmult 1.0,1s:1.130660 Info:Lattice Sieving: Total time: 1.44117e+06s Info:HTTP server: Shutting down HTTP server Info:Complete Factorization / Discrete logarithm: Total cpu/elapsed time for entire factorization: 3.0591e+06/514.566 42305656640758803952386916502522386594673701339711076773807650316499751071879989 8057734420595692608988255268415951236165750403862305780821243222099``` `cado-nfs-3.0.0` ```Linux Ubuntu 20.04.1 LTS [5.4.0-89-generic|libc 2.31 (Ubuntu GLIBC 2.31-0ubuntu9.3)] GenuineIntel Intel(R) Core(TM) i5-3470S CPU @ 2.90GHz [Family 6 Model 58 Stepping 9] (4 processeurs)```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 27, 2012 23:12:49 UTC 2012 年 10 月 28 日 (日) 8 時 12 分 49 秒 (日本時間)
403e61950950Warut RoonguthaiOctober 27, 2012 23:12:49 UTC 2012 年 10 月 28 日 (日) 8 時 12 分 49 秒 (日本時間)
1000Dmitry DomanovNovember 2, 2012 07:08:09 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 9 秒 (日本時間)
4511e622441000Dmitry DomanovNovember 2, 2012 07:08:09 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 9 秒 (日本時間)
850Serge BatalovNovember 8, 2013 17:10:28 UTC 2013 年 11 月 9 日 (土) 2 時 10 分 28 秒 (日本時間)
394KTakahashiJanuary 5, 2014 04:05:49 UTC 2014 年 1 月 5 日 (日) 13 時 5 分 49 秒 (日本時間)
5043e66975500Dmitry DomanovNovember 8, 2012 09:00:51 UTC 2012 年 11 月 8 日 (木) 18 時 0 分 51 秒 (日本時間)
6475Ignacio SantosJune 13, 2021 19:09:24 UTC 2021 年 6 月 14 日 (月) 4 時 9 分 24 秒 (日本時間)
14×10236-419
c220
name 名前 Warut Roonguthai October 27, 2012 14:50:48 UTC 2012 年 10 月 27 日 (土) 23 時 50 分 48 秒 (日本時間) `3138172030127883276612790630385493347845059461709311461038933803802222109362057008320108019411561840249251407465972792913431216544644937197665946964167009818804227623866827711389039449503276176001707124319659921700280049<220>` ```273152116237590028580454386531106597397<39> 11488734092026124430004646388765926119405240035420199189957370600714723250188566806420071181215740818253418698238236541454537424648123769510269269402474372043237745192620442280311917<182>``` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=1709323484 Step 1 took 11544ms Step 2 took 6599ms ********** Factor found in step 2: 273152116237590028580454386531106597397 Found probable prime factor of 39 digits: 273152116237590028580454386531106597397 Probable prime cofactor 11488734092026124430004646388765926119405240035420199189957370600714723250188566806420071181215740818253418698238236541454537424648123769510269269402474372043237745192620442280311917 has 182 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10237-419
c206
name 名前 Serge Batalov October 27, 2012 09:44:20 UTC 2012 年 10 月 27 日 (土) 18 時 44 分 20 秒 (日本時間) `92371251126048380892137789399013662217537021085763394979475798539463517183682713759869487984811451566292056803789587650978357499781464320223688570251956429357564671872539089179008832927681577243823697094299<206>` `224243355963186242371247231869228993<36>` `411924138083327901465095405522935517879284050494035637032474363726675511918407117954003146719662702546949228180598015394536195041688549978104326813690730500460459731793243<171>` ```Using B1=2000000, B2=2853999340, polynomial Dickson(6), sigma=2580349337 Step 1 took 12108ms Step 2 took 5189ms ********** Factor found in step 2: 224243355963186242371247231869228993 Found probable prime factor of 36 digits: 224243355963186242371247231869228993 Composite cofactor 411924138083327901465095405522935517879284050494035637032474363726675511918407117954003146719662702546949228180598015394536195041688549978104326813690730500460459731793243 has 171 digits```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e6118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:08:27 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 27 秒 (日本時間)
4511e61000 / 2500Dmitry DomanovNovember 2, 2012 07:08:27 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 27 秒 (日本時間)
5043e6500 / 7291Dmitry DomanovNovember 8, 2012 09:00:37 UTC 2012 年 11 月 8 日 (木) 18 時 0 分 37 秒 (日本時間)
14×10238-419
c175
composite cofactor 合成数の残り `1807541029108099103968877958194744864912060375535501212268083337509693082469883338652212275905141598359077472993422973624738267517600195290989476852186389692587355048167032221<175>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:41:01 UTC 2012 年 10 月 28 日 (日) 22 時 41 分 1 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:08:56 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 56 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:08:56 UTC 2012 年 11 月 2 日 (金) 16 時 8 分 56 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:00:25 UTC 2012 年 11 月 8 日 (木) 18 時 0 分 25 秒 (日本時間)
14×10239-419
c233
composite cofactor 合成数の残り `17974218525593085338675168369319452049785658086018678371375060712737593010001022091097730332517008521538486801126916740171591390703675120416672171271090129549051635936979770748856913524455509886917900279009942554525979479682065346581<233>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:41:10 UTC 2012 年 10 月 28 日 (日) 22 時 41 分 10 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:09:17 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 17 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:09:17 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 17 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:08:24 UTC 2012 年 11 月 12 日 (月) 22 時 8 分 24 秒 (日本時間)
14×10240-419
c231
composite cofactor 合成数の残り `582774526210827899271307919450118989730968661637987118187613472972405443400849828130553162848817272588204068965484649067999647639702329266948191128618641934898640060331113419908099906409212570942038571143730923666532606921353954577<231>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:41:20 UTC 2012 年 10 月 28 日 (日) 22 時 41 分 20 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:09:29 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 29 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:09:29 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 29 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:08:13 UTC 2012 年 11 月 12 日 (月) 22 時 8 分 13 秒 (日本時間)
14×10241-419
c237
name 名前 Warut Roonguthai October 27, 2012 08:02:56 UTC 2012 年 10 月 27 日 (土) 17 時 2 分 56 秒 (日本時間) `144050261193991457819511196306551302986059021507732926700024591437446688541727759411369475543866905790100248692486647055252535541829623524642369503324988707488452828169645934747289539996069485730278233079496194500778385875666103841717573<237>` `114195085626445854234043512546413<33>` `1261440108422945974718739761310862303178595485798570723923141069684801572603146172621751575071357293425744393738717912283596603292280374410121233484995560387581608803956669440757236828865549079078773163321<205>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=549781686 Step 1 took 12979ms Step 2 took 6942ms ********** Factor found in step 2: 114195085626445854234043512546413 Found probable prime factor of 33 digits: 114195085626445854234043512546413 Composite cofactor 1261440108422945974718739761310862303178595485798570723923141069684801572603146172621751575071357293425744393738717912283596603292280374410121233484995560387581608803956669440757236828865549079078773163321 has 205 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:41:27 UTC 2012 年 10 月 28 日 (日) 22 時 41 分 27 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:09:42 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 42 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:09:42 UTC 2012 年 11 月 2 日 (金) 16 時 9 分 42 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:07:59 UTC 2012 年 11 月 12 日 (月) 22 時 7 分 59 秒 (日本時間)
14×10243-419
c147
name 名前 Warut Roonguthai October 27, 2012 08:04:47 UTC 2012 年 10 月 27 日 (土) 17 時 4 分 47 秒 (日本時間) `155024342792109865155088314312358567247073400534250002081908842478086212467569184101367319768408468743662003281281165973801742463795870977014589899<147>` ```68551815113603138645786557856064831847<38> 2261418498331599622168678754381448319386753156538389562374372266058317600217556218865068041619756634764075517<109>``` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3846359381 Step 1 took 4306ms Step 2 took 3603ms ********** Factor found in step 2: 68551815113603138645786557856064831847 Found probable prime factor of 38 digits: 68551815113603138645786557856064831847 Probable prime cofactor 2261418498331599622168678754381448319386753156538389562374372266058317600217556218865068041619756634764075517 has 109 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10244-419
c210
composite cofactor 合成数の残り `463268830048090095427786308522431051723378029530313808978071582857280220153377416630581617928148972578025954668682950475783938603791366310986618624346480507191799289034537436885183000568178278043833194164658891<210>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:41:49 UTC 2012 年 10 月 28 日 (日) 22 時 41 分 49 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:57:37 UTC 2012 年 11 月 2 日 (金) 16 時 57 分 37 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:57:37 UTC 2012 年 11 月 2 日 (金) 16 時 57 分 37 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 12, 2012 13:07:48 UTC 2012 年 11 月 12 日 (月) 22 時 7 分 48 秒 (日本時間)
14×10246-419
c242
name 名前 Warut Roonguthai October 27, 2012 05:41:24 UTC 2012 年 10 月 27 日 (土) 14 時 41 分 24 秒 (日本時間) `14206376024508028124565563947464821461369311995356544522275090235856284241171498356626716307803460876147799077194403093742801680005439012535097359339119387339886531645209965163936505617099605976013548823762802227965657100701896449725157360983<242>` ```6737527267697920053932641382164447<34> 2108544494152684315583512497636132157796205703313855897177019508753270779491080999327296504990149374229885188176679783061174504141853883762017513612326281807113424624078799785697992746691292995409643640631689<208>``` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3648350481 Step 1 took 13198ms Step 2 took 7176ms ********** Factor found in step 2: 6737527267697920053932641382164447 Found probable prime factor of 34 digits: 6737527267697920053932641382164447 Probable prime cofactor 2108544494152684315583512497636132157796205703313855897177019508753270779491080999327296504990149374229885188176679783061174504141853883762017513612326281807113424624078799785697992746691292995409643640631689 has 208 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10248-419
c212
name 名前 Warut Roonguthai October 27, 2012 06:45:06 UTC 2012 年 10 月 27 日 (土) 15 時 45 分 6 秒 (日本時間) `59596638836329399125234875235036604741730872655242860980106518960308150505951064160915084040117139371070027477929181319708148635120452975129226782587309461206568575636580197648549305077064819333438170762694378809<212>` `50271346306525277927796529840686277<35>` `1185499160355561983840194223978284897041693701107045646589607972306813940584854138542774585091124677151102323302104850279347007332650434050996261492146671643256625283181795577317<178>` ```Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=4261158317 Step 1 took 10249ms Step 2 took 5881ms ********** Factor found in step 2: 50271346306525277927796529840686277 Found probable prime factor of 35 digits: 50271346306525277927796529840686277 Composite cofactor 1185499160355561983840194223978284897041693701107045646589607972306813940584854138542774585091124677151102323302104850279347007332650434050996261492146671643256625283181795577317 has 178 digits``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
351e61118118Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
1000Warut RoonguthaiOctober 28, 2012 13:42:03 UTC 2012 年 10 月 28 日 (日) 22 時 42 分 3 秒 (日本時間)
403e61000Dmitry DomanovNovember 2, 2012 07:57:48 UTC 2012 年 11 月 2 日 (金) 16 時 57 分 48 秒 (日本時間)
4511e61000 / 2454Dmitry DomanovNovember 2, 2012 07:57:48 UTC 2012 年 11 月 2 日 (金) 16 時 57 分 48 秒 (日本時間)
5043e6500 / 7285Dmitry DomanovNovember 8, 2012 09:00:09 UTC 2012 年 11 月 8 日 (木) 18 時 0 分 9 秒 (日本時間)
14×10250-419
c146
name 名前 Warut Roonguthai October 28, 2012 07:18:33 UTC 2012 年 10 月 28 日 (日) 16 時 18 分 33 秒 (日本時間) `70476228494636718767747614371594781901060881239092783290814737375574418122961577127810696392671016080554650516478234056424745117506721595407041723<146>` ```59108348596734089361496117525008048667<38> 1344499096665215170127917227005493088088917784991083<52> 886815579639241291722346417859445521710355103108594646243<57>``` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=1395421991 Step 1 took 19781ms Step 2 took 12386ms ********** Factor found in step 2: 59108348596734089361496117525008048667 Found probable prime factor of 38 digits: 59108348596734089361496117525008048667 Composite cofactor 1192322745733599099528372466167683786674686544210517619930295004126996678527676623723913282336366793294451169 has 109 digits N = 1192322745733599099528372466167683786674686544210517619930295004126996678527676623723913282336366793294451169 (109 digits) Divisors found: r1=1344499096665215170127917227005493088088917784991083 (pp52) r2=886815579639241291722346417859445521710355103108594646243 (pp57) Version: Msieve v. 1.49 (SVN unknown) Total time: 2.63 hours. Factorization parameters were as follows: n: 1192322745733599099528372466167683786674686544210517619930295004126996678527676623723913282336366793294451169 Y0: -1459603680048880733954 Y1: 141144622871 c0: -146820632669345282687400375 c1: 2254590896087290628435 c2: 739099967176065287 c3: -960882397215 c4: -228966468 c5: 180 skew: 58172.72 type: gnfs Factor base limits: 3200000/3200000 Large primes per side: 3 Large prime bits: 27/27 Sieved algebraic special-q in [0, 0) Total raw relations: 7437862 Relations: 664020 relations Pruned matrix : 384723 x 384950 Polynomial selection time: 0.00 hours. Total sieving time: 2.43 hours. Total relation processing time: 0.04 hours. Matrix solve time: 0.14 hours. time per square root: 0.02 hours. Prototype def-par.txt line would be: gnfs,108,5,61,2000,0.00015,0.3,250,15,50000,2400,3200000,3200000,27,27,50,50,2.6,2.6,100000 total time: 2.63 hours. Intel64 Family 6 Model 42 Stepping 7, GenuineIntel Windows-7-6.1.7601-SP1 processors: 8, speed: 2.19GHz``` `GMP-ECM 6.3`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e40--
351e6118 / 904Makoto KamadaOctober 27, 2012 01:00:00 UTC 2012 年 10 月 27 日 (土) 10 時 0 分 0 秒 (日本時間)
14×10251-419
c208
composite cofactor 合成数の残り `3112816233863202994985946197989475651544320285181123972044227916119433371937731602072185648858556790977499165360974816540577975558565693466491846484393507480001230827534919273873377110254876736838021295858727<208>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:04:10 UTC 2015 年 11 月 20 日 (金) 23 時 4 分 10 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:31:29 UTC 2015 年 11 月 21 日 (土) 18 時 31 分 29 秒 (日本時間)
4511e6478 / 4331CypApril 21, 2017 09:15:58 UTC 2017 年 4 月 21 日 (金) 18 時 15 分 58 秒 (日本時間)
14×10252-419
c239
composite cofactor 合成数の残り `15393153963145396186577758917537701541707621120472528036740272655468325791458366516199909621636632421592066460578842130845643853765765384627422140411518061565802515451611063313990295147473113887839076489073341516882376842428943528768086963<239>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:04:21 UTC 2015 年 11 月 20 日 (金) 23 時 4 分 21 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:10:18 UTC 2015 年 11 月 22 日 (日) 5 時 10 分 18 秒 (日本時間)
4511e6478 / 4331CypApril 21, 2017 12:43:22 UTC 2017 年 4 月 21 日 (金) 21 時 43 分 22 秒 (日本時間)
14×10253-419
c226
name 名前 KTakahashi November 20, 2015 14:05:18 UTC 2015 年 11 月 20 日 (金) 23 時 5 分 18 秒 (日本時間) `1448187704612551389021307206332308393948777551503684542253472278320008696284670915014790452790141479621093228547124951246609392925907015610203200012262493328523627762979085445318618085414404275947127091107102228991498965308641<226>` `188834745255469739218397762416031<33>` `7669074367925960660474656733957583672960868433132962546909226433652687373862174799195161063771705885821468654353353545263262287322501563787101274846555874276399306361360108616634330686087933311<193>` ```Input number is 1448187704612551389021307206332308393948777551503684542253472278320008696284670915014790452790141479621093228547124951246609392925907015610203200012262493328523627762979085445318618085414404275947127091107102228991498965308641 (226 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3875532521 Step 1 took 7052ms Step 2 took 1466ms ********** Factor found in step 2: 188834745255469739218397762416031 Found probable prime factor of 33 digits: 188834745255469739218397762416031 Composite cofactor 7669074367925960660474656733957583672960868433132962546909226433652687373862174799195161063771705885821468654353353545263262287322501563787101274846555874276399306361360108616634330686087933311 has 193 digits``` `GMP-ECM 6.4.4`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:04:44 UTC 2015 年 11 月 20 日 (金) 23 時 4 分 44 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:10:43 UTC 2015 年 11 月 22 日 (日) 5 時 10 分 43 秒 (日本時間)
4511e6600 / 4331Dmitry DomanovJanuary 9, 2017 11:52:14 UTC 2017 年 1 月 9 日 (月) 20 時 52 分 14 秒 (日本時間)
14×10255-419
c196
composite cofactor 合成数の残り `4106891657937826208116065517463322220632870346297613477831948876624687189573837596457031439533279559488337696741074985363410408593358917723218030995905018140985250015604486371494806906276508305489<196>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:08:20 UTC 2015 年 11 月 20 日 (金) 23 時 8 分 20 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:31:43 UTC 2015 年 11 月 21 日 (土) 18 時 31 分 43 秒 (日本時間)
4511e6600 / 4331Dmitry DomanovJanuary 9, 2017 11:52:43 UTC 2017 年 1 月 9 日 (月) 20 時 52 分 43 秒 (日本時間)
14×10256-419
c210
name 名前 Dmitry Domanov November 21, 2015 16:44:45 UTC 2015 年 11 月 22 日 (日) 1 時 44 分 45 秒 (日本時間) `116754790655891099800867116631756175397006753925620632052529377598561031215962059884559464515695057169450111977721530752616382833626667251813905751235193318018501581199781681264963232165848220519179113031709353<210>` ```286134591205873665035310291611393<33> 408041509989563263439104959983333592314834871250397614962269381726297072534886981872314225410839098800673279421059874236579228251298880728651185979804288846049496525118513015721<177>``` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=2901009947 Step 1 took 35802ms Step 2 took 11984ms ********** Factor found in step 2: 286134591205873665035310291611393 Found probable prime factor of 33 digits: 286134591205873665035310291611393```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:08:32 UTC 2015 年 11 月 20 日 (金) 23 時 8 分 32 秒 (日本時間)
403e6600 / 2247Dmitry DomanovNovember 21, 2015 09:31:57 UTC 2015 年 11 月 21 日 (土) 18 時 31 分 57 秒 (日本時間)
14×10257-419
c230
name 名前 KTakahashi November 20, 2015 14:09:34 UTC 2015 年 11 月 20 日 (金) 23 時 9 分 34 秒 (日本時間) `20707456639827067490128754183845609354421671970778237652743144271911185394703203423705598366379144011225276650581011525580250598690376755140323217822451821918834697092727800416365556116990164946184917383741015162537533337324504947<230>` `1136126963133274811515370193587<31>` `18226357891127660067316820749341995462622287743279825746842507529216059282556615272694205337956264302335786567672115022063145014449901827047393523235180842326758219087634808017375895831208609549961281<200>` ```Input number is 20707456639827067490128754183845609354421671970778237652743144271911185394703203423705598366379144011225276650581011525580250598690376755140323217822451821918834697092727800416365556116990164946184917383741015162537533337324504947 (230 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=118141309 Step 1 took 7035ms ********** Factor found in step 1: 1136126963133274811515370193587 Found probable prime factor of 31 digits: 1136126963133274811515370193587 Composite cofactor 18226357891127660067316820749341995462622287743279825746842507529216059282556615272694205337956264302335786567672115022063145014449901827047393523235180842326758219087634808017375895831208609549961281 has 200 digits``` `GMP-ECM 6.4.4`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:08:50 UTC 2015 年 11 月 20 日 (金) 23 時 8 分 50 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:11:06 UTC 2015 年 11 月 22 日 (日) 5 時 11 分 6 秒 (日本時間)
4511e6600 / 4331Dmitry DomanovJanuary 9, 2017 11:53:00 UTC 2017 年 1 月 9 日 (月) 20 時 53 分 0 秒 (日本時間)
14×10258-419
c256
composite cofactor 合成数の残り `1942016923290331530031904563739769732279095574975724788458870855874601192953252878346511305312803440144264114301567485088084339020668608683589956998196698571230406436398945762241642391455125537522541267859619919545013177971979470106810930780968234151754751<256>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:22:22 UTC 2015 年 11 月 21 日 (土) 18 時 22 分 22 秒 (日本時間)
4511e6800 / 4305Dmitry DomanovFebruary 5, 2016 14:30:58 UTC 2016 年 2 月 5 日 (金) 23 時 30 分 58 秒 (日本時間)
14×10259-419
c254
composite cofactor 合成数の残り `74237042056875118978116510795391576534943640828464178771281506333215084330628453679532476319709245322138387391156565391433363505388283591863832296400935174624082178284498616274562518459835904321179138754864514746923272305182116720780167680267423036072309<254>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:20:50 UTC 2015 年 11 月 21 日 (土) 18 時 20 分 50 秒 (日本時間)
4511e6800 / 4305Dmitry DomanovFebruary 4, 2016 18:12:54 UTC 2016 年 2 月 5 日 (金) 3 時 12 分 54 秒 (日本時間)
14×10261-419
c226
composite cofactor 合成数の残り `2024579133856333260349679379697995092914275824489059473872456991412273719672942587800946079421101733944537191688579688074948133324055810151525080908790491883381918556981112864741358517920358506114796417540589332851711976879839<226>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:10:37 UTC 2015 年 11 月 20 日 (金) 23 時 10 分 37 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:11:26 UTC 2015 年 11 月 22 日 (日) 5 時 11 分 26 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 21:57:52 UTC 2017 年 4 月 21 日 (金) 6 時 57 分 52 秒 (日本時間)
14×10262-419
c206
composite cofactor 合成数の残り `71766977888581900810103015328426004395464874652545415585928501816952225361699309585321538218126629174966697175924360964061153246243451814473229410358570325346026359277273759343723146277464024628165448270227<206>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:10:58 UTC 2015 年 11 月 20 日 (金) 23 時 10 分 58 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:33:30 UTC 2015 年 11 月 21 日 (土) 18 時 33 分 30 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 16:47:47 UTC 2017 年 4 月 21 日 (金) 1 時 47 分 47 秒 (日本時間)
14×10263-419
c237
name 名前 KTakahashi November 20, 2015 14:11:51 UTC 2015 年 11 月 20 日 (金) 23 時 11 分 51 秒 (日本時間) `322384523525902623056955904766519162856657900488323081199879664132445835573311077798304561958151968015434827312519998752960995880444503191014233309174021934877809387806764631256486689882708470426012929580068317601061705868616229775725239<237>` `30523597587524175628050870611891<32>` `10561812794232025406048464695744154984967121005252918376443776335373911366295255896446476490029936317272530936444326469387399154659586370482599784979385012983206369432803056455591466405889623331904984264429<206>` ```Input number is 322384523525902623056955904766519162856657900488323081199879664132445835573311077798304561958151968015434827312519998752960995880444503191014233309174021934877809387806764631256486689882708470426012929580068317601061705868616229775725239 (237 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=2157278795 Step 1 took 7862ms Step 2 took 3994ms ********** Factor found in step 2: 30523597587524175628050870611891 Found probable prime factor of 32 digits: 30523597587524175628050870611891 Composite cofactor 10561812794232025406048464695744154984967121005252918376443776335373911366295255896446476490029936317272530936444326469387399154659586370482599784979385012983206369432803056455591466405889623331904984264429 has 206 digits``` `GMP-ECM 6.4.4`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:11:17 UTC 2015 年 11 月 20 日 (金) 23 時 11 分 17 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:13:48 UTC 2015 年 11 月 22 日 (日) 5 時 13 分 48 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 15:37:56 UTC 2017 年 4 月 21 日 (金) 0 時 37 分 56 秒 (日本時間)
14×10264-419
c255
name 名前 KTakahashi November 20, 2015 14:13:12 UTC 2015 年 11 月 20 日 (金) 23 時 13 分 12 秒 (日本時間) `219354992162949628065892388226943358072433303558505038441033016459769078748446150556989134673645992813681363479955429333723107600476164868039579376234004021677270571130879571336765117210590001140016567101967659759825640406171439498526775232515832222359093<255>` `30968737572007807401283636159346833<35>` `7083110561188048656289760877917901575552852277880967971793341057725659448035796375721168430466536356934449797356617801523409001610907762515986406370240197351146794723727295490409844741719041207709125352227021569086373221<220>` ```Input number is 219354992162949628065892388226943358072433303558505038441033016459769078748446150556989134673645992813681363479955429333723107600476164868039579376234004021677270571130879571336765117210590001140016567101967659759825640406171439498526775232515832222359093 (255 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3282559747 Step 1 took 8846ms Step 2 took 4430ms ********** Factor found in step 2: 30968737572007807401283636159346833 Found probable prime factor of 35 digits: 30968737572007807401283636159346833 Composite cofactor 7083110561188048656289760877917901575552852277880967971793341057725659448035796375721168430466536356934449797356617801523409001610907762515986406370240197351146794723727295490409844741719041207709125352227021569086373221 has 220 digits``` `GMP-ECM 6.4.4`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:12:34 UTC 2015 年 11 月 20 日 (金) 23 時 12 分 34 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 20:14:05 UTC 2015 年 11 月 22 日 (日) 5 時 14 分 5 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 14:41:19 UTC 2017 年 4 月 20 日 (木) 23 時 41 分 19 秒 (日本時間)
14×10265-419
c229
composite cofactor 合成数の残り `2083944074273421498008758636237254305317306820953947140066550157892492230400845888599706603986490060149756051619591892445634434978381339328827759264566353434633944116653478278774577111042197932310080908489094616308543290104556409<229>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:13:35 UTC 2015 年 11 月 20 日 (金) 23 時 13 分 35 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 11:17:50 UTC 2015 年 11 月 25 日 (水) 20 時 17 分 50 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 13:31:04 UTC 2017 年 4 月 20 日 (木) 22 時 31 分 4 秒 (日本時間)
14×10266-419
c222
name 名前 Cyp April 19, 2017 12:26:26 UTC 2017 年 4 月 19 日 (水) 21 時 26 分 26 秒 (日本時間) `218255526815461514178944885003125254163365868656352164129288723201555835682674192330504441303274986473135486360402459771004330991361809616613115520318792666692958189606213915631572622438417617031465659540502434760480040267<222>` ```283773982734976295944310316245921753<36> 769117467048752930074407996810548360226830245239432703514449484209614788978311666276005063294714744368964671235611953787689846355522489208157145949100445698847183182466940975923778022339<186>``` ```Run 25 out of 478: Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=2177997359 Step 1 took 84658ms Step 2 took 25582ms ********** Factor found in step 2: 283773982734976295944310316245921753 Found probable prime factor of 36 digits: 283773982734976295944310316245921753 Probable prime cofactor 769117467048752930074407996810548360226830245239432703514449484209614788978311666276005063294714744368964671235611953787689846355522489208157145949100445698847183182466940975923778022339 has 186 digits``` `GMP-ECM 6.4.4` `Gentoo Linux, Intel(R) Core(TM) i7-3770K CPU`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:13:55 UTC 2015 年 11 月 20 日 (金) 23 時 13 分 55 秒 (日本時間)
403e6600 / 2161Dmitry DomanovNovember 21, 2015 09:34:53 UTC 2015 年 11 月 21 日 (土) 18 時 34 分 53 秒 (日本時間)
4511e625 / 4331CypApril 19, 2017 12:26:26 UTC 2017 年 4 月 19 日 (水) 21 時 26 分 26 秒 (日本時間)
14×10268-419
c269
composite cofactor 合成数の残り `15555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555551<269>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 20, 2015 22:24:19 UTC 2015 年 11 月 21 日 (土) 7 時 24 分 19 秒 (日本時間)
4511e61700Serge BatalovDecember 19, 2015 07:51:31 UTC 2015 年 12 月 19 日 (土) 16 時 51 分 31 秒 (日本時間)
5043e65150600Dmitry DomanovFebruary 8, 2016 14:27:25 UTC 2016 年 2 月 8 日 (月) 23 時 27 分 25 秒 (日本時間)
150Dmitry DomanovFebruary 12, 2016 20:22:34 UTC 2016 年 2 月 13 日 (土) 5 時 22 分 34 秒 (日本時間)
2400Dmitry DomanovOctober 14, 2016 22:24:44 UTC 2016 年 10 月 15 日 (土) 7 時 24 分 44 秒 (日本時間)
2000Dmitry DomanovOctober 21, 2016 17:09:39 UTC 2016 年 10 月 22 日 (土) 2 時 9 分 39 秒 (日本時間)
5511e7840 / 15820Dmitry DomanovOctober 27, 2016 15:36:56 UTC 2016 年 10 月 28 日 (金) 0 時 36 分 56 秒 (日本時間)
14×10270-419
c223
composite cofactor 合成数の残り `4918559301848812959482329259737821397831503921188597817503107329449481212544866833115503012703263273646878650150540631999337421948256128665204708447021582660472874230222953607293615113401813554800658989851148774974648892403<223>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:14:45 UTC 2015 年 11 月 20 日 (金) 23 時 14 分 45 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:34:36 UTC 2015 年 11 月 21 日 (土) 18 時 34 分 36 秒 (日本時間)
4511e6478 / 4331CypApril 20, 2017 11:19:02 UTC 2017 年 4 月 20 日 (木) 20 時 19 分 2 秒 (日本時間)
14×10271-419
c268
composite cofactor 合成数の残り `7812935989731569842067079636140409621072604498018862659746637647190133377978681846085161002288074111278531168033930464869691389028405602991238350354372453819967632122328254924940007812935989731569842067079636140409621072604498018862659746637647190133377978681846085161<268>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 20, 2015 20:44:39 UTC 2015 年 11 月 21 日 (土) 5 時 44 分 39 秒 (日本時間)
4511e6800 / 4305Dmitry DomanovJanuary 23, 2016 15:48:04 UTC 2016 年 1 月 24 日 (日) 0 時 48 分 4 秒 (日本時間)
14×10272-419
c197
composite cofactor 合成数の残り `31617629848920798175010622290862827182295841089884624210315321610504554547892271316665002898263285608795083102388022497754016432886271856551254586963637140541526965728377244693376682117170474400183<197>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:15:13 UTC 2015 年 11 月 20 日 (金) 23 時 15 分 13 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:33:52 UTC 2015 年 11 月 21 日 (土) 18 時 33 分 52 秒 (日本時間)
4511e6600 / 4331Dmitry DomanovJanuary 9, 2017 11:53:30 UTC 2017 年 1 月 9 日 (月) 20 時 53 分 30 秒 (日本時間)
14×10273-419
c239
composite cofactor 合成数の残り `13160177607420748696515718411707278667582297009739017492205305213036002657216629204168855319864304085231090908508928117220227821764638911607131182389511091701912410669590995778849158287155779449701460808644221786644408553792428723986355869<239>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 11:41:29 UTC 2015 年 11 月 25 日 (水) 20 時 41 分 29 秒 (日本時間)
4511e6432 / 4305CypApril 20, 2017 10:24:47 UTC 2017 年 4 月 20 日 (木) 19 時 24 分 47 秒 (日本時間)
14×10274-419
c208
name 名前 KTakahashi November 20, 2015 14:16:09 UTC 2015 年 11 月 20 日 (金) 23 時 16 分 9 秒 (日本時間) `8870044469642847465917287304082027695096941293413358667225719599502703904698497535004988336505429782543560614593613970448876862080434347863661722104734300360134089225723200983857082343093673651950060579145311<208>` ```2457056766140667794567749963<28> 3610028303731519382448089141304067772911868629239177659647377785993984894056328144127075533308179493585912406934159675661094519011744554442300639069322259943014536106961361495070397<181>``` ```Input number is 8870044469642847465917287304082027695096941293413358667225719599502703904698497535004988336505429782543560614593613970448876862080434347863661722104734300360134089225723200983857082343093673651950060579145311 (208 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=239827847 Step 1 took 6303ms ********** Factor found in step 1: 2457056766140667794567749963 Found probable prime factor of 28 digits: 2457056766140667794567749963 Probable prime cofactor 3610028303731519382448089141304067772911868629239177659647377785993984894056328144127075533308179493585912406934159675661094519011744554442300639069322259943014536106961361495070397 has 181 digits``` `GMP-ECM 6.4.4`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
14×10275-419
c268
composite cofactor 合成数の残り `8044909396535198883463114673672817361921240670297023973674849850816636741614938904860619904745859272204284660131683329311740589643934091482147044497675311375775988256638884778802141837602459319608338642829875950197896438927176623934349034175010717399566244918612553549<268>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 12:58:06 UTC 2015 年 11 月 25 日 (水) 21 時 58 分 6 秒 (日本時間)
4511e6432 / 4305CypApril 20, 2017 09:56:49 UTC 2017 年 4 月 20 日 (木) 18 時 56 分 49 秒 (日本時間)
14×10276-419
c240
composite cofactor 合成数の残り `685595872286690055257430214439247525964680959273226819105161586223954115103867870296080258976931336639263632595051607921018947509055817014595937597488089766829632286660456043922124707031906188843003259361375108740325566461025459646688798149<240>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 12:58:35 UTC 2015 年 11 月 25 日 (水) 21 時 58 分 35 秒 (日本時間)
4511e6432 / 4305CypApril 19, 2017 20:44:51 UTC 2017 年 4 月 20 日 (木) 5 時 44 分 51 秒 (日本時間)
14×10278-419
c238
composite cofactor 合成数の残り `1195874280932312208330622218397355635561604274394348465052310111959378113296617464574170790037738035748868667578701424946330878243983518724123150802678946883037524195229235388137689498351285394851505858212827703533363596315210244890928427<238>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:17:26 UTC 2015 年 11 月 20 日 (金) 23 時 17 分 26 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:39:59 UTC 2015 年 11 月 25 日 (水) 22 時 39 分 59 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 16:26:27 UTC 2017 年 4 月 20 日 (木) 1 時 26 分 27 秒 (日本時間)
14×10281-419
c263
composite cofactor 合成数の残り `10102241429421137720857911549942254946822872575785570836757201986096680905073304952087317193078778598707129243002621680622409920994270677666538502278873611994189231603395438870871288128755169622505899642808775955765186177560333055118751568403205742656985880473241<263>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:18:09 UTC 2015 年 11 月 20 日 (金) 23 時 18 分 9 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:40:28 UTC 2015 年 11 月 25 日 (水) 22 時 40 分 28 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 17:34:30 UTC 2017 年 4 月 20 日 (木) 2 時 34 分 30 秒 (日本時間)
14×10282-419
c267
name 名前 KTakahashi November 20, 2015 14:19:11 UTC 2015 年 11 月 20 日 (金) 23 時 19 分 11 秒 (日本時間) `517069671634855985537862106644279198361785828104399127903182033968911499215544330929732572775253773320192352994965488202487319880387538756911766876028602386800494504250696388354784671451718139723051293642902607914179060306760778524304532701497423728995320744464688637<267>` `94866642671269508651208144263<29>` `5450489835785568878486065953473529744663898516778693834955272251899053323284628290896768578774330955952398598409641365222998966872591268125714255296587177725019330299188376082909126993457092010317012178970954696081187355315129356572158299<238>` ```Input number is 517069671634855985537862106644279198361785828104399127903182033968911499215544330929732572775253773320192352994965488202487319880387538756911766876028602386800494504250696388354784671451718139723051293642902607914179060306760778524304532701497423728995320744464688637 (267 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=1111111111 Step 1 took 9126ms Step 2 took 4586ms ********** Factor found in step 2: 94866642671269508651208144263 Found probable prime factor of 29 digits: 94866642671269508651208144263 Composite cofactor 5450489835785568878486065953473529744663898516778693834955272251899053323284628290896768578774330955952398598409641365222998966872591268125714255296587177725019330299188376082909126993457092010317012178970954696081187355315129356572158299 has 238 digits``` `GMP-ECM 6.4.4`
c238
name 名前 Cyp April 19, 2017 09:04:10 UTC 2017 年 4 月 19 日 (水) 18 時 4 分 10 秒 (日本時間) `5450489835785568878486065953473529744663898516778693834955272251899053323284628290896768578774330955952398598409641365222998966872591268125714255296587177725019330299188376082909126993457092010317012178970954696081187355315129356572158299<238>` `116562361801642231987212231482059939<36>` `46760289955867878052804013277556232561191260348496508896746503619132154898579508477786469316555715895535488621825938717467857424416026536332844056303651024121195586898152320097128467042105436324928267241<203>` ```Run 116 out of 478: Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=1683676770 Step 1 took 96766ms Step 2 took 27151ms ********** Factor found in step 2: 116562361801642231987212231482059939 Found probable prime factor of 36 digits: 116562361801642231987212231482059939 Composite cofactor 46760289955867878052804013277556232561191260348496508896746503619132154898579508477786469316555715895535488621825938717467857424416026536332844056303651024121195586898152320097128467042105436324928267241 has 203 digits``` `GMP-ECM 6.4.4` `Gentoo Linux, Intel(R) Core(TM) i7-3770K CPU`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:18:33 UTC 2015 年 11 月 20 日 (金) 23 時 18 分 33 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:40:45 UTC 2015 年 11 月 25 日 (水) 22 時 40 分 45 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 09:04:10 UTC 2017 年 4 月 19 日 (水) 18 時 4 分 10 秒 (日本時間)
14×10283-419
c265
composite cofactor 合成数の残り `9252286309071406621050107799846263667808462372213990705837000693097233208843047712660826623948191092136795894501439729615915351071780071725507286890765699011700132339611804113701267416436077935598474600626855121245870852577667144609722736419094309888866262835146051<265>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:19:59 UTC 2015 年 11 月 20 日 (金) 23 時 19 分 59 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:41:01 UTC 2015 年 11 月 25 日 (水) 22 時 41 分 1 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 13:36:51 UTC 2017 年 4 月 19 日 (水) 22 時 36 分 51 秒 (日本時間)
14×10284-419
c254
composite cofactor 合成数の残り `27149499520364171926494265386326048081925064872022576138802935375657281627117938310181134266412855070739764980771725055323132927595757478410810381407981380567708956688509929231100504945426606811647324504935025905387975814302555117591783704599698401744713<254>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:20:12 UTC 2015 年 11 月 20 日 (金) 23 時 20 分 12 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:41:24 UTC 2015 年 11 月 25 日 (水) 22 時 41 分 24 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 10:14:19 UTC 2017 年 4 月 19 日 (水) 19 時 14 分 19 秒 (日本時間)
14×10285-419
c264
name 名前 Dmitry Domanov November 25, 2015 16:57:42 UTC 2015 年 11 月 26 日 (木) 1 時 57 分 42 秒 (日本時間) `325878390339545721735124382223808307852948443078166369566537987126552102146410790754776972228893498980601383509373788298205468230909847396468617644125828982411186478803645011025225875401476486296096400390812052495121724793234523447965371453314983768251173441645871<264>` ```29320105063394118446442929079671237<35> 11114502817604221186881790288576271629784876820983595467868088552723978517608727132386326065790549446387173157361124885559831348393316866090561086224744898956778838017275508467404918711348399172409102090207322610386410530199043683<230>``` ```Using B1=3000000, B2=5706890290, polynomial Dickson(6), sigma=1228472304 Step 1 took 34653ms Step 2 took 11115ms ********** Factor found in step 2: 29320105063394118446442929079671237 Found probable prime factor of 35 digits: 29320105063394118446442929079671237```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:20:23 UTC 2015 年 11 月 20 日 (金) 23 時 20 分 23 秒 (日本時間)
403e6600 / 2247Dmitry DomanovNovember 25, 2015 13:41:42 UTC 2015 年 11 月 25 日 (水) 22 時 41 分 42 秒 (日本時間)
14×10286-419
c273
composite cofactor 合成数の残り `257940671539919528229674943764168773547868772570264982552363310158906655736966747874098643010881666706534478747938277748267278864878421437616394920791050963049741145738021069997647421137028957863175495831629941410030676342785979080126821253168907724548632703046012185769523<273>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:20:38 UTC 2015 年 11 月 20 日 (金) 23 時 20 分 38 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:42:05 UTC 2015 年 11 月 25 日 (水) 22 時 42 分 5 秒 (日本時間)
4511e6478 / 4331CypApril 19, 2017 11:39:39 UTC 2017 年 4 月 19 日 (水) 20 時 39 分 39 秒 (日本時間)
14×10287-419
c244
composite cofactor 合成数の残り `4234580712376618455262452410381470965168430116465395248198006061208940611183387357459790543303650568447588600400622233305806764022627199777972534050745643474425516310992960620819364033424101899991465597793935605571008384698494631654599570656263<244>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:20:49 UTC 2015 年 11 月 20 日 (金) 23 時 20 分 49 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:42:39 UTC 2015 年 11 月 25 日 (水) 22 時 42 分 39 秒 (日本時間)
4511e6478 / 4331CypApril 18, 2017 15:08:28 UTC 2017 年 4 月 19 日 (水) 0 時 8 分 28 秒 (日本時間)
14×10289-419
c276
name 名前 Cyp April 18, 2017 08:14:08 UTC 2017 年 4 月 18 日 (火) 17 時 14 分 8 秒 (日本時間) `425915581197697045096985880153781405381728240798332460009045247309535337060769668788745318269140165442648364144990262195983822780011776390041531173606864299004444994729197297035224401224256481015066601980581623096049242871262661347733744977849046212282818146431215362418728717<276>` ```12701481519589667054267156719693516463<38> 33532748171211497887735910941781162553022528585334826352582581717557542438294418189335472841788554383210827771538397385955993428793651349773046794263839275535715462647187205005215208326276490674619491423723105166810053920993505889197849859<239>``` ```Run 268 out of 478: Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=2076019466 Step 1 took 123355ms Step 2 took 32505ms ********** Factor found in step 2: 12701481519589667054267156719693516463 Found probable prime factor of 38 digits: 12701481519589667054267156719693516463 Probable prime cofactor 33532748171211497887735910941781162553022528585334826352582581717557542438294418189335472841788554383210827771538397385955993428793651349773046794263839275535715462647187205005215208326276490674619491423723105166810053920993505889197849859 has 239 digits``` `GMP-ECM 6.4.4` `Gentoo Linux, Intel(R) Core(TM) i7-3770K CPU`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:21:05 UTC 2015 年 11 月 20 日 (金) 23 時 21 分 5 秒 (日本時間)
403e6600 / 1323Dmitry DomanovNovember 25, 2015 17:03:41 UTC 2015 年 11 月 26 日 (木) 2 時 3 分 41 秒 (日本時間)
4511e6268 / 4331CypApril 18, 2017 08:14:07 UTC 2017 年 4 月 18 日 (火) 17 時 14 分 7 秒 (日本時間)
14×10290-419
c267
name 名前 Cyp April 18, 2017 07:05:17 UTC 2017 年 4 月 18 日 (火) 16 時 5 分 17 秒 (日本時間) `536934303616391624954960766590338338047154867198821991460169173556052674796346981514171000716772196060355497659758690341566910594667097728896695519783100613010590250649693691576373302950984038635116000760730084873052063853341433922271437833912696222977659627018327041<267>` ```381994026431576960046054525742898059<36> 1014268708012921978316106475265432579646613<43>``` `1385835086412835673703721808646086244552024076259457489575750259123563584018491175450620361129931288599576480194424392343049183694880126712850547881949818692026118151441913165987245388913623<190>` ```Run 45 out of 478: Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=1188496405 Step 1 took 110437ms ********** Factor found in step 1: 381994026431576960046054525742898059 Found probable prime factor of 36 digits: 381994026431576960046054525742898059 Composite cofactor 1405609162614922924320109699605231457767877944467560416061202840287603630422471559061189992872697828979018848228339675979560183291434935419107114584834815783459449752478797931763252053685112185364569231756142868692636663785321508899 has 232 digits -- Run 60 out of 478: Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=2023009989 Step 1 took 87206ms Step 2 took 25021ms ********** Factor found in step 2: 1014268708012921978316106475265432579646613 Found probable prime factor of 43 digits: 1014268708012921978316106475265432579646613 Composite cofactor 1385835086412835673703721808646086244552024076259457489575750259123563584018491175450620361129931288599576480194424392343049183694880126712850547881949818692026118151441913165987245388913623 has 190 digits``` `GMP-ECM 6.4.4` `Gentoo Linux, Intel(R) Core(TM) i7-3770K CPU`
c190
name 名前 Dmitry Domanov April 20, 2017 06:46:13 UTC 2017 年 4 月 20 日 (木) 15 時 46 分 13 秒 (日本時間) `1385835086412835673703721808646086244552024076259457489575750259123563584018491175450620361129931288599576480194424392343049183694880126712850547881949818692026118151441913165987245388913623<190>` `7909613819786497592916141503753952440293<40>` `175208944202315455779551271697609420005592888019094524786760710186180442358883118890653871898924092620415546779536327033515688084062105080983942252811<150>` ```Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=3785126025 Step 1 took 62986ms Step 2 took 22154ms ********** Factor found in step 2: 7909613819786497592916141503753952440293 Found probable prime factor of 40 digits: 7909613819786497592916141503753952440293```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:21:17 UTC 2015 年 11 月 20 日 (金) 23 時 21 分 17 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 17:03:21 UTC 2015 年 11 月 26 日 (木) 2 時 3 分 21 秒 (日本時間)
4511e64378478CypApril 18, 2017 07:05:16 UTC 2017 年 4 月 18 日 (火) 16 時 5 分 16 秒 (日本時間)
2000Dmitry DomanovApril 20, 2017 00:23:08 UTC 2017 年 4 月 20 日 (木) 9 時 23 分 8 秒 (日本時間)
1900Lionel DebrouxDecember 29, 2017 15:02:15 UTC 2017 年 12 月 30 日 (土) 0 時 2 分 15 秒 (日本時間)
14×10291-419
c288
composite cofactor 合成数の残り `596684140987938456292886672633508076546051229595533393001747432127178962622000596684140987938456292886672633508076546051229595533393001747432127178962622000596684140987938456292886672633508076546051229595533393001747432127178962622000596684140987938456292886672633508076546051229595533393<288>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 20, 2015 13:43:27 UTC 2015 年 11 月 20 日 (金) 22 時 43 分 27 秒 (日本時間)
4511e6800 / 4305Dmitry DomanovDecember 31, 2015 00:27:37 UTC 2015 年 12 月 31 日 (木) 9 時 27 分 37 秒 (日本時間)
14×10292-419
c284
composite cofactor 合成数の残り `20996801579262072666695858519902314726295173918766886854830634846965928269460937698294403396739698208348605375709153869085733732720515103290856562275267998814771944545197971765418844101167855406812128485620520336038757568900224876089861351314708649311894400903399506101804993452242663<284>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 17:02:57 UTC 2015 年 11 月 26 日 (木) 2 時 2 分 57 秒 (日本時間)
4511e6432 / 4305CypApril 18, 2017 14:05:47 UTC 2017 年 4 月 18 日 (火) 23 時 5 分 47 秒 (日本時間)
14×10293-419
c242
name 名前 KTakahashi November 20, 2015 14:23:07 UTC 2015 年 11 月 20 日 (金) 23 時 23 分 7 秒 (日本時間) `27139290676975296447567583002670247237758786702242921509488836438134704762046823748724684383872582298783548191090060988608826356765499928970708285895299008284381805471873849778479311912293607333223671217870348572444439937900371187278023336841<242>` `540644499510751614689592659509<30>` `50198033461053618901414634779631927059274077161687913901931115543562208759765901061475667529776063988007503567882597626322608693341559073443818806711644490181527738248674313467847236423451988931896018594670230149<212>` ```GMP-ECM 6.4.4 [configured with GMP 6.0.0] [ECM] Input number is 27139290676975296447567583002670247237758786702242921509488836438134704762046823748724684383872582298783548191090060988608826356765499928970708285895299008284381805471873849778479311912293607333223671217870348572444439937900371187278023336841 (242 digits) Using B1=1000000, B2=1045563762, polynomial Dickson(6), sigma=3825366648 Step 1 took 8237ms Step 2 took 4337ms ********** Factor found in step 2: 540644499510751614689592659509 Found probable prime factor of 30 digits: 540644499510751614689592659509 Composite cofactor 50198033461053618901414634779631927059274077161687913901931115543562208759765901061475667529776063988007503567882597626322608693341559073443818806711644490181527738248674313467847236423451988931896018594670230149 has 212 digits```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:21:54 UTC 2015 年 11 月 20 日 (金) 23 時 21 分 54 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 13:43:06 UTC 2015 年 11 月 25 日 (水) 22 時 43 分 6 秒 (日本時間)
4511e6478 / 4331CypApril 17, 2017 17:50:51 UTC 2017 年 4 月 18 日 (火) 2 時 50 分 51 秒 (日本時間)
14×10294-419
c287
composite cofactor 合成数の残り `96277779328537723391010393825838049758240650669439686774824161186552659746893832986674873384576631222208516392636381102028059099729729705567352678166208415383491896990450832873967978712047720603974991791803538424327628826769227808953554271993955515966132695660843697172081614370949494967<287>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6904Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
403e6600Dmitry DomanovNovember 20, 2015 13:42:30 UTC 2015 年 11 月 20 日 (金) 22 時 42 分 30 秒 (日本時間)
4511e6800 / 4305Dmitry DomanovDecember 31, 2015 00:25:46 UTC 2015 年 12 月 31 日 (木) 9 時 25 分 46 秒 (日本時間)
14×10297-419
c250
composite cofactor 合成数の残り `5768815625540571644572137620822833052858122992566403180289250152591571506107115744749798681912116970194216881600792303402542343646140770183178442303720157909750880714435308020324015174593507679887932266678003456098387568383888571325134527993804218539<250>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:23:50 UTC 2015 年 11 月 20 日 (金) 23 時 23 分 50 秒 (日本時間)
403e6600Dmitry DomanovNovember 25, 2015 17:02:20 UTC 2015 年 11 月 26 日 (木) 2 時 2 分 20 秒 (日本時間)
4511e6478 / 4331CypApril 17, 2017 17:11:20 UTC 2017 年 4 月 18 日 (火) 2 時 11 分 20 秒 (日本時間)
14×10298-419
c200
name 名前 Serge Batalov November 21, 2015 08:43:41 UTC 2015 年 11 月 21 日 (土) 17 時 43 分 41 秒 (日本時間) `93840498690936745738731667768332036488257898577574160070126664250474819507970408480812946880731949066427110870032271141563271907181397934027315829368217070504519515449542738352289620971081828569958019<200>` ```216973557808762754045022283996411909886481269801<48> 432497395713289434916494867661000862923369071795206227008664798138705028451123453069993850631366473953469973511515429378691729335520566261512513478191819<153>``` ```Using B1=11000000, B2=58553269330, polynomial Dickson(12), sigma=725542522 Step 1 took 75244ms Step 2 took 35244ms ********** Factor found in step 2: 216973557808762754045022283996411909886481269801 Found probable prime factor of 48 digits: 216973557808762754045022283996411909886481269801 Probable prime cofactor ```
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:24:21 UTC 2015 年 11 月 20 日 (金) 23 時 24 分 21 秒 (日本時間)
403e60--
4511e6660 / 4464Serge BatalovNovember 20, 2015 22:37:31 UTC 2015 年 11 月 21 日 (土) 7 時 37 分 31 秒 (日本時間)
14×10299-419
c220
composite cofactor 合成数の残り `1867395695458633729242852523684446028116277733060220597832780785146623193534123636879784022505945564436428793449286766055054815238592704994802396833971337566236442288914998836388085102644047944839610289525100765807159721<220>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:24:32 UTC 2015 年 11 月 20 日 (金) 23 時 24 分 32 秒 (日本時間)
403e6600Dmitry DomanovNovember 21, 2015 09:34:15 UTC 2015 年 11 月 21 日 (土) 18 時 34 分 15 秒 (日本時間)
4511e6478 / 4331CypApril 17, 2017 12:58:28 UTC 2017 年 4 月 17 日 (月) 21 時 58 分 28 秒 (日本時間)
14×10300-419
c292
composite cofactor 合成数の残り `1412205317881466122780280159704390904599695443861668768068024038614198729330559354163420882964692825710176349271212517112419374687955015043694178756095156705897214942472768822368034700996353893766027208306815534496029630595013184933399871162229146803551252013981435053466399480930862426295881<292>`
ECM
level レベルB1reported runs 報告された回数name 名前date 日付
3025e4430Makoto KamadaNovember 20, 2015 11:00:00 UTC 2015 年 11 月 20 日 (金) 20 時 0 分 0 秒 (日本時間)
351e6325KTakahashiNovember 20, 2015 14:24:50 UTC 2015 年 11 月 20 日 (金) 23 時 24 分 50 秒 (日本時間)
403e61200Dmitry DomanovNovember 21, 2015 09:36:39 UTC 2015 年 11 月 21 日 (土) 18 時 36 分 39 秒 (日本時間)
4511e61200Dmitry DomanovNovember 22, 2015 22:46:23 UTC 2015 年 11 月 23 日 (月) 7 時 46 分 23 秒 (日本時間)
5043e6720 / 6947600Dmitry DomanovDecember 2, 2015 19:24:05 UTC 2015 年 12 月 3 日 (木) 4 時 24 分 5 秒 (日本時間)
120Dmitry DomanovDecember 4, 2015 06:44:48 UTC 2015 年 12 月 4 日 (金) 15 時 44 分 48 秒 (日本時間)
5511e7120 / 17428Dmitry DomanovDecember 4, 2015 06:45:23 UTC 2015 年 12 月 4 日 (金) 15 時 45 分 23 秒 (日本時間) | 140,443 | 333,267 | {"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} | 3.03125 | 3 | CC-MAIN-2022-33 | latest | en | 0.33104 |
http://www.talkstats.com/threads/probability-of-energy-shortage.72870/#post-211891 | 1,642,412,711,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00030.warc.gz | 116,655,204 | 10,383 | # Probability of Energy Shortage
#### lcolli71
##### New Member
To anyone who can help,
I am trying to ascertain the likelihood that the UK will experience a power shortage this winter, ie, the probable amount of time that the UK energy production capacity will be less than demand. I have one constant and two variables. The constant is fixed energy capacity - derived from sources such as coal plants, nuclear plants, etc. Then there is a variable capacity - wind. Total capacity is therefore fixed capacity plus wind capacity. The other variable is demand.
I have a data set that gives the wind capacity and demand for every 5 minutes of the past 8 years. From this I have been able to establish the probable frequency distribution of wind capacity and demand capacity for this coming winter, attached in a pdf. How do I calculate the probability that for any given 5min time stamp the wind capacity will be sufficiently small and the demand sufficiently high that total capacity is smaller than demand?
what is the probability that F+w<d, where:
F - constant
w - unevenly distributed variable between wmin and wmax
d -
unevenly distributed variable between dmin and dmax
F+wmin<dmax
Any help would be very much appreciated.
#### Attachments
• 250.8 KB Views: 2
#### Miner
##### TS Contributor
The easiest way that I could come up with was a Monte Carlo simulation using your empirically derived distributions. I came up with a probability that demand would exceed capacity approximately 0.55% of the time.
#### Attachments
• 131.9 KB Views: 6
• 94.8 KB Views: 3
#### lcolli71
##### New Member
That's really helpful thank you very much, it's exactly what I needed.
I will need to do this calculation for a range of different datasets. I'm a bit of an amateur when it comes to statistics and I've never modelled a monte carlo simulation before. I was just wondering what software you used to conduct this analysis, and if you know of anywhere I might be able to find a step by step guide as to how to do it?
#### Miner
##### TS Contributor
I used Quantum XL. It is an Excel add-in, and is currently priced at \$549 USD.
This link has a step-by-step for Minitab. While there are some Minitab-specific instructions, the actual steps are fairly generic. There are also a lot of YouTube videos that may be helpful. | 520 | 2,330 | {"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} | 3.109375 | 3 | CC-MAIN-2022-05 | latest | en | 0.945346 |
https://www.teacherspayteachers.com/Browse/Core-Standard/7.SP.C.6 | 1,659,946,207,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00319.warc.gz | 844,138,158 | 88,312 | EASEL BY TPT
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## 7th Grade Math Spiral Review & Quizzes | Homework or Warm Ups
This 7th Grade math spiral review resource can easily be used as math HOMEWORK, WARM UPS, or a DAILY MATH REVIEW! This resource was designed to keep math concepts fresh all year and to help you easily track student progress. All pages are 100% EDITABLE and easy to differentiate to fit your students' needs.✪ CLICK HERE to see more SUBJECTS & GRADE LEVELS!✪ CLICK HERE to see the DIGITAL VERSION!★★SAVE \$\$ with these BUNDLES★★7th Grade Math BUNDLE: Spiral Review, Quizzes & Games!Digital &
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## Probability Activities - Hands-On Theoretical & Experimental Probability
Make teaching probability exciting with these hands-on fun probability activities! This set includes 6 low-prep probability stations perfect for exploring theoretical and experimental probability and critical thinking. Quick and easy setup plus clear student directions make these activities perfect for centers or substitute days, too!Lead your students to discover concepts of probability by exploring it on their own before teaching probability rules. Students will love exploring probability conc
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## 7th Grade Math Warm-Ups (CCSS-Aligned Math Bell Ringers)
These 120 daily math warm-ups can be used as bell work to immediately engage your students. They are designed to be used as spiral review bell ringers throughout the year! However, eight additional ideas for use have been included. These warm-ups/bell ringers allow you to reinforce each of the 7th Grade Math Common Core State Standards throughout the year and refresh your student's memory by continually spiraling the content. What is included?Printable PDF (120 Warm-Ups)16 spiral review bell
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## Seventh Grade Math Interactive Notebook Bundle- All Standards
This Bundle contains lessons for an Interactive Notebook for 7th Grade Math for each of the Common Core Standards as well as Texas Essential Knowledge and Skills (TEKS). This unit not only provides the Flippables (foldable activities) but also items necessary to complete your INB lessons and FULL COLOR photo examples of every page (inside and outside) to guide you in your teaching.The Seventh Grade Interactive Notebook Bundle Includes:7th Grade INB Unit for Operations with Rational Numbers7th
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## 7th Grade Math Probability Unit: 7.SP.5, 7.SP.6, 7.SP.7, 7.SP.8
A 9 day CCSS-Aligned Probability Unit - including simple probability, experimental and theoretical probability, making predictions, and independent and dependent probability.Students will practice with both skill-based problems, real-world application questions, and error analysis to support higher level thinking skills. You can reach your students and teach the standards without all of the prep and stress of creating materials!Standards: 7.SP.5, 7.SP.6, 7.SP.7, 7.SP.8; Texas Teacher? Grab the T
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## 7th Grade Math End of Year Escape Room Activity
About this resource : This escape room style activity provides students with a collaborative way to review 7th Grade Math skills. Included are eight challenges that require students to use skills from each group of standards to find codes that will unlock each puzzle. This escape room is NOT…Even though this is an “End of Year” escape it is not a comprehensive, complete review of all the 7th grade standards. Math escapes do focus on content but the primary focus is on the collaboration and the p
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## ⭐7th Grade SPIRAL MATH REVIEW Common Core | Distance Learning | Google Classroom
- Great for online learning and distance learning!Need a SPIRAL REVIEW of the 7th Grade Common Core standards? This resource contains 20 WEEKS of review specifically written for the common core math standards for 7th grade. The review sheets are organized into 5 boxes. Each box contains problems from the 5 domains of the 7th grade CCSS standards. Be sure to download the sample for a full overview of what you get.What topics does this resource cover?✔ Ratios and Proportional Reasoning (7RP)✔ T
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## 7th Grade Math Standard Based Assessments BUNDLE Common Core ⭐ Distance Learning
Assess your students' understanding of the 7th Grade Common Core Math Standards with this bundle of standard-based assessments.What is included:► 1-page quick assessment for every common core math standard for 7th grade► a student checklist for each domain to track your students' mastery of these standards► a class tracking sheet to look at all of your students progress at a glancePrint and Digital (updated September 2020)✔️ PRINT VERSION - a PDF file that you may PRINT✔️ DIGITAL VERSION - made
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## Wiffleball -- Theoretical & Experimental Probability - 21st Century Math Project
Theoretical and Experimental Probability gets a 21st Century Math Project facelift in this interactive game. Tired of flipping coins and spinning spinners? Students play the role of team managers and with only player statistics they must build a team capable of winning the championship. Students will plays a series of Wiffleball games where they will keep track of statistics using a baseball/softball inspired scorebook and after the game crunch the number to see if their players performed up to
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## 6th, 7th, 8th Grade SPIRAL MATH REVIEW BUNDLE | Google | Distance Learning
Need a SPIRAL REVIEW of the 6th, 7th and 8th Grade Common Core standards? This BUNDLE contains 300 QUESTIONS of review specifically written for the common core math standards for 6th, 7th and 8th grade. The review sheets are organized by grade level into 5 boxes. Each box contains problems from the 5 domains of that grades CCSS standards.This BUNDLE RESOURCE includes:✔ 6th Grade Math WEEKLY/DAILY REVIEW WARM-UPS (20 weeks / 100 questions)✔ 7th Grade Math WEEKLY/DAILY REVIEW WARM-UPS (20 weeks
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## 7th Grade I CAN Math Games BUNDLE | Test Prep Review
I CAN Math Games are the perfect way to make math fun! This bundle of 7th Grade Math Games focuses on ALL 7th Grade Math Standards, and provides students with practice in the form of multiple choice or short answer questions. QR codes (optional) make these games even more interactive as students get immediate feedback on their work! These games can be used for independent practice, a small group activity, a whole group review, or for progress monitoring. The possibilities are endless!✪ CLICK HE
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## End of Year 7th Grade Math ERROR ANALYSIS (Find the Error) BUNDLE
Have your students engage in higher-order thinking with these math ERROR ANALYSIS activities. This GROWING BUNDLE resource includes 8 different sets (77 questions) of 7th Grade Common Core word problems that are solved incorrectly. Students have to IDENTIFY THE ERROR, provide the CORRECT SOLUTION and share a helpful STRATEGY for solving the problem. An ANSWER KEY has also been provided with examples of possible answers. Be sure to download the sample for a full overview of what you get.Benefit
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## Probability Activities Print and Digital Worksheets |Google Classroom
Are you teaching your students all about chance and probability and looking for fun hands-on probability activities? This print and digital (google slides) bundle of probability worksheets will reinforce the understanding of theoretical probability, experimental probability, language of chance, sample space, predictions, lists, tree diagrams, dependent probability events and independent probability events.Both US and British English spelling and terminology included.Overview of Activities1. The
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## 7th Grade Math Foldables + Activities Bundle for Interactive Notebooks
This bundle includes all of the 7th grade math foldable and activity unit bundles listed below full of foldables and notes for interactive notebooks, task cards, scavenger hunts, math bingo games, puzzles, and more!PLEASE NOTE: Almost every single resource in this bundle has been updated during the 2021-2022 school year. I am going to continue editing and adding resources through 2023. If you have purchased this bundle, please re-download (for FREE) to access the new items! Click the links bel
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7th
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## Probability Project - Design Your Own Amusement Park
This project is designed to be a cumulative activity for students to complete after a probability unit. Skills in this unit include probability of simple events, theoretical vs. experimental probability, probability of compound events, and tree diagrams. This project can be given over the course of a couple of days or you can split each task up into a separate day. In this project, students will complete different tasks that relate to designing their own amusement park. The project is spit into
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## Math Interactive Notebook Seventh Grade with Scaffolded Notes + Google Slides
Interactive Math Notebook for Seventh Grade - 473 pages - answer keys included! Great for distance learning! This huge bundle has everything you need for a full year of teaching seventh grade math:A 245-page hands-on interactive math notebook including answer keys100 pages of scaffolded notes including answer keys86 pages of interactive printables including answer keys42 Google Slides Real Feedback from Teachers"I love TPT, but I may have found the ultimate product! Looking forward to using this
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## End of Year 7th GRADE MATH Assessments, Warm-Ups, Task Cards, Worksheets BUNDLE
* Featured in the TPT Newsletter * Best Seller *7th Grade Math STANDARD BASED Assessments, Warm-Ups, Task Cards, Error Analysis, Homework Practice Worksheets, Problem Solving Graphic Organizers, Mazes, Riddles, Coloring Activities, Guided Notes available to PRINT AND DIGITALThis is a GROWING BUNDLE of all of the 7th Grade MATH RESOURCES currently in my store { 61 Resources / OVER 630 PAGES } PLUS ALL FUTURE 7th GRADE RESOURCES!As a current middle school math teacher, I am using the activities
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## 7th Grade Math Review and Test Prep: CCSS
Teachers use this COMPLETE review unit to organize your 7th Grade Math Common Core Review and Test Prep to best suit your students’ needs. This resource is aligned to the Common Core State Standards. All 7th grade CCSS standards are covered in the pack. However, it is NOT written for a specific state or test. This file is jam packed with teacher guides, warm ups, cheat sheets, class activities, and 8-10 question mini assessments. There is more than enough for a 10 day review, so think of it like
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## Probability Unit Activity Bundle
This Probability Activity Bundle includes 7 classroom activities to determine simple probability, compare experimental and theoretical probability, make predictions, determine independent probability, and find dependent probability. These hands-on and engaging activities are all easy to prep! Students are able to practice and apply concepts with these probability activities, while collaborating and having fun! Math can be fun and interactive!Standards: CCSS (7.SP.5, 7.SP.6, 7.SP.7, 7.SP.8) and
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7th
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## 7th Grade Math Common Core WORD PROBLEM TASK CARDS { BUNDLE }
I use these task cards with my students to help them practice PROBLEMS SOLVING. This resource includes 7 sets (280 cards/40 cards per set) of 7th Grade Common Core task cards, student answer sheets, and answer keys. These TASK CARDS are perfect for reinforcing concepts through individual student practice, pair-share, early finishers, and assessment prep.I place one task card on each of my students' desks and give them 2-3 minutes to complete the problem and record their answer on their answer s
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## ⭐NO PREP 7th Grade Math Escape Rooms Bundle⭐Distance Learning
7th Grade Math Escape Room Bundle: This bundle includes 24 Tasks/Puzzles (6 Total Escape Rooms) for all the 7th Grade Math Common Core Standards. These escape rooms will have your students using mathematical skills from the 7th Grade math standards to find codes that will unlock each puzzle. This is a great way to get students working together in a fun and engaging way to review 7th Grade Math skills. Plus there is NO PREP all you need is an internet browser as well as either a laptop, chromeboo
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6th - 8th
0
CCSS:
## Probability Digital Math Activity Bundle | 7th Grade Math Distance Learning
These supplemental digital math activities can be used to support probability concepts. They are interactive (drag and match, using the typing tool, using the shape tool) and paperless through Google Slides™ and PowerPoint™. Topics Included:Simple Probability (4 interactive slides + exit ticket)Experimental and Theoretical Probability (4 interactive slides + exit ticket)Independent Events (4 interactive slides + exit ticket)Dependent Events (4 interactive slides + exit ticket)What is included?P
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7th
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CCSS:
## End of Year 7th Grade Math Mazes, Riddles, Color by Number BUNDLE Print, Digital
Have your students apply their understanding of 7th GRADE MATH with these mazes, riddles and color by number pages. This BUNDLE includes 14 of my top-selling 7th Grade MATH FUN ACTIVITIES packets (44 activities in all). Many more activities will be added to this BUNDLE, however, your purchase today will entitle you to FREE updates for LIFE!!Print or Digital: ✔️ PRINTED VERSION - a PDF file that you may PRINT✔️ DIGITAL VERSION (Easel by TpT) - There is also a digital version using TpT's new digi
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6th - 8th
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CCSS:
Subjects: | 3,197 | 14,203 | {"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} | 2.96875 | 3 | CC-MAIN-2022-33 | latest | en | 0.892724 |
https://www.pcb.com/resources/technical-information/driving-long-cables?utm_source=Cal+Nation+%23111+%28Jan+2017%29+-+FINAL&utm_campaign=Cal+Nation+%23111+Jan+17&utm_medium=email | 1,619,189,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00529.warc.gz | 1,048,155,167 | 49,467 | # Driving Long Cables
Operation over long cables may affect frequency response and introduce noise and distortion when an insufficient current is available to drive cable capacitance.
Unlike charge output systems, where the system noise is a function of cable length, ICP® sensors provide a high voltage, low impedance output well-suited for driving long cables through harsh environments.
While there is virtually no increase in noise with ICP® sensors, the capacitive loading of the cable may distort or filter higher frequency signals depending on the supply current and the output impedance of the sensor.
Figure 12: Long Extension Cable Schematic
Generally, this signal distortion is not a problem with lower frequency testing within a range up to 10,000 Hz. However, for higher frequency vibration, shock or transient testing over cables longer than 100 ft. (30 m.), the possibility of signal distortion exists.
The maximum frequency that can be transmitted over a given cable length is a function of both the cable capacitance and the ratio of the peak signal voltage to the current available from the signal conditioner according to:
where, fmax= maximum frequency (hertz)
V= maximum peak output from sensor (volts)
Ic= constant current from signal conditioner (mA)
109 = scaling factor to equate units
Note that in this equation, 1 mA is subtracted from the total current supplied to sensor (Ic). This is done to compensate for powering the internal electronics. Some specialty sensor electronics may consume more or less current. Contact the manufacturer to determine the correct supply current.
When driving long cables, the equation above shows that as the length of cable, peak voltage output or maximum frequency of interest increases, a greater constant current will be required to drive the signal.
The nomograph below provides a simple, graphical method for obtaining the expected maximum frequency capability of an ICP® measurement system. The maximum peak signal voltage amplitude, cable capacitance and supplied constant current must be known or presumed.
For example, when running a 100 ft. (30 m) cable with a capacitance of 30 pF/ft (98 pF/m), the total capacitance is 3000 pF. This value can be found along the diagonal cable capacitance lines. Assuming the sensor operates at a maximum output range of 5 volts and the constant current signal conditioner is set at 2 mA, the ratio on the vertical axis can be calculated to equal 5. The intersection of the total cable capacitance and this ratio result in a maximum frequency of approximately 10.2 kHz.
The nomograph does not indicate whether the frequency amplitude response at a point is flat, rising or falling. For precautionary reasons, it is good general practice to increase the constant current (if possible) to the sensor (within its maximum limit) so that the frequency determined from the nomograph is approximately 1.5 to 2 times greater than the maximum frequency of interest.
Note that higher current levels will deplete battery-powered signal conditioners at a faster rate. Also, any current not used by the cable goes directly to power the internal electronics and will create heat. This may cause the sensor to exceed its maximum temperature specification. For this reason, do not supply excessive current over short cable runs or when testing at elevated temperatures.
To more accurately determine the effect of long cables, it is recommended to experimentally determine the high frequency electrical characteristics.
The method illustrated in Figure 13 involves connecting the output from a standard signal generator into a unity gain, low-output impedance (<5 ohm) instrumentation amplifier in series with the ICP® sensor. The extremely low output impedance is required to minimize the resistance change when the signal generator/amplifier is removed from the system.
Figure 13: Long Cable Testing
In order to check system frequency/amplitude response, set the signal generator to supply the maximum amplitude of the expected measurement signal. Observe the ratio of the amplitude from the generator to that shown on the scope. If this ratio is 1:1, the system is adequate for your test. (If necessary, be certain to factor in any gain in the signal conditioner or scope.) If the output signal is rising (1:1.3 for example), add series resistance to attenuate the signal. Use of a variable 100 ohm resistor will help set the correct resistance more conveniently. Note that this is the only condition that requires the addition of resistance. If the signal is falling (1:0.75 for example), the constant current level must be increased or the cable capacitance reduced.
It may be necessary to physically install the cable during cable testing to reflect the actual conditions encountered during data acquisition. This will compensate for potential inductive cable effects that are partially a function of the geometry of the cable route. | 964 | 4,946 | {"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} | 2.921875 | 3 | CC-MAIN-2021-17 | latest | en | 0.872338 |
https://www.coursehero.com/file/6178598/Unit-3-Symmetric-Key-Crypto-Part13/ | 1,493,174,091,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121121.5/warc/CC-MAIN-20170423031201-00564-ip-10-145-167-34.ec2.internal.warc.gz | 882,306,521 | 128,502 | Unit 3-Symmetric Key Crypto_Part13
# Unit 3-Symmetric Key Crypto_Part13 - To encrypt, the...
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Unit 3 - 25 Block Ciphers An iterated block cipher splits the plaintext into fixed sized blocks and generates fixed sized blocks of ciphertext. The ciphertext is obtained from the plaintext by iterating a function F over some number of rounds. The function F , which depends on the output of the previous round and the key K , is known as a round function . Usually implemented in software.
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Unit 3 - 26 Feistel Cipher A Feistel cipher was developed by Horst Feistel. Feistel, is a general cipher design principle, not a specific cipher
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Unformatted text preview: To encrypt, the plaintext P is split into left and right halves, For each round i = 1, 2, . . . , n new left and right halves are computed according to the rule K i is the subkey for round i . The subkey is derived from the key K according to a key schedule algorithm . Finally, the ciphertext C is the output of the final round P = (L , R ) L i = R i −1 R i = L i −1 ⊕ F(R i −1 , K i ) C = (L n , R n )...
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## This note was uploaded on 03/17/2011 for the course ECE 4472 taught by Professor Ouda during the Fall '10 term at UWO.
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Unit 3-Symmetric Key Crypto_Part13 - To encrypt, the...
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Ask a homework question - tutors are online | 417 | 1,664 | {"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} | 2.890625 | 3 | CC-MAIN-2017-17 | longest | en | 0.882773 |
http://stackoverflow.com/questions/2274937/code-readability-vs-conciseness | 1,398,290,014,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00030-ip-10-147-4-33.ec2.internal.warc.gz | 330,988,390 | 21,296 | # Code readability vs conciseness [closed]
What are the merits and demerits of the following two code snippets:
``````return n==0 ? 0 : n==1 ? 1 : fib(n-1) + fib(n-2);
``````
and
`````` if(n==0)
return 0;
if(n==1)
return 1;
return fib(n-1) + fib(n-2);
``````
for calculating the nth letter in the Fibonacci sequence?
Which one would you favour and why?
-
community wiki? – falstro Feb 16 '10 at 17:39
I prefer `return round(pow(GOLDEN_RATIO,n) / sqrt(5));` – KennyTM Feb 16 '10 at 17:39
The nth letter? Is this the Roman version of the Fibonacci sequence? – Mark Byers Feb 16 '10 at 17:40
The second one, first of all because the first one is wrong (fib(n-1) + fib(n+1)?) and second because there's no point in doing that, it just makes the code unreadable. – IVlad Feb 16 '10 at 17:41
## closed as not constructive by falstro, Jørn Schou-Rode, Earlz, 0xA3, YOUFeb 17 '10 at 1:55
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.
The first one is the devil and must be purged with fire.
-
And the second is little better. – Clifford Feb 16 '10 at 17:41
What is wrong with first one ? – Hannoun Yassir Feb 16 '10 at 17:49
@Yassir - it doesn't calculate a fibonacci number – mocj Feb 16 '10 at 17:56
@mocj - Edited, thanks. Sorry for the silly mistake – Tom Feb 16 '10 at 18:10
@tom - that's exactly the point, isn't it? It won't be your first mistake when you write your code like that. – Hans Passant Feb 16 '10 at 18:26
I would favour:
``````return n <= 1 ? n : fib(n-1)+fib(n-2);
``````
-
n would have to be unsigned, since fib(-1) is undefined – Tom Feb 16 '10 at 17:42
Indeed; neither of the Tom's suggestions have much merit. – Clifford Feb 16 '10 at 17:43
Tom: Sure. But yours are even worse. They wouldn't terminate until they've made an overflow on IntXX.MinValue and back to 0. Might even Stack Overflow before then... – Sani Huttunen Feb 16 '10 at 18:09
I would prefer writing:
``````if (n == 0) {
return 0;
}
else if (n == 1) {
return 1;
}
else {
return fib(n-1) + fib(n-2);
}
``````
This is very readable code. I don't even like omitting braces as the code is not that readable and when you maintain code that omits braces, you easily make bugs.
-
I agree, in my opinion, brackets should never ever be omitted. – dassouki Feb 16 '10 at 18:07
I agree, don't omit braces. (Although I prefer the opening brace on a separate line) – Liz Albin Feb 16 '10 at 18:53
I disagree, please omit braces. Adding braces on one-line if statements is equivalent to adding semicolons to the end of every line in Python. – Cory Petosky Feb 16 '10 at 19:05
This is just an matter of opinion. The only exception that I allow myself to do is that I can test is parameters are valid like this: int foo(int bar) { if (bar < 0) throw new IllegalArgumentException(); } It requires that the if statement really is on-line statement and does not have new line before the throw statement. One of my favorite quotes is from Kent Beck: "I'm not a great programmer, I'm a pretty good programmer with great habits." I have seen programmers to fail to implement working code/fixing bugs because of this just too many times. – Lauri Feb 17 '10 at 8:09
Everything in life is a matter of equilibrium. Finding the right compromise between two opposite ends of the spectrum. Optimality is a scoring function that is highly dependent on the evaluator, and the situation, but you should strive for the sweet spot in everything.
Programming is not different. you should evaluate
• simplicity
• terseness
• efficiency
• practicality
• artistic freedom of expression
• time constraints
and find the sweet spot.
Your first construct is clearly powerful and geeky, but definitely not easy to understand.
-
I prefer the second over the first, mostly for readability.
The second "reads" well - it has the code broken up, so it reads most like English. This will make it easier to understand for many developers.
Personally, I find multiple, chained ternary operations difficult to follow at times.
Also, I personally find "conciseness" to be a poor goal, in most cases. Modern IDEs make "longer" code much more manageable than it used to be. I'm not saying you want to be overly verbose, but trying to be concise often causes an increase in the maintenance effort, in my experience.
-
If you're asking for readability, I'd prefer the second option because it doesn't contain the (double!) ternary operator. Usually you're writing code that other people also have to read, and from the second snippet, it's clear at first sight what the function does. In the first snippet though, one has to resolve both ternary operators "in your head" and additionally think about associativity (I'd think about that automatically because parentheses are missing).
But anyway, you could reduce the two if statements to one:
``````if(n <= 1) return n;
return fib(n-1) + fib(n-2);
``````
-
I'd prefer neither because they are both too slow. Readability should not come at the cost of an exponential explosion in runtime, especially when there exists a simple way that runs in linear time.
I'd do this something like this (pseudo-code):
``````a = 0;
b = 1;
n.times { a, b = b, a + b; }
``````
In C you'd have to use a temporary variable unfortunately, but the principle is the same.
-
Wow, how delightfully on topic, yet completely off topic. – Earlz Feb 16 '10 at 17:43
Earlz: Hypocrite? stackoverflow.com/questions/2274937/… – Mark Byers Feb 16 '10 at 17:59
umm? I don't get it. – Earlz Feb 16 '10 at 18:12
Of the two, the second is easier to understand at a glance. However, I'd consolidate it as
``````if (n <= 1)
return n;
else
return fib(n-1) + fib(n-2);
``````
Or, if you're not into multiple returns:
``````if (n > 1)
n = fib(n-1) + fib(n-2);
return n;
``````
-
I often find that indentation can make the multiple-ternary operators a lot more readable:
``````return n == 0 ? 0 :
n == 1 ? 1 :
fib(n-1) + fib(n-2);
``````
-
I prefer the second one in most situations, but there are times where it seems a bit of a waste to not do it in one line. For instance, I'd prefer
``````text="my stuff_"+id==null ? "default" : id;
``````
to
``````text="my stuff_";
if(id==null){
text+="default";
}else{
text+=id;
}
``````
Note, this also helps with DRY because now if you need to change the name of `text` then you only change it in one place, compared to 3.
-
if you use the ?: operator two or three times you will get used to it so i would go with the first
-
I'd say even more than @Lauri:
``````if (n == 0) {
tmp = 0;
}
else if (n == 1) {
tmp = 1;
}
else {
tmp = fib(n-1) + fib(n-2);
}
return tmp;
``````
It's good to have just one exit point.
-
So you'd prefer to use a temporary variable than having 3 points of return all at the end of a function and all in one place? – Earlz Feb 16 '10 at 17:58
What's the problem with that? The compiler should be able to optimise that without problems and it helps to write maintainable and debuggable code. – fortran Feb 16 '10 at 18:37
Introducing a temporary variable strictly for the purpose of restricting yourself to having only one exit point is the opposite of "maintainable" coding. – mocj Feb 16 '10 at 20:16
yeah, I bet you're a lot smarter than Dijkstra was... – fortran Feb 16 '10 at 20:23
When did I make such a claim? You can't seriously claim this version is somehow more maintainable than Lauris can you? – mocj Feb 16 '10 at 21:37
I'd bet there's no difference in the compiled code. I'd at least try to make it a little more readable:
``````return n==0 ? 0 : ( n==1 ? 1 : fib(n-1) + fib(n+1) );
``````
-
I'd try to make your answer more readable as well ;) – Earlz Feb 16 '10 at 17:47
Apart from that, the first expression is buggy (it uses `fib(n+1)` instead of `fib(n-2)`) and both exhibit the exponential explosion which makes Fibonacci sequence a classical tool for teaching some important practical aspects of algorithmics. | 2,287 | 8,300 | {"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-2014-15 | latest | en | 0.892619 |
https://socratic.org/questions/how-do-you-evaluate-log-0-8-12 | 1,586,278,575,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00288.warc.gz | 696,827,314 | 5,952 | # How do you evaluate log_0.8 (12)?
Jul 18, 2018
${\log}_{0.8} 12 \approx - 11.1359106$
#### Explanation:
Here we will use the logarithm change of base relationship:
${\log}_{a} b = \ln \frac{b}{\ln} a$
Hence, ${\log}_{0.8} 12 = \ln \frac{12}{\ln} 0.8$
$\approx \frac{2.48490665}{-} 0.22314355$
$\approx - 11.1359106$
As a check: ${0.8}^{-} 11.1359106 \approx 11.9999999 \approx 12$ | 166 | 391 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 3.625 | 4 | CC-MAIN-2020-16 | longest | en | 0.571867 |
https://cs.nyu.edu/pipermail/fom/1998-August/001965.html | 1,516,270,065,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00546.warc.gz | 684,979,952 | 2,698 | # FOM: what is computability theory?
Wed Aug 19 17:41:49 EDT 1998
```Stephen G Simpson wrote:
> Martin Davis writes:
> > The short answer to this question is it is the same subject hat
> > used to be called "recursive function theory" and then
> > metamorphosed to "recursion theory". ....
>
> Are you sure that "computability theory" (as discussed by Soare in his
> 1995 position paper) is the same subject as recursion theory? For
> example, does it include asymptotic computational complexity theory,
> or doesn't it? Soare chose to waffle on this important point.
What about the following (roughly formulated) theorems
related to so called descriptive complexity theory?
THEOREM 1. [Sazonov 80, Gurevich 83] "General" recursive functions
defined by systems of (least fixed point) functional equations
f(x)=T(f,x) with the terms T (constructed from 0, Successor,
If-Then-Else, f, x) having *arbitrary* form define exactly the class
of Polynomial Time computable (global) functions, *provided* these
equations are interpreted over a finite row of natural numbers
0,1,2,...,\Box with the largest number \Box whose value is not
specified in advance (i.e. \Box is a parameter "resource bound";
let, for definiteness, Succ(\Box)=\Box.)
Call such functions \Box-recursive. Thus,
\Box-recursion = PTIME-computability.
THEOREM 2. [Gurevich 83]
\Box-primitive recursion = LOGSPACE-computability.
(Strictly speaking, we should consider here *relative recursion*,
i.e. right-hand sides T(f,g,x) of recursive equations may actually
contain finite (non-global) functional parameters
g:{0,1,2,...,\Box}а.{0,1,2,...,\Box}.)
By the way, we may even consider \Box as having a *fixed* value,
say = 2^1000 or even = 1000. Even in this case it makes a sense
to distinguish between \Box-recursive (i.e. having "feasible"
recursive description) and non-\Box-recursive finite functions.
This may lead to non-asymptotic analogue of "asymptotic"
("infinitistic") complexity/recursion theory. Recall chess and
how playing this finite (8x8) game is complex
(non-feasibly-8-recursive?).
> (I
> take it as a given that recursion theory *does not* include asymptotic
> complexity theory. I learned this from my thesis advisor, Gerald
> Sacks, in the late 1960's.)
Would you recall the arguments of Sacks (or yourself)? | 619 | 2,310 | {"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} | 2.9375 | 3 | CC-MAIN-2018-05 | latest | en | 0.882681 |
https://kupdf.net/download/auto-cad-for-electrical-eng-student_5982e60ddc0d6017602bb17f_pdf | 1,604,036,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00239.warc.gz | 389,523,409 | 16,091 | # Auto Cad for Electrical Eng Student
August 3, 2017 | Author: Sabikan Sulaiman | Category: Technical Drawing, Auto Cad, Autodesk, Computer Aided Design, Rectangle
#### Short Description
4 Days Modules in Introduction to AutoCAD for Electrical Engineering Students, Bengkel SEM KHAS, FKE, Universiti Teknika...
#### Description
Module Title
Introduction to AutoCAD For Electrical Engineering Students
4/29/2010 FKE UTeM sulaimansabikan
Contents
CHAPTER 1. CHAPTER 2. CHAPTER 3. CHAPTER 4. CHAPTER 5. CHAPTER 6. CHAPTER 7. CHAPTER 8. CHAPTER 9. CHAPTER 10. CHAPTER 11. CHAPTER 12. CHAPTER 13. CHAPTER 14.
Introduction Exploring the AutoCAD Interface Drawing Setup in AutoCAD Starting A New Drawing Basic Drawing Organizing A Drawing Through Layers Modifying AutoCAD Drawing Text Dimensioning Block Creation in AutoCAD Plotting Your Drawing Template Drawing Creation Isometric Drawing Engineering Drawing
6 7 11 15 18 22 23 28 30 32 35 37 38 39
2
Thanks to:
3
Module Planning: Introduction to AutoCAD for Electrical Engineering Students DAY 1 CHAPTER 1 – CHAPTER 9 DAY 2
Exercise 1 : Setup Drawing, Geometric Drawing, Basic Drawing, Basic Editing Exercise 2 : Basic Drawing, Editing, Layer, Text, Dimension Chapter 10 – CHAPTER 12 Exercise 3 : Basic Drawing, Editing, Layer, Text, Dimension, Block Exercise 4 : Drawing Setup, 2D Drawing, Editing, Layer, Text, Dimension, Block
DAY 3
Exercise 5 : Setup Drawing, Geometric Drawing, Basic Drawing, Basic Editing Exercise 6 : Basic Drawing, Editing, Layer, Text CHAPTER 13 Exercise 7 : Simple Isometric Drawing Exercise 8 : Using Template, Title Block, Basic Drawing, Editing, Layer, Text, Dimension
DAY 4 CHAPTER 14 Assignment 1 : Electrical Engineering Drawing I using AutoCAD (Assessment 1) Assignment 2 : Electrical Engineering Drawing II using AutoCAD (Assessment 2)
4
Learning Outcomes After completing this course, students (or participants) will be able to:
1. Navigate and configure AutoCAD Interface.
the
2. Use the drawing commands, editing commands and dimensions command in the development of an engineering drawing. 3. Use the precision drafting tools, layer functions, text functions, block functions, plot functions, title block, template, isometric functions and layout views in the development of an engineering drawing. 4. Develop an electrical schematics drawing, electrical layout drawing, instrumentation/electrical circuits and also simple isometric drawings.
5
CHAPTER 1. Introduction 1.1
AutoCAD is a CAD (Computer Aided Design or Computer Aided Drafting) software application for 2D and 3D design and drafting.
1.2
Developer?
1.3
Sold by?
Autodesk.
1.4
Version?
1.5
Application Areas
# Architecture, Engineering & Construction # Automotive & Transportation # Education # Government
1.6
1.7
# Manufacturing # Media & Entertainment # Utilities & Telecommunications
Website?
www.autodesk.com.my
1.8
File format?
.dwg
6
CHAPTER 2. Exploring the AutoCAD Interface 2.1
Just double-click on the AutoCAD short cut in the Windows desktop.
2.2
2.2.1
2.2.2
3D Modeling
Available in AutoCAD 2007 and above.
7
2.3
The AutoCAD program window is divided into six parts: Menu bar Toolbars Drawing area 2.3.1
2.3.2
Command window Status bar Dashboard
Dialogue boxes
Toolbars
8
Drawing area
2.3.4
Command window and Status bar
2.3.5
Dashboard
Only available in AutoCAD 2007 and above.
9
2.4 Opening an Existing Drawing Open a drawing Select objects Delete objects Use dynamic input to enter commands Zoom and pan Find more information on using AutoCAD (F1) Note : DWG Free Viewer Software: Convert drawings to other AutoCAD version.
10
CHAPTER 3. Drawing Setup in AutoCAD 3.1
Use a drawing template to start a drawing
When you use a template file, new drawings created from the template automatically use the settings defined in the template. This saves you setup time and helps to make sure each drawing you create follows your company’s CAD standards. Drawing template files have a
.dwt file extension.
Some of the settings stored in drawing template files include; Unit type and scale (precision) Grid limits Title blocks/borders, blocks, and logos Annotation styles Layer names Linetypes Snap, grid, and ortho settings
3.2
Set the drawing units
Before you start a drawing, you must first decide what drawing units to use. In AutoCAD, distances are measured in drawing units. In a drawing, one drawing unit may equal one inch, one millimeter, one meter, or one mile. Before you begin drawing, you decide what one drawing unit will represent. AutoCAD does not include a setting that determines the length of a drawing unit.
After you decide what drawing units to use, you can set the format of the drawing units. The format settings available for linear units include ■ Architectural. A length of 15.5 units displays as 1’-3 1/2” ■ Decimal. A length of 15.5 units displays as 15.5000 ■ Engineering. A length of 15.5 units displays as 1’-3.5” ■ Fractional. A length of 15.5 units displays as 15 1/2 ■ Scientific. A length of 15.5 units displays as 1.5500E+01 For example, a mechanical engineer who works in millimeters would set the format for linear units to decimal. An architect who works in feet and inches, would set the format to architectural. The drawing unit format controls only the display style of the drawing units on-screen, such as in the display of coordinates and values in the dialog boxes and prompts.
11
Introduction to AutoCAD To set the format of the drawing units
3.3
Drawing Limits A0 : 841 mm x 1189 mm A1 : 594 mm x 841 mm A2 : 420 mm x 594 mm A3 : 297 mm x 420 mm A4 : 210 mm x 297 mm
3.4 Set the plot scale When you plot a drawing, you either specify
a precise scale or fit the image
design to the current paper size. For example, a distance of one drawing unit typically represents one millimeter or one meter in a metric drawing, while one inch or one foot in real-world units are common in an imperial drawing.
12
Introduction to AutoCAD To plot using a scale.
3.5
Understanding the Model Space and Paper Space
There are two distinct working environments or “spaces,” in which you can create objects in a drawing, model space and paper space. Use the tabs at the bottom of the drawing area to access model space and paper space. Use the Model tab for model space, and use one or more of the Layout tabs for paper space.
Model Space
Paper Space
In model space, you draw a model of your subject at 1:1 scale. In paper space, you can create one or more layout viewports, dimensions, notes, and a title block to represent a drawing sheet. You can switch between model space and paper space.
13
You can use layout viewports to access model space from within paper space. A layout viewport is like a picture frame containing a “photograph” of the model in model space. You can set a viewport to a specific scale.
14
CHAPTER 4.Starting A New Drawing 4.1
To start a drawing from scratch with the Create Drawing dialog box
The Create New Drawing dialog box is displayed under the following conditions:
You set the STARTUP system variable to 1 (on). You set the FILEDIA system variable to 1 (on). You click New on the File menu or Qnew on the Standard toolbar when no drawing template file is specified in the Options dialog box.
Start from Scratch Use a Template Use a wizard
15
4.2 Specify Drafting Setting Snap Grid Polar Tracking Object Snap Dynamic Input
16
17
CHAPTER 5. Basic Drawing 5.1
Command
A commad is an AutoCAD function used to perform some task. Command can be activated by; Menu Toolbar button Keyboard Command line Command abbreviation Alt key
5.2
Line
With LINE, you can create a series of contiguous line segments. Coordinate System point. Absolute Coordinate
x,y Polar Coordinate
@distance,direction Relative Coordinate
@x,y Direct distance Entry
5.3
Circle
You can create circles in several ways. The default method is to specify the center and the radius
5.4 Polygon You can create rectangles and regular polygons quickly. Creating polygons is a simple way to draw equilateral triangles, squares, pentagons, hexagons, and so on.
I
C 18
5.5
Hatch Pattern
You can hatch an area using a predefined hatch pattern, define a simple line pattern using the current linetype, or create more complex hatch patterns. One type of pattern is called solid, which fills an area with a solid color.
19
5.6 Arc You can create arcs in several ways. With the exception of the first method, arcs are drawn counterclockwise from the start point to the endpoint.
20
5.7
Ellipse
The shape of an ellipse is determined by two axes that define its length and width. The longer axis is called the major axis, and the shorter one is the minor axis.
5.8 Rectangle Use RECTANG to create closed polylines in a rectangular shape. You can specify the length, width, area, and rotation parameters. You can also control the type of corners on the rectangle—fillet, chamfer, or square.
5.9 Polyline A polyline is a connected sequence of line segments created as a single object. You can create straight line segments, arc segments, or a combination of the two. Multisegmented lines provide editing capabilities unavailable for single lines. For example, you can adjust their width and curvature. After you've created a polyline, you can edit it with PEDIT or use EXPLODE to convert it to individual line and arc segments.
21
CHAPTER 6.Organizing A Drawing Through Layers Layers are like transparent overlays on which you organize and group different kinds of drawing objects. The objects you create have common properties which include colors, linetypes, and lineweights. An object can inherit these properties from the layer it is drawn or placed on, or properties can be specifically assigned to individual objects. You can group objects with similar properties together on to a layer. This allows you to modify and manage multiple objects in your drawing from one central point. Changing the property for a layer, will update the property for all objects on the layer.
22
7.1
Erase
You can erase objects from your drawing using several methods. Erase them with ERASE. Select them and then cut them to the Clipboard with CTRL+X. Select them and press DELETE.
7.2
Copy
You can create duplicates of objects at a specified distance and direction from the originals. Use coordinates, grid snap, object snaps, and other tools to copy objects with precision.
7.3
Mirror
You can flip objects about a specified axis to create a symmetrical mirror image. Mirroring is useful for creating symmetrical objects because you can quickly draw half the object and then mirror it instead of drawing the entire object. You flip objects about an axis called a mirror line to create a mirror image. To specify this temporary mirror line, you enter two points. You can choose whether to erase or retain the original objects.
23
7.4 Offset OFFSET creates a new object whose shape parallels the shape of a selected object. Offsetting a circle or an arc creates a larger or smaller circle or arc, depending on which side you specify for the offset. You can offset Lines 2D polylines Arcs Construction lines (xlines) and rays Circles Splines Ellipses and elliptical arcs (resulting in an oval-shaped spline)
7.5
Array
You can create copies of objects in a rectangular or polar (circular) pattern called an array. For rectangular arrays, you control the number of rows and columns and the distance between each. For polar arrays, you control the number of copies of the object and whether the copies are rotated. To create many regularly spaced objects, arraying is faster than copying. 7.5.1
Rectangular Arrays
7.5.2
Polar Arrays
24
7.6 Move You can move objects at a specified distance and direction from the originals. Use coordinates, grid snap, object snaps, and other tools to move objects with precision.
7.7
Rotate
You can rotate objects in your drawing around a specified base point. To determine the angle of rotation, you can enter an angle value, drag using the cursor, or specify a reference angle to align to an absolute angle. To rotate an object 1. 2. 3. 4.
Click Modify menu » Rotate. Select the object to rotate. Specify the base point for the rotation. Do one of the following: Enter the angle of rotation. Drag the object around its base point and specify a point location to which you want to rotate the object. Enter c to create a copy of the selected objects. Enter r to rotate the selected objects from a specified reference angle to an absolute angle.
7.8 Scale and Stretch You can resize objects to make them longer or shorter in only one direction or to make them proportionally larger or smaller. You can also stretch certain objects by moving an endpoint, vertex, or control point. 7.8.1
Scale with Reference
To scale an object by reference 1.
25
Introduction to AutoCAD 2. 3. 4. 5.
Select the object to scale. Select the base point. Enter r (Reference). Select the first and second reference points, or enter a value for the reference length.
7.9 Trim and Extend You can shorten or lengthen objects to meet the edges of other objects. This means you can first create an object such as a line and then later adjust it to fit exactly between other objects. Objects you select as cutting edges or boundary edges are not required to intersect the object being trimmed. You can trim or extend an object to a projected edge or to an extrapolated intersection; that is, where the objects would intersect if they were extended. If you do not specify a boundary and press ENTER at the Select Objects prompt, all displayed objects become potential boundaries.
7.10 Break to Point and Break You can break an object into two objects with or without a gap between them. You can also join objects to make a single object.
26
7.11 Chamfer and Fillet A chamfer connects two objects with an angled line. It is usually used to represent a beveled edge on a corner. A fillet connects two objects with an arc that is tangent to the objects and has a specified radius.
27
CHAPTER 8.Text 8.1.1
Single-Line Text
Command : TEXT For short entries that do not require multiple fonts or lines, create single-line text. Single-line text is most convenient for labels.
%%d %%p %%c %%% %%o %%u
: : : : : :
simbol darjah () simbol campur/tolak () simbol diameter () simbol peratus (%) buat garisan atas teks Buat garisan bawah teks
28
Multiline Text
For long, complex entries, create multiline, or paragraph text. Multiline text consists of any number of text lines or paragraphs that fit within a width you specify; it can extend vertically to an indefinite length. Regardless of the number of lines, each set of paragraphs created in a single editing session forms a single object, which you can move, rotate, erase, copy, mirror, or scale. There are more editing options for multiline text than there are for single-line text. For example, you can apply underlining, fonts, color, and text height changes to individual characters, words, or phrases within a paragraph.
29
CHAPTER 9.Dimensioning Dimension styles are the main method used to control the way dimensions look. Using styles you can change the text font, the arrow head style, the relative position of the text, the scale of dimensions and many other parameters. Styles are created using the DIMSTYLE command.
9.1
Dimension Styles
Here is a list of the parts of a dimension along with their descriptions.
Dimensions have several distinct elements: dimension text, dimension lines, arrowheads, and extension lines.
30
9.2 The Dimension Style Manager Creates new styles, sets the current style, modifies styles, sets overrides on the current style, and compares styles.
Lines Symbols and Arrows Text Fit Primary Units Alternate Units Tolerances
31
CHAPTER 10.
10.1 Creating and Inserting a Symbol To save a drawing as a symbol, you use the Block tool, that can be saved and recalled at any time. You can also use entire existing files as blocks. Block usually consists of smaller components of a larger drawing. Typical examples include doors and windows for floor plants, nuts and bolts for mechanical assemblies and resistor and transistor for electrical schematics.
BLOCK -
Creates a block definition from objects you select.
WBLOCK - Writes objects or a block to a new drawing file. Write Block command creates a block similar to to the process used in the BLOCK command. However, this will write the objects directly to a location on the hard disk, which allows the block to be inserted in any type of drawing file.
32
Ps:
EXPLODE -
INSERT –
Breaks a compound object into its component objects.
Places a drawing or named block into the current drawing.
33
GRIP You can specify whether a block displays a single grip or multiple grips. You can specify whether a selected block reference displays a single grip at its insertion point or displays multiple grips associated with the objects grouped within the block. Menu > Tools > Options > Selection > Grips > Tick the “ Enable grips within blocks”
34
CHAPTER 11.
11.1 Configuring a Plotter
11.2 Plotting in Paper Space
35
11.3 Plotting in Model Space
36
CHAPTER 12.
Template Drawing Creation
A template is a drawing file that includes some of the following settings: Unit type and precision Drawing limits Snap, Grid, and Ortho settings Layer organization Title blocks Dimension and text styles Linetypes Common blocks Plotter setting Template files have the extention
.dwt.
AutoCAD has template which conform to several drawing conventions, including; a. b. c. d. e.
ANSI DIN Gb ISO JIS
37
CHAPTER 13.
Isometric Drawing
Isometric drawings simulate a 3D object from a particular viewpoint by aligning along three major axes. By setting the Isometric Snap/Grid, you can easily align objects along one of three isometric planes; however, although the isometric drawing appears to be 3D, it is actually a 2D representation. Therefore, you cannot expect to extract 3D distances and areas, display objects from different viewpoints, or remove hidden lines automatically. If the snap angle is 0, the axes of the isometric planes are 30 degrees, 90 degrees, and 150 degrees. Once you set the snap style to Isometric, you can work on any of three planes, each with an associated pair of axes: Left. Aligns snap and grid along 90- and 150-degree axes. Top. Aligns snap and grid along 30- and 150-degree axes. Right. Aligns snap and grid along 30- and 90-degree axes.
To turn on an isometric plane Click Tools menu » Drafting Settings. In the Drafting Settings dialog box, Snap and Grid tab, under Snap Type, select Isometric Snap. Click OK. You can cycle through the three isometric planes by pressing
F5
.
F5
38
CHAPTER 14.
Engineering Drawing
An engineering drawing, a type of technical drawing, is created within the technical drawing discipline, and used to fully and clearly define requirements for engineered items. Technical Drawing disciplines,
Lines style and Types Dimension Techniques and tolerance Scales Multiple views and projections Sectional View Isometric Projection Orthographic projection Isometric projection Auxiliary projection Symbols and Conventions Computer-Aided Drafting
ISO 128 is an international standard (ISO), about the general principles of presentation in technical drawings. Engineering drawings created by computer-aided design (CAD).
Electrical Engineering Drawing 1. 2. 3. 4. 5.
Wiring schematic drawing Electrical floor plants and panel layouts Motor electrical schematics Interconnection drawings Cable/conduit plans
39
Introduction to AutoCAD Sample of Engineering Drawing for Electrical Services
40
Introduction to AutoCAD Example of Schematic Wiring Diagram for Single Storey House printed with Title Block.
41
Introduction to AutoCAD Example of Schematic Wiring Diagram for Single Storey House
42
Example of Electrical Fitting Layout for Single Storey House
43
Example of TV Schematic Diagram for Single Storey House
44
Example of Telephone Schematic Diagram for Single Storey House
45
Pelan Susunatur Lampu dan Kipas Tingkat Bawah Rumah Banglo
46
Schematic Consumer Unit 60A TPN 4 Way, Tingkat Bawah
47
DC Machine
48
Exercise 1 BASIC GEOMETRICAL OBJECT AND ELECTRICAL SYMBOLS LAYER : Layer 0 Objek Cline Dimension Hline Text
Color White White Red Cyan Yellow White
Linetype Continuous Continuous Center Continuous Hidden Continuous
Use as scale : L = 5 cm.
49
Exercise 2-A TELEPHONE SCHEMATIC DIAGRAM
50
Exercise 2-B TV SCHEMATIC DIAGRAM
51
Exercise 03 DC CURRENT MACHINE
52
Exercise 04 Ceiling Fan
53
Exercise 5 Drawing Legend Draw and make block, apply layer.
54
Exercise 6 CADANGAN MEMBINA DAN MENYIAPKAN 4 UNIT BENGKEL BERKEMBAR 1 TINGKAT DAN 12 UNIT BENGKEL TERES 1 TINGKAT, DIKAWASAN IKS PULAU SEBANG LOT PT 1611 DAN 1612 MUKIM PULAU SEBANG DAERAH ALOR GAJAH MELAKA.
SCHEMATIC DIAGRAM
LIGHTING & POWER
55
Example: Complete Drawing
56
57
Exercise 7 Isometric
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##### What are differences between real and virtual images formed by lenses?
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##### (2 min ) Identify thc measure of center that probably serves the basis for the following items:The average credit card debt for U.S. households 515,480MEANMODEMEDIANThe average age of attendees Jt Justin Bleber concert 18 vears oldMEANMODEMEDIAN
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##### Math Review: Unconstrained OptimizationSuppose f(x)43 x2 344 X 14792? Find the value of x which maximizes this function_
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##### A) What is the molarity of a solution of KMnO4 if 40.00 ml will oxidize 0.3 g NazCz04 (134 g/mol)?b) What is the value of 1.00 ml of the above solution in terms of FeSO4 TH-O ( 278 g/mol)What is the value of [.00 ml ofthe above permanganate solution in terms 0f grams AszO3 (198 g/mol)A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.2SM Ce4- leading the reactionCH_(COOH)26Cet+2HzOHCOOH2C026Cej+6H-After 'standing for 10 minutes at 60PC, the solution was cooled and the x&
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##### QUESTION (62236-a) Use Gauss Jordan method to solve the linear system 235-2 -46-b) Let A= -2 2 -4 22find its eigen values and eigen vectors 5
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##### Explain how you would use both normative and descriptive %BF data when discussing %BF results to...
Explain how you would use both normative and descriptive %BF data when discussing %BF results to a potential client. How do these numbers related to individual health?... | 1,912 | 6,873 | {"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": 2, "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} | 2.96875 | 3 | CC-MAIN-2023-06 | latest | en | 0.670416 |
https://www.numbersaplenty.com/18384234 | 1,685,771,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00351.warc.gz | 1,004,567,495 | 3,155 | Search a number
18384234 = 2311278549
BaseRepresentation
bin100011000100…
…0010101101010
31021121000102120
41012020111222
514201243414
61454012110
7312156231
oct106102552
937530376
1018384234
11a417390
1261a7036
133a68b5b
142627b18
1519322a9
hex118856a
18384234 has 16 divisors (see below), whose sum is σ = 40111200. Its totient is φ = 5570960.
The previous prime is 18384221. The next prime is 18384239. The reversal of 18384234 is 43248381.
It is a Harshad number since it is a multiple of its sum of digits (33).
It is a junction number, because it is equal to n+sod(n) for n = 18384195 and 18384204.
It is not an unprimeable number, because it can be changed into a prime (18384239) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 139209 + ... + 139340.
It is an arithmetic number, because the mean of its divisors is an integer number (2506950).
Almost surely, 218384234 is an apocalyptic number.
18384234 is an abundant number, since it is smaller than the sum of its proper divisors (21726966).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
18384234 is a wasteful number, since it uses less digits than its factorization.
18384234 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 278565.
The product of its digits is 18432, while the sum is 33.
The square root of 18384234 is about 4287.6839902213. The cubic root of 18384234 is about 263.9257998542.
The spelling of 18384234 in words is "eighteen million, three hundred eighty-four thousand, two hundred thirty-four". | 493 | 1,659 | {"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} | 3.046875 | 3 | CC-MAIN-2023-23 | latest | en | 0.878395 |
thesciencepundit.blogspot.com | 1,679,761,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00307.warc.gz | 652,727,280 | 20,745 | ## Saturday, March 14, 2015
### Happy Pi Day!
Before I get to my post, let me say happy birthday Wacław Sierpiński!
Recently, I watched an Aperiodical video where a group of mathematicians used a variety of methods to estimate the decimal value of π under the constraints of a π-hour time limit and a ban on electronic calculating devices. Needless to say, none of these methods was particularly efficient. This got me thinking of how I might celebrate this mathematical "holiday" by estimating the value of π in my own inefficient manner. What could be more tedious than performing Newton's method on the sine curve?
Newton's Method: first step
The concept is fairly straightforward. Imagine that we're driving along a winding road at night. Since the headlights are affixed to the front of our car and facing forward, the beam of light always points in the direction that the car is moving. Now think of that road as the sine curve, our car as the point on the curve corresponding to our current guess of the value of π, and the headlight beam is what's called the tangent line, which always points in the direction of movement along the curve. Since the sine curve is continuous and crosses the x-axis at π, then as our x-coordinates approach π, our y-coordinates will likewise approach 0. Newton's method basically says that if we're close enough to π, then our car's/guess's tangent line will point towards an x value that's closer to π than our current guess is.
So if our initial guess is 3, then the tangent line through the point (3, sin(3)) will cross the x-axis at a value closer to π than 3 is. That value becomes our next guess, and then we repeat the process until satisfied. The closer that the sine of our guess is to 0, the closer that our guess is to π.
Determining the iteration formula makes for a good first semester calculus problem. I've gone ahead and done that as well as writing a short MATLAB routine* to perform it.
* see Appendix A
The guesses converged to 10 digits of accuracy in only two iterations.
initial guess 3
1st iteration 3.14254654307428
2nd iteration 3.14159265330048
pi 3.14159265358979
Of course, this is sort of like putting the cart before the horse since the procedure presumes that evaluating trigonometric functions is more basic than evaluating π. We can keep with the spirit of the original challenge by evaluating the sines and cosines with means which can be replicated with paper and pencil. This is where second semester calculus gives us the Maclaurin series. I've also taken the liberty of writing another MATLAB routine* that calculates the trigonometric functions using the first eight terms of their respective series. What else would I be doing on Pi Day? :-)
* see Apendix B
This time the accuracy was to 6 digits.
initial guess 3
1st iteration 3.14254589805078
2nd iteration 3.14159188052770
pi 3.14159265358979
I certainly hope that my Pi Day diversions have amused you. Thank you for reading!
### Appendix A:
% Script File : Newtons_Pi
% by Javier Pazos
% March 12, 2015
% Estimates the value of pi using Newton's method on the sine curve.
% Initial guess is p = 3.
PiGes = 3;
eps = 1e-4;
SiPi = sin(PiGes);
CoPi = cos(PiGes);
format long
% Main loop.
% Each iteration finds the x-intercept of the tangent line of the
% sine function at the previous estimate.
while abs(SiPi) > eps
% Iteration loop.
% Sets PiGes = (PiGes*cos(PiGes) - sin(PiGes))/cos(PiGes)
PiGes = (PiGes*CoPi - SiPi)/CoPi;
SiPi = sin(PiGes);
CoPi = cos(PiGes);
PiGes % Prints latest guess.
end
### Appendix B:
% Script File : Newtons_Pi_Taylor
% by Javier Pazos
% March 12, 2015
% Estimates the value of pi using Newton's method on the sine curve.
% Sine and Cosine are estimated using MacLaurin series.
% Initial guess is p = 3.
PiGes = 3;
eps = 1e-4;
SiPi = sinTay(PiGes);
CoPi = cosTay(PiGes);
format long
% Main loop.
% Each iteration finds the x-intercept of the tangent line of the
% sine function at the previous estimate.
while abs(SiPi) > eps
% Iteration loop.
% Sets PiGes = (PiGes*cos(PiGes) - sin(PiGes))/cos(PiGes)
PiGes = (PiGes*CoPi - SiPi)/CoPi;
SiPi = sinTay(PiGes);
CoPi = cosTay(PiGes);
PiGes % Prints latest guess.
end
function SiEst = sinTay(x)
% by Javier Pazos
% March 13, 2015
% Estimates sine using the MacLaurin series.
% Set number of iterations and initial sum.
n = 7;
sum = 0;
format long
% Main loop
% Sums first n+1 terms of the MacLaurin series for sine.
for j=0:n
sum = sum + ((-1)^j)*(x^(2*j+1))/factorial(2*j+1);
end
% Returns value.
SiEst = sum;
end
function CoEst = cosTay(x)
% by Javier Pazos
% March 13, 2015
% Estimates cosine using the MacLaurin series.
% Set number of iterations and initial sum.
n = 7;
sum = 0;
format long
% Main loop
% Sums first n terms of the MacLaurin series for cosine.
for j=0:n
sum = sum + ((-1)^j)*(x^(2*j))/factorial(2*j);
end
% Returns value.
CoEst = sum;
end
## Friday, April 11, 2014
### Fibonacci, the Golden Ratio and 2048
Like many people, when I first discovered the Fibonacci sequence, I was immediately enamored. It has some wonderful properties, some of which I will discuss below. But first, we need to beware of the large amount of Golden Garbage and Fibonacci Flim-Flam that pervades the internet.
A few years back, I ran across this excellent article by Donald Simanek where he meticulously takes apart many of the more popular myths associated with the Fibonacci numbers and the Golden Ratio. If you have not read it, go read it now! The link is here. I highly recommend it! More recently, Keith Devlin made a video where he stomped out much of the pseudoscientific nonsense going around about the Golden Ratio. I have embedded the video below as it is well worth the watch.
Debunking 10 Golden Ratio Myths from Keith Devlin on Vimeo.
What really irritates me the most about all this--as well as Simanek and Devlin--is that the Golden Ratio really is a special number with some wonderful and beautiful properties. The nonsense is completely unnecessary. Anyway, now that we've visited the debunking, let's talk about some Fibonacci and Golden Ratio facts that I really like.
### Some interesting facts about Fibonacci-like sequences
As Dr. Devlin pointed out in his video, it has been known for a long time that the ratio of consecutive Fibonacci numbers is the golden ratio Φ. There are many ways to show this. Here's one that I came up.
Let γn = Fn/Fn-1 be some ratio of consecutive Fibonacci numbers. Then γn+1 = Fn+1/Fn = (Fn+Fn-1)/Fn = 1+Fn-1/Fn = 1+1/γn. We can verify that γ2 = F2/F1 = 1/1 and γ3 = F3/F2 = 2/1 = 2 = 1+1/1 = 1+1/γ2. Now assume as an inductive hypothesis that γn can be represented by the continued fraction 1+1/(1+...+1/(1+1/1) with n red 1's. Then γn+1 = 1+1/γn = 1+1/(1+1/(1+...+1/(1+1/1)) which is the continued fraction with n+1 red 1's. Hence as n --> ∞, γn --> 1+1/(1+1/(1+1/(1+...) = Φ. This identity can be verified since Φ is the positive root of the quadratic equation x=1+1/x. (The negative root of this quadratic, sometimes referred to as its algebraic conjugate is Ψ = -1/Φ. More on Ψ later).
Φ as a continued fraction
As an interesting side note, the fact that Φ can be represented by the infinite continued fraction of all 1's, is why it is considered to be the "most irrational number." So is Φ the most irrational number? (In my best Keith Devlin voice) That's TRUE!
Leonardo of Pisa originally described the Fibonacci sequence
Several years back, I was playing around with Fibonacci-like sequences--sequences with the same recursion relation as the Fibonacci sequence, but with a different starting point. I began to wonder how many of the Fibonacci-like sequences also had the property that the ratio of consecutive elements tended towards Φ. If one were to allow non-integer seeds, then it is apparent that if γ1 = Φ, then γn = Φ for all n, and likewise for γ1 = Ψ. This is true since both Φ and Ψ are roots to the quadratic x=1+1/x. But which other sequences go to Φ and which to Ψ? Does it go to whichever limit is closer to γ1? Do most go Φ but those close enough to Ψ go to Ψ? Or is it something different? At the time, I conjectured that with the exception of the two above where γn doesn't vary with n, all Fibonacci-like sequences tended towards Φ. Unfortunately, I did not have the mathematical tools back then to prove it. I was determined to figure it out on my own though (what's the point of doing a puzzle if you just look up the answer without solving it yourself?).
Recently I was reminded of my old puzzle and finally worked out a proof. I'll just give you the gist of it without going into the details. Basically, I worked out a formula for the difference between γn and Φ (provided that γ1 > 0). Then using some known inequalities involving the Fibonacci numbers, I was able to show that for any given ε > 0, there is an N s.t. for all n > N, |γn-Φ| < ε. Thus for any γ1 > 0, the sequence of ratios tends towards Φ. Suppose now that γ1 < -1. Then γ2 = 1+1/γ1 > 0 and the sequence tends towards Φ. And if γ1 > -½, then γ2 < -1 and the sequence tends towards Φ. If we continue in this vein, we find that if γ1 < -1/F2n or > -1/F2n-1 for some n, then the sequence tends towards Φ. In other words, the set of initial ratios that spawn a sequence that tends towards Φ is R\{Ψ}.
This is pretty amazing. It means that no matter how close to Ψ the ratio γ1 is (just as long as it's not equal), there is some finite n for which γn is even closer to Φ. This naturally led me to wonder why and keep playing around. I guess there are many ways to explain this, but here's the one I came up with. Let's consider the Fibonacci sequence in reverse. We are used to thinking about the Fibonacci sequence as beginning with 1, 1 and moving forward following the recursion relation Fn+1 = Fn-1 + Fn. However, there is no reason why we can't follow the sequence backwards using the equivalent relation Fn-1 = Fn+1 - Fn. If we do this we get that F0 = 0, F-1 = 1, F-2 = -1, F-3 = 2, F-4 = -3, ... , F-n = (-1)n+1Fn. Now consider what happens to Fn+1/Fn as n --> -∞. This is equivalent to looking at the limit as n --> ∞ of F(-n)+1/F-n = F-(n-1)/F-n = -(Fn-1/Fn) = -1/Φ = Ψ. Furthermore, using essentially the same proof as above, we can show that with the exception of the two sequences whose ratio doesn't vary, this happens for every Fibonacci-like sequence. So of course they all tend towards Φ no matter how close to Ψ we begin; looking at it this way, arbitrarily close to Ψ is where they all ultimately "originate." This at once made the original conjecture seem a little less amazing, while simultaneously being much more astounding (I can't think of a better way of explaining my emotions). When I realized this result, I pictured in my mind a graph of all the sequences as something akin to the magnetic field lines of a bar magnet with Φ and Ψ as the fixed poles. Quite lovely!
I showed my work to a Number Theorist here at the Math Department where I go, and after a few days he came back to me with three alternative ways to prove this. I will show you my favorite below which involves a linear equation. It is known that for Fibonacci-like sequences, the nth term can be expressed as Fn = aΦn + bΨn where a and b are constants defined by the starting points of the Fibonacci-like sequence. For example, for the actual Fibonacci sequence a = 1/√5 and b = -1/√5 , for the Lucas sequence (which begins with 1, 3) a = b = 1 , for the Φ sequence b = 0, and a = 0 for the Ψ sequence. Consider, for a, b ≠ 0, what happens to Fn+1/Fn as n --> ∞. Note that |Ψ| < 0 and so the second term disappears in the limit while the a's in the numerator and denominator cancel leaving the limit as Φ. Now consider, for a, b ≠ 0, what happens as n --> -∞. We have Fn = an+bn, so the opposite happens and we're left with a limit of Ψ.
Beautiful!
### Musings on a variation of 2048
In case you've been living under a rock and haven't heard of the game 2048, go check it out now. (WARNING: Game is highly addictive!) It's a fun little game. The basic idea is that you move tiles around on a 4X4 board. The starting position consists of two tiles randomly placed on the board, valued either 2 or 4 (randomly chosen). On each move you move all tiles on the board either left, right, up or down provided they have room to move. If two tiles of the same value are pushed together, they merge into a single tile with double the value of the original tiles. On each move a single tile, either a 2 or 4, is randomly placed on an open space. The game end when you run out of room to play. You WIN if you create the 2048 tile, although you may continue to play to see how high you can go.
Binary Exponential
One of the attributes of the game is that it's fairly easy at first. In fact, it's fairly easy for a while. You think that you're doing so well then, seemingly suddenly, the world caves in on you. This is due to the exponential nature of the game. Each level is "as big" as all the levels before it. For example, in order to make the 512 tile, you must first make two 256 tiles. In other words, once you reach the 256 tile, to get to the next level, you have to do everything that you've already done over again to make a second 256 tile, but with less room to work with (since you didn't have that first 256 tile taking up space the first time around). Of course it's actually a little messier than that. On the first go there were 14 empty spots to work with; It is quite unusual to clear the board that efficiently when creating the higher valued tiles. Naturally, many of the tiles cluttering the playing board at this time can be used to make the 256 tile, so it's not quite as dire as that. Bear in mind that at each level, the maximum number of free spaces is reduced by one. So (for simplicity, let's ignore the fact that the game sometimes gives 4's and pretend that it always deals a 2) to make the 2048 tile, you must first make two 1024 tiles. But to make that second 1024 tile, you must first make two 512 tiles. But to make the second 512 tile, you must first make two 256 tiles, etc. Therefore, while constructing the 2048 tile, you must pass through a position where the playing board only has 16 - 11 = 5 free spaces to work with (because 2048 = 211). There is no escaping this since making duplicate tiles in parallel rather than in series leaves you with less free spaces to work with. And getting the 4096 tile is even harder yet. Knowing what I know about the exponential nature of the game and how long it took me to get good enough to finally make that tile, I'm not too optimistic about being able to make the 8192 tile. Although if I stop and carefully plan out each move rather than use the heuristic algorithm that has been this successful so far, who knows? What I do know is that 2048 has been a real time hoover and I sure hope I'm getting over it soon.
The game's popularity has spawned several variants. There is the pi variant. If you, like me, enjoy British sci-fi, then the Doctor Who variant is for you. The last two aren't "true" variants as the only difference is what's on the tiles; the next two are quite clever. If you're in the mood for a good laugh, I recommend the Numberwang! variant. (Note: If you are unfamiliar with the show That Mitchell and Webb Look, then watch the video below or else that variant won't make any sense.)
My favorite though, has to be the Fibonacci variant (via). The one thing that I don't like about that particular version of the variant is that it stops play after achieving the 2584 tile. Perhaps one of the other versions available are more like the app I have on my Android and let you keep playing (although that one might stop after the 6765 tile--not that I could do much better--although I have made the 4181 tile a handful of times). Anyhow, allow me to rewind a bit and relate a story.
I first became aware of 2048 when I saw this xkcd cartoon which I didn't understand at the time (late to the game, as usual). I soon became hooked and when I saw that the Garden State Mathematics Conference was looking for speakers, I thought that perhaps all this play wasn't a complete waste of time after all. I prepared an outline and abstract for the talk and submitted it and was approved. Then, a couple of days before my talk, I discovered the Fibonacci variant. That changed everything. How could I not make half my talk about that? So I did.
There are some key differences between the games. For one, instead of merging like numbers to get double, you merge consecutive Fibonacci numbers to get the next Fibonacci number. Except for 1, like tiles don't mesh, but each tile does have two different values that it can merge with. One needs to be a little more careful here. For example, if you're trying to merge a 13 with a 21 to get 34, you could accidentally bump into an 8 and get stuck with a pair of 21's. But don't fret it; not all is lost. If you make another 13 then hit the two 21's in sequence, you'll make a 55. Another thing to be note is that making the F18 tile takes about the same amount of room as making the 211 tile does in regular 2048. (see figures below)
Set-up for the win in 2048
Similar situation for 2584
As you can see, in the Fibonacci variant, each Fn is not adjacent to the Fn-1 and Fn+1 tiles but rather to the Fn-2 and Fn+2 tiles. Also, Unlike regular 2048, for each level you don't need to repeat everything you've done, but merely to the level below. Hence the game's increase in difficulty is not as fast as the binary exponential growth of regular 2048. However, it is exponential. Can you guess what the base is? (hint: Φ)
I should note here that since I figured all this out the day before my presentation, I made a mistake. I incorrectly reasoned that since the ratio of the number of levels needed to reach each level was n(n-1)/n, that the game's difficulty would approach 2n in the limit. This didn't quite sit right with me and when I realized that that last level is the "biggest" of the levels, I recalculated to get the result below. I will be giving the same talk again in a week or two and the mistake has been corrected.
There are many to show that it approaches Φ in the limit including continued fractions (which I did). There is also a simpler way to look at it. Since the game gives you nothing but 1's (F1 = F2 = 1, unlike in regular 2048 where 21 = 2 ≠ 22 = 4), every tile is built up from 1 tiles in the quantity of the face value of the tile in question. In other words, each level's "size" is equal to its value. So the increase in the difficulty of achieving each level relative to the previous level is just the ratio of consecutive Fibonacci numbers, which we've already shown is Φ. There you have it: the difficulty of each level of the Fibonacci variant of 2048 basically increases exponentially in Φ.
WONDERFUL!
Warning: NSFW!
## Monday, April 23, 2012
### Famous Blue Raincoat
Inspired by this Paul Krugman post, I give you First: the original
The cover (which, like Krugman, is how I first heard the song)
I absolutely love this song! I'm not sure if it's the haunting melody or the cryptic story, but it's awesome.
## Wednesday, February 16, 2011
### Jeffrey Sachs speaks the truth!
Here's a video (you'll have to click on the word 'Here' to watch the video on YouTube, as BloombergTV has disabled embedding) from a couple of days ago of (one of my all time favorite economists) Jeffrey Sachs being interviewed on Bloomberg Television. He really hammers home the point that (many others have also made) that before we can start seeing any real progress in this country, we're going to need to go to the public financing of campaigns. Well, he didn't exactly use those words, but the case he laid out and the evidence he presented clearly support that position. I was also amused by the way he handled the question about the Laffer Curve: he answered it the way any good biologist would answer a similar question about Intelligent Design. | 5,084 | 20,014 | {"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} | 4.15625 | 4 | CC-MAIN-2023-14 | latest | en | 0.951111 |
https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/11/lesson/11.2.4/problem/11-92 | 1,723,582,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00873.warc.gz | 231,705,860 | 15,509 | ### Home > A2C > Chapter 11 > Lesson 11.2.4 > Problem11-92
11-92.
For the polynomial function $f\left(x\right) = x^{3} − 5x^{2} + 11x − 15$, which of the following are possible factors?
1. $\left(x + 1\right)$
1. $\left(x − 2\right)$
1. $\left(x + 4\right)$
1. $\left(x − 3\right)$
Which of the constants in these polynomials are factors of $15$.
(a), (d) | 137 | 363 | {"found_math": true, "script_math_tex": 6, "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-2024-33 | latest | en | 0.585956 |
http://us.metamath.org/qleuni/ska7.html | 1,656,452,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00271.warc.gz | 66,042,580 | 2,358 | Quantum Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > QLE Home > Th. List > ska7 GIF version
Theorem ska7 235
Description: Soundness theorem for Kalmbach's quantum propositional logic axiom KA7.
Assertion
Ref Expression
ska7 ((a ∩ (ab)) ≡ a) = 1
Proof of Theorem ska7
StepHypRef Expression
1 anabs 121 . 2 (a ∩ (ab)) = a
21bi1 118 1 ((a ∩ (ab)) ≡ a) = 1
Colors of variables: term Syntax hints: = wb 1 ≡ tb 5 ∪ wo 6 ∩ wa 7 1wt 8 This theorem was proved from axioms: ax-a1 30 ax-a2 31 ax-a5 34 ax-r1 35 ax-r2 36 ax-r4 37 ax-r5 38 This theorem depends on definitions: df-b 39 df-a 40 df-t 41 df-f 42 This theorem is referenced by: (None)
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https://docs.trifacta.com/pages/viewpage.action?pageId=155386107 | 1,656,418,258,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00718.warc.gz | 257,473,638 | 21,822 | ##### Page tree
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Version 3
This example illustrates how you can apply the following functions to generate new and random data in your dataset:
• `RANDBETWEEN` - Generate a random Integer value between two specified Integers. See RANDBETWEEN Function.
• `PI` - Generate the value of pi to 15 decimal points. See PI Function.
• `ROUND` - Round a decimal value to the nearest Integer or to a specified number of digits. See ROUND Function.
• `TRUNC` - Round a value down to the nearest Integer value. See TRUNC Function.
Source:
In the following example, a company produces 10 circular parts, the size of which is measured in each product's radius in inches.
p0011
p0022
p0033
p0044
p0055
p0066
p0077
p0088
p0099
p01010
Based on the above data, the company wants to generate some additional sizing information for these circular parts, including the generation of two points along each part's circumference where quality stress tests can be applied.
Transformation:
To begin, you can use the following steps to generate the area and circumference for each product, rounded to three decimal points:
Transformation Name `New formula` `Single row formula` `ROUND(PI() * (POW(radius_in, 2)), 3)` `'area_sqin'`
Transformation Name `New formula` `Single row formula` `ROUND(PI() * (2 * radius_in), 3)` `'circumference_in'`
For quality purposes, the company needs two tests points along the circumference, which are generated by calculating two separate random locations along the circumference. Since the `RANDBETWEEN` function only calculates using Integer values, you must first truncate the values from `circumference_in`:
Transformation Name `New formula` `Single row formula` `TRUNC(circumference_in)` `'trunc_circumference_in'`
Then, you can calculate the random points using the following:
Transformation Name `New formula` `Single row formula` `RANDBETWEEN(0, trunc_circumference_in)` `'testPt01_in'`
Transformation Name `New formula` `Single row formula` `RANDBETWEEN(0, trunc_circumference_in)` `'testPt02_in'`
Results:
After the `trunc_circumference_in` column is dropped, the data should look similar to the following:
p00113.1426.28355
p002212.56612.56633
p003328.27418.8501313
p004450.26525.1332424
p005578.54031.41600
p0066113.09737.6991515
p0077153.93843.9821111
p0088201.06250.26511
p0099254.46956.5492929
p01010314.15962.8322121
• No labels | 648 | 2,440 | {"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} | 3.5625 | 4 | CC-MAIN-2022-27 | latest | en | 0.69644 |
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# CS61A_fa94_f - CS61A Final-Fall 1994 CS61A Fall 1994 Final...
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CS61A, Fall 1994 Final Question 1 (5 points): Write a function multicompose that takes as its argument a list of any length, whose elements are one-argument functions. It should return a one-argument function that carries out the composition of all of the given functions, like this: > (define third (multicompose (list first bf bf))) THIRD > (third '(the long and winding road)) AND > ((multicompose (list square square 1+)) 2) 81 > ((multicompose (list square 1+ sqrt 1+)) 8) 16 Your solution must use functional programming style --- no mutation! Question 2 (5 points): Imagine that you are working in a version of Scheme in which the only numbers available are integers. You want to be able to represent amounts of money, such as \$49.95, but you can't use the obvious representation 49.95 because that isn't an integer. You decide to create an abstract data type called a price that has two components, called dollars and cents . You write these constructors and selectors: (define (make-price d c) (+ (* d 100) c)) (define (dollars p) (quotient p 100)) (define (cents p) (remainder p 100)) (a) Write the procedure +price that takes two prices as arguments and returns their sum, as a new price. Respect the data abstraction. (b) Now we want to change the internal representation so that instead of representing a price as a single number of cents, we represent it internally as a list of two numbers, one for the number of dollars and one for the number of leftover cents. Rewrite the constructors and selectors as needed. Don't change the interface to procedures that use the abstraction . Question 3 (5 points): Here is a partial implementation of a telephone system, modeled using objects. Each telephone is an object, and the telephone company's switching equipment is another kind of object. When someone wants to place a call, they ask their telephone to call the number of the other person's phone: CS61A: Final--Fall 1994 1
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(ask my-phone 'call 6428311) The telephone asks the phone company to connect itself to that number. The result from this request can be one of three things: The word NOT-IN-SERVICE if there is no such telephone number. The word BUSY if the desired telephone is in use already. The word CONNECTED if the connection succeeds. In the third of these cases, the calling phone should note the fact that it's now in use, by setting its BUSY? variable true. The phone company must handle the connect request by finding the desired number in its phones variable, which is an association list of numbers and phones. If the phone exists, the phone company asks it if it's busy. If the phone isn't busy, the connection succeeds. The phone company must ask the called telephone to ring , and must also remember all current connections by maintaining a list of caller-callee pairs in its
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Ask a homework question - tutors are online | 793 | 3,377 | {"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} | 2.640625 | 3 | CC-MAIN-2017-51 | latest | en | 0.905055 |
https://www.fmaths.com/tips/often-asked-what-is-mathematics-of-investment.html | 1,624,562,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00548.warc.gz | 691,301,423 | 11,963 | # Often asked: What Is Mathematics Of Investment?
## What are the 4 types of investments?
There are four main investment types, or asset classes, that you can choose from, each with distinct characteristics, risks and benefits.
• Growth investments.
• Shares.
• Property.
• Defensive investments.
• Cash.
• Fixed interest.
## What is the investment equation?
Investment is often modeled as a function of income and interest rates, given by the relation I = f (Y, r), with the interest rate negatively affecting investment because it is the cost of acquiring funds with which to purchase investment goods, and with income positively affecting investment because higher income
## What type of math is used in finance?
While you won’t need to learn complex advanced mathematical theories, you will need to develop strong analytical abilities and enough of a background in algebra, calculus and statistics to apply concepts of these math branches to the finance field.
## What do you mean by investment?
Investment is an asset acquired or invested in to build wealth and save money from the hard earned income or appreciation. Investment is primarily made to obtain an additional source of income or gain profit from the investment over a specific period of time.
You might be interested: Readers ask: Why Is Mathematics Important To Children's Learning?
## What should a beginner invest in?
6 ideal investments for beginners
• 401(k) or employer retirement plan.
• Target-date mutual fund.
• Index funds.
• Investment apps.
## What investments will make you rich?
3 Cheap Investments That Could Make You Rich
• S&P 500 index funds. An S&P 500 index fund is a collection of all the stocks within the S&P 500 index, all bundled together into a single investment.
• Dividend stocks. Dividend stocks are investments that essentially pay you to own them.
• Fractional shares. Fractional shares are small portions of a single share of stock.
## How do you calculate cost of investment?
When you divide the investment’s gain minus the cost by the cost of the investment, you get 0.53 (\$35,000/\$65,000). To make this a percentage, multiply the number by 100. Example.
Item Amount
Cost of goods solds \$45,000
Operating expenses \$20,000
## How do I calculate investment needs?
In this case, it is the future value of the higher studies you want to reach as calculated in the last step. You need to divide the rate by 12 and multiply Nper with 12 since we are calculating monthly savings required. This gives you the monthly investments required at the assumed rate of return.
## How do you calculate profit?
When calculating profit for one item, the profit formula is simple enough: profit = price – cost. total profit = unit price * quantity – unit cost * quantity. Depending on the quantity of units sold, our profit calculator can also determine the total cost, profit per unit and total profit.
## Can I study finance without maths?
You don’t need a numerical degree to get a job in retail banking or insurance. The type of mathematical ability needed to apply for a role in retail banking or insurance is often at the level you will have already experienced during your GCSEs, which means that you shouldn’t require any degree-level knowledge.
You might be interested: Readers ask: What Is The Meaning Of M In Mathematics?
## Do you need maths for finance?
Do I need to be good at maths to study finance? Maths is important when applying for a finance degree, as well as while studying it. You should put your time and energy into achieving a good grade, but don’t give up if you don’t get what you were hoping for.
## What kind of jobs are in finance?
Financial Career Options for Professionals
• The Financial Services Industry.
• Financial Planning.
• Corporate Finance.
• Commercial Banking.
• Investment Banking.
• Hedge Funds.
• Private Equity, Venture Capital.
• Insurance.
## What is an example of investment?
An investment can refer to any mechanism used for generating future income. This includes the purchase of bonds, stocks, or real estate property, among other examples. Additionally, purchasing a property that can be used to produce goods can be considered an investment.
## What are the advantages of investing?
Benefits of Investing
• Potential for long-term returns. While cash is undoubtedly safer than shares, it’s unlikely to grow much, or find opportunities to grow, in the long run.
• Outperform inflation.
• Provide a regular income.
• Tailor to your changing needs.
• Invest to fit your financial circumstances.
## What is the importance of investment?
Investing is how you take charge of your financial security. It allows you to grow your wealth but also generate an additional income stream if needed ahead of retirement. Various investments such as stocks, ETFs, bonds, or real estate will provide either growth or income but in some cases both. | 1,000 | 4,912 | {"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-2021-25 | longest | en | 0.937676 |
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Contributor III
## Convert 6 digit number to date
I tried to convert a 6 digit number like 010160 to a date in DDMMYYYY format to get 01-01-1960. The exact format of the column to convert to a new column is:
zoeknaam 010160AC-1 180572BU-4 170258HO-5 190580EN-8 300551CA-11 190477DA-14 281069AD-15
So far I removed the last charactars exept the number by using the following code and I extracted the number 1 as a new column patiëntId:
NAV:
LOAD zoeknaam, geslacht, H_PC, geboortedatum, land, SubField(zoeknaam, '-', -1) as patiëntId,
Date(Date#(PurgeChar (SubField(zoeknaam, '-', -2), 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'),'DD-MM-YY'),'DD/MM/YYYY') as geboortedatumcheck
to convert to a date but it doesn't work. The output in geboortedatumcheck should be:
geboortedatumcheck 01-01-1960 18-05-1972 17-02-1958 19-05-1980 30-05-1951 19-04-1977 28-10-1969
The code:
PurgeChar (SubField(zoeknaam, '-', -2), 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz') as geboortedatumcheck
works properly and gives the numeric values as:
geboortedatumcheck 010160 180572 170258 190580 300551 190477 281069
How can I convert the 6 digit numbers to a date as shown?
1 Solution
Accepted Solutions
MVP
To handle years after 1999, you would need to include a cut-off so years after 30 are processed as 19xx and years 30 or lower are processed as 20xx
Something like
Date(Date#(Left(zoeknaam,4) &
if( Mid(zoeknaam,5,2) > 30, '19' & Mid(zoeknaam,5,2), '20' & Mid(zoeknaam,5,2) ), 'DDMMYYYY')) as Date
17 Replies
Try it with:
date(date#(left(zoeknaam, 6), 'DDMMYY'), 'DD-MM-YYYY')
- Marcus
Date(Date#(Left(Field,6), 'DDMMYY'), 'DD-MM-YYYY') as geboortedatumcheck
Best Anil, When applicable please mark the correct/appropriate replies as "solution" (you can mark up to 3 "solutions". Please LIKE threads if the provided solution is helpful
MVP
If your Year of date is fixed to be 1900+.. then try like:
Date(Date#(Left(zoeknaam,4)&'19'&Mid(zoeknaam,5,2), 'DDMMYYYY')) as Date
Haha, Nice
Best Anil, When applicable please mark the correct/appropriate replies as "solution" (you can mark up to 3 "solutions". Please LIKE threads if the provided solution is helpful
MVP
To handle years after 1999, you would need to include a cut-off so years after 30 are processed as 19xx and years 30 or lower are processed as 20xx
Something like
Date(Date#(Left(zoeknaam,4) &
if( Mid(zoeknaam,5,2) > 30, '19' & Mid(zoeknaam,5,2), '20' & Mid(zoeknaam,5,2) ), 'DDMMYYYY')) as Date
MVP
Hi Colin,
...years after 30 are processed as 19xx and years 30 or lower are processed as 20xx
Didn't know that. But always wondered it would be something so. Is this documented somewhere? Could you please redirect?
Thanks
MVP
This is a cut-off that you determine based on the value used for comparison in the if statement.
There is no standard for this, I just used 30 as an example.
If your data spans more than 100 years (which is possible if it is a date of birth) then there is no way to determine if the correct date is 19xx or 20xx from a 2-digit year!
MVP
I wonder if this is static or if it keeps on changing... because although it is still far away, we will eventually be in 2031 and at that time it would somehow need to read the date correctly
MVP
That's true. I just wanted to understand the logic Qlik follows. See, when I try to parse the data in YY format, qlik reads somewhere as 19XX and somewhere 20XX.
Date(Date#(Left(zoeknaam,6), 'DDMMYY')) as Date
It could be similar to what you explain, only the number could be different, may be 50 instead of your 30?
Community Browser | 1,097 | 3,655 | {"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} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.465663 |
http://www.chegg.com/homework-help/contemporary-abstract-algebra-7th-edition-chapter-0-solutions-9780547165097 | 1,474,881,417,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660712.3/warc/CC-MAIN-20160924173740-00198-ip-10-143-35-109.ec2.internal.warc.gz | 383,047,480 | 18,539 | # Contemporary Abstract Algebra (7th Edition) View more editions
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Chapter: Problem:
90% (10 ratings)
This software checks the validity of a Postal Service money order number. Use it to verify that 39539881642 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with the 9 in position 2 replaced with a 0. Was the error detected? Explain why or why not. Enter the number with two digits transposed. Was the error detected? Explain why or why not.
SAMPLE SOLUTION
Chapter: Problem:
90% (10 ratings)
• Step 1 of 6
If input is“39539881642” then the output is same
The money order number is valid.
• Step 2 of 6
If input is“39539881632” which is the same number with one digit is incorrect
Then the output is your input number is 39539881632
This number is not valid we see the error is detected.
• Step 3 of 6
If input is“30539881642”, then the output is
The money order number is valid we see the error is undetected.
• Step 4 of 6
The reason why the error is undetected is that the difference between “39539881642” and “30539881642” is divisible by 9. Recall the proposition, Test for Divisibility by 9
“An integer with decimal representation is divisible by 9 if and only if the sum of digits of ,, is divisible by 9”. We see the replacement, from 9 to 0, of one digit in a number will not affect divisibility by 9 of the sum of the digits and hence not affect the divisibility by 9 of the number.
• Step 5 of 6
We transpose two digits in “39539881642” and then obtain “39539886142”. If we input “39539886142”, then the output is
The money order number is valid”.
This number is not valid”. We see the error is undetected.
• Step 6 of 6
The reason why the error is undetected is that the difference between “39539881642” and “39539886142” is divisible by 9. Recall the proposition, Test for Divisibility by 9,
“An integer with decimal representation is divisible by 9 if and only if the sum of digits of ,, is divisible by 9”. We see the transposition of two digits in a number will not affect the sum of the digits and hence not affect the divisibility by 9 of the number.
Corresponding Textbook
Contemporary Abstract Algebra | 7th Edition
9780547165097ISBN-13: 0547165099ISBN: Joseph A. GallianAuthors:
Alternate ISBN: 9780495788256, 9780547165387, 9780547165394, 9781111785031, 9781133168782 | 681 | 2,564 | {"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} | 4.09375 | 4 | CC-MAIN-2016-40 | latest | en | 0.84108 |
https://es.mathworks.com/matlabcentral/answers/524748-multi-objective-optimization-with-discrete-variables?s_tid=prof_contriblnk | 1,603,274,406,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00469.warc.gz | 306,328,026 | 23,524 | # Multi Objective Optimization with discrete variables ?
10 views (last 30 days)
Ugur Acar on 11 May 2020
Commented: Ugur Acar on 13 May 2020
hi,
Can we use discrete values instead of just integers in @gamultiobj?
like in the example of "Stepped Cantilever Beam Design Problem" cant we add a function like "cantileverMapVariables.m" for "CreationFcn", "MutationFcn" and "CrossoverFcn" options ?
Thank you
#### 1 Comment
Ugur Acar on 12 May 2020
Can any kind of code be generated as in "cantileverMapVariables.m"
"%cantileverMapVariables Map integer variables to a discrete set V = cantileverMapVariables(x) maps the integer variables in the second problem solved in the "Solving a Mixed Integer Engineering Design Problem Using the Genetic Algorithm" example to the required discrete values.
% The possible values for x(3) and x(5)
allX3_5 = [2.4, 2.6, 2.8, 3.1];
% The possible values for x(4) and x(6)
allX4_6 = 45:5:60;
% Map x(3), x(4), x(5) and x(6) from the integer values used by GA to the
% discrete values required.
x(3) = allX3_5(x(3));
x(4) = allX4_6(x(4));
x(5) = allX3_5(x(5));
x(6) = allX4_6(x(6));"
Alan Weiss on 12 May 2020
Yes, if you are careful to define the creation, mutation, and crossover functions so that all integer variables are integers, and then map them to the values you want, then you can try to solve the problem that way. Not easy to do correctly, but theoretically possible. Good luck,
Alan Weiss
MATLAB mathematical toolbox documentation
Ugur Acar on 12 May 2020
Thank you for your answer, so it is possible theoretically :) I wish that mapping to variables was explained in the users manual as in genetic algoritm for single objective
Ugur Acar on 13 May 2020
"MathWorks Support Team" actually demostrated how to use integer in @gamultiobj with an easy example. Is it possible to add discrete variables options to this example also ? | 532 | 1,875 | {"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-2020-45 | latest | en | 0.772879 |
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https://howkgtolbs.com/convert/94.82-kg-to-lbs | 1,680,219,357,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00682.warc.gz | 345,761,717 | 12,174 | # 94.82 kg to lbs - 94.82 kilograms to pounds
Do you need to learn how much is 94.82 kg equal to lbs and how to convert 94.82 kg to lbs? You are in the right place. This whole article is dedicated to kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to emphasize that whole this article is dedicated to a specific number of kilograms - that is one kilogram. So if you need to learn more about 94.82 kg to pound conversion - keep reading.
Before we move on to the practice - that is 94.82 kg how much lbs calculation - we will tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 94.82 kg to lbs? 94.82 kilograms it is equal 209.0423168284 pounds, so 94.82 kg is equal 209.0423168284 lbs.
## 94.82 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, in formal International System of Units (in abbreviated form SI).
From time to time the kilogram can be written as kilogramme. The symbol of this unit is kg.
First definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but totally impractical to use.
Then, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was substituted by another definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 94.82 kilogram to pounds
You learned something about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to point out that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Naturally, this unit is used also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is equal 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 94.82 kg?
94.82 kilogram is equal to 209.0423168284 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 94.82 kg in lbs
Theoretical section is already behind us. In next section we want to tell you how much is 94.82 kg to lbs. Now you know that 94.82 kg = x lbs. So it is high time to get the answer. Just see:
94.82 kilogram = 209.0423168284 pounds.
This is an exact result of how much 94.82 kg to pound. You can also round off this result. After it your result is as following: 94.82 kg = 208.604 lbs.
You learned 94.82 kg is how many lbs, so look how many kg 94.82 lbs: 94.82 pound = 0.45359237 kilograms.
Naturally, in this case you can also round off the result. After it your result will be exactly: 94.82 lb = 0.45 kgs.
We also want to show you 94.82 kg to how many pounds and 94.82 pound how many kg outcomes in charts. Look:
We want to begin with a chart for how much is 94.82 kg equal to pound.
### 94.82 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
94.82 209.0423168284 208.6040
Now look at a chart for how many kilograms 94.82 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
94.82 0.45359237 0.45
Now you know how many 94.82 kg to lbs and how many kilograms 94.82 pound, so it is time to go to the 94.82 kg to lbs formula.
### 94.82 kg to pounds
To convert 94.82 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 209.0423168284 result in pounds
The first version of a formula give you the most exact outcome. In some cases even the smallest difference could be considerable. So if you want to get an exact result - this formula will be the best for you/option to calculate how many pounds are equivalent to 94.82 kilogram.
So let’s go to the shorer version of a formula, which also enables conversions to know how much 94.82 kilogram in pounds.
The another version of a formula is down below, let’s see:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, this formula is simpler. It can be better option if you want to make a conversion of 94.82 kilogram to pounds in easy way, for instance, during shopping. Just remember that your result will be not so exact.
Now we are going to show you these two formulas in practice. But before we will make a conversion of 94.82 kg to lbs we are going to show you easier way to know 94.82 kg to how many lbs without any effort.
### 94.82 kg to lbs converter
Another way to know what is 94.82 kilogram equal to in pounds is to use 94.82 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on first formula which we gave you in the previous part of this article. Due to 94.82 kg pound calculator you can effortless convert 94.82 kg to lbs. You only have to enter number of kilograms which you want to convert and click ‘convert’ button. The result will be shown in a second.
So try to calculate 94.82 kg into lbs using 94.82 kg vs pound converter. We entered 94.82 as an amount of kilograms. This is the result: 94.82 kilogram = 209.0423168284 pounds.
As you see, our 94.82 kg vs lbs converter is intuitive.
Now we are going to our chief issue - how to convert 94.82 kilograms to pounds on your own.
#### 94.82 kg to lbs conversion
We are going to begin 94.82 kilogram equals to how many pounds calculation with the first version of a formula to get the most correct outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 209.0423168284 the result in pounds
So what need you do to check how many pounds equal to 94.82 kilogram? Just multiply amount of kilograms, this time 94.82, by 2.20462262. It gives 209.0423168284. So 94.82 kilogram is equal 209.0423168284.
It is also possible to round it off, for instance, to two decimal places. It gives 2.20. So 94.82 kilogram = 208.6040 pounds.
It is time for an example from everyday life. Let’s convert 94.82 kg gold in pounds. So 94.82 kg equal to how many lbs? As in the previous example - multiply 94.82 by 2.20462262. It gives 209.0423168284. So equivalent of 94.82 kilograms to pounds, when it comes to gold, is equal 209.0423168284.
In this example you can also round off the result. This is the outcome after rounding off, this time to one decimal place - 94.82 kilogram 208.604 pounds.
Now we can move on to examples calculated using a short version of a formula.
#### How many 94.82 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 208.604 the result in pounds
So 94.82 kg equal to how much lbs? And again, you have to multiply amount of kilogram, in this case 94.82, by 2.2. Look: 94.82 * 2.2 = 208.604. So 94.82 kilogram is exactly 2.2 pounds.
Let’s do another conversion with use of shorer formula. Now convert something from everyday life, for instance, 94.82 kg to lbs weight of strawberries.
So convert - 94.82 kilogram of strawberries * 2.2 = 208.604 pounds of strawberries. So 94.82 kg to pound mass is 208.604.
If you learned how much is 94.82 kilogram weight in pounds and are able to convert it with use of two different formulas, let’s move on. Now we want to show you all results in tables.
#### Convert 94.82 kilogram to pounds
We realize that results shown in charts are so much clearer for most of you. We understand it, so we gathered all these results in tables for your convenience. Due to this you can easily compare 94.82 kg equivalent to lbs results.
Let’s begin with a 94.82 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
94.82 209.0423168284 208.6040
And now see 94.82 kg equal pound table for the second version of a formula:
Kilograms Pounds
94.82 208.604
As you see, after rounding off, when it comes to how much 94.82 kilogram equals pounds, the results are the same. The bigger number the more significant difference. Please note it when you need to make bigger number than 94.82 kilograms pounds conversion.
#### How many kilograms 94.82 pound
Now you know how to calculate 94.82 kilograms how much pounds but we will show you something more. Are you curious what it is? What do you say about 94.82 kilogram to pounds and ounces conversion?
We will show you how you can convert it step by step. Let’s begin. How much is 94.82 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 94.82, by 2.20462262. So 94.82 * 2.20462262 = 209.0423168284. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 94.82 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So final outcome is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final outcome is exactly 2 pounds and 33 ounces.
As you can see, conversion 94.82 kilogram in pounds and ounces simply.
The last conversion which we want to show you is conversion of 94.82 foot pounds to kilograms meters. Both of them are units of work.
To convert it you need another formula. Before we give you this formula, see:
• 94.82 kilograms meters = 7.23301385 foot pounds,
• 94.82 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to calculate 94.82 foot pounds to kilograms meters you have to multiply 94.82 by 0.13825495. It is 0.13825495. So 94.82 foot pounds is exactly 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 94.82 foot pounds is 0.14 kilogram meters.
We hope that this calculation was as easy as 94.82 kilogram into pounds conversions.
This article was a big compendium about kilogram, pound and 94.82 kg to lbs in calculation. Due to this calculation you learned 94.82 kilogram is equivalent to how many pounds.
We showed you not only how to make a conversion 94.82 kilogram to metric pounds but also two other conversions - to know how many 94.82 kg in pounds and ounces and how many 94.82 foot pounds to kilograms meters.
We showed you also another way to make 94.82 kilogram how many pounds calculations, it is using 94.82 kg en pound calculator. This is the best solution for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you are able to make 94.82 kilogram equal to how many pounds calculation - on your own or using our 94.82 kgs to pounds calculator.
It is time to make your move! Calculate 94.82 kilogram mass to pounds in the way you like.
Do you need to do other than 94.82 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 94.82 kilogram equal many pounds.
### How much is 94.82 kg in pounds
We want to sum up this topic, that is how much is 94.82 kg in pounds , we prepared one more section. Here you can see all you need to remember about how much is 94.82 kg equal to lbs and how to convert 94.82 kg to lbs . Let’s see.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. How does 94.82 kg to pound conversion formula look? . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 94.82 kilogram to pounds? The exact answer is 209.0423168284 lb.
You can also calculate how much 94.82 kilogram is equal to pounds with second, easier version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 94.82 kg equal to how much lbs ? The result is 209.0423168284 pounds.
How to convert 94.82 kg to lbs quicker and easier? You can also use the 94.82 kg to lbs converter , which will do all calculations for you and give you an exact result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,612 | 13,792 | {"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} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.939835 |
https://moorestat.com/blogs/pearsons-correlation-coefficient/ | 1,695,752,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00025.warc.gz | 451,787,507 | 20,092 | ## Welcome to Moore Statistics Consulting LLC
The Pearson’s Correlation Coefficient is typically used to evaluate two measurements to each other. The objective is to determine if there is a relationship between the two variables. You want to know if the two measures are moving in the same direction. The variables could be increasing at the same time; one could be decreasing while the other is increasing; or no relationship at all. And the association between the two variables is evaluated on a scale from -1 to 1. The close you are to either -1 to 1 , the stronger the relationship. A negative correlation represents one variable decreasing while the other is increasing over time. A positive correlation represents both variables increasing over time. Are you dizzy yet? Because I am…Okay here is the deal. If you report the Pearson’s Correlation Coefficient you need to include confidence intervals. Confidence intervals indicate whether or not zero is in the interval. If zero is in the interval then your Pearson’s correlation is not statistically significant.You should generate a p-value from your statistical program. I would report the p-value as well.
So the moral of this story is make sure you report more than your Pearson’s Correlation Coefficient and your test statistic needs to be statistically significant. One last thing, if you are trying to evaluate whether or not two measures are in agreement that is not the Pearson’s Correlation. There are specific tests to evaluate agreement which speaks more towards the magnitude of your values.The real last thing, there are correlation statistics specifically for variables that are not continuous. So you are not limited to Pearson’s Correlation Coefficient.
I hope you enjoyed this short blog, there are more to follow-Amy | 343 | 1,795 | {"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-2023-40 | latest | en | 0.928743 |
https://socratic.org/questions/in-5-49-seconds-1-20-g-of-argon-35-decays-to-leave-only-15-g-what-is-the-half-li | 1,580,308,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00146.warc.gz | 654,205,309 | 6,386 | # In 5.49 seconds, 1.20 g of argon-35 decays to leave only .15 g. What is the half life of argon-35?
May 22, 2016
1.83 sec
#### Explanation:
${m}_{i} = {m}_{r} \cdot {2}^{n}$
${m}_{i} \text{ is the initial mass present of the radioisotope.}$
${m}_{r} \text{ is the mass remaining after a certain time }$
$n \text{ is the number of periods," = (time)/T" where T is the half-life.}$
${m}_{i} = {m}_{r} \cdot {2}^{n}$
${m}_{i} / {m}_{r} = {2}^{n}$
${2}^{n} = \frac{1.20 \setminus g}{0.15 \setminus g}$
${2}^{n} = 8$
${2}^{n} = {2}^{3}$
$\text{ } \textcolor{red}{\underline{\overline{\textcolor{b l a c k}{n = 3}}}}$
n= (time)/T"
$T \text{(half-life)} = \frac{t i m e}{n}$
$T = \frac{5.49 \sec}{3}$
color(blue)(ulbar(color(red)(T = 1.83" sec")) | 321 | 758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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} | 3.921875 | 4 | CC-MAIN-2020-05 | latest | en | 0.676119 |
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