text:
# Python program to print all permutations of a string in lexicographic order using recursion

An assignment at school required me to print all permutations of a string in lexicographic or dictionary order.

Here is my solution to the task -

from math import factorial

def print_permutations_lexicographic_order(s):

seq = list(s)
for _ in range(factorial(len(seq))):
print(''.join(seq))
nxt = get_next_permutation(seq)
# if seq is the highest permutation
if nxt is None:
# then reverse it
seq.reverse()
else:
seq = nxt

def get_next_permutation(seq):
"""
Return next greater lexicographic permutation. Return None if cannot.

This will return the next greater permutation of seq in lexicographic order. If seq is the highest permutation then this will return None.

seq is a list.
"""
if len(seq) == 0:
return None

nxt = get_next_permutation(seq[1:])

# if seq[1:] is the highest permutation
if nxt is None:
# reverse seq[1:], so that seq[1:] now is in ascending order
seq[1:] = reversed(seq[1:])

# find q such that seq[q] is the smallest element in seq[1:] such that
# seq[q] > seq[0]
q = 1
while q < len(seq) and seq[0] > seq[q]:
q += 1

# if cannot find q, then seq is the highest permutation
if q == len(seq):
return None

# swap seq[0] and seq[q]
seq[0], seq[q] = seq[q], seq[0]

return seq
else:
return [seq[0]] + nxt

s = input('Enter the string: ')
print_permutations_lexicographic_order(s))

Here are some example inputs/outputs:

Enter the string: cow
>>> cow
cwo
ocw
owc
wco
woc

Enter the string: dogs
>>> dogs
dosg
dsgo
dsog
gdos
gdso
gods
gosd
gsdo
gsod
odgs
odsg
ogds
ogsd
osdg
osgd
sdgo
sdog
sgdo
sgod
sodg
sogd
ogds
ogsd

So, I would like to know whether I could make my program shorter and more efficient.

Any help would be highly appreciated.

• Why the downvote :( Is anything wrong with the question? – Justin May 27 at 16:16
• I didn't downvote, but I notice that the 2nd example is wrong. "dgos" and "dgso" are missing while "ogds" and "ogsd" occur twice. Also the tag "dictionary" refers to the data structure, not dictionary order. – Sedsarq May 27 at 16:25
• I've edited the tags, performance doesn't look like an aim, since your code doesn't look to care too much about it. – Peilonrayz May 27 at 16:28
• Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers. The moment answers arrive, you stop touching the code. That's the rules. It gets awfully messy otherwise. – Mast May 27 at 18:03

And in the third time I will recommend you the itertools module:

As I told in another answers, your code is very C/C++ styled, it is not Pythonic. Try to avoid manual iteration with indices as much as possible. Python has an enormous standard library that contains many useful modules. I already recommended you an itertools module. It contains pair of dozens generic functions to work with iterators. One of them - permutations - do 90% of your work:

And in the second time I will recommend you the itertools.permutations function that can solve your problem with literally one line of code:

from itertools import permutations

def get_lexicographic_permutations(s):
return sorted(''.join(chars) for chars in permutations(s))

print(get_lexicographic_permutations('dogs'))

['dgos', 'dgso', 'dogs', 'dosg', 'dsgo', 'dsog', 'gdos', 'gdso', 'gods', 'gosd', 'gsdo', 'gsod', 'odgs', 'odsg', 'ogds', 'ogsd', 'osdg', 'osgd', 'sdgo', 'sdog', 'sgdo', 'sgod', 'sodg', 'sogd']

• Title says he's supposed to use recursion though. – Sedsarq May 27 at 16:01
• @Sedsarq The question isn't tagged with reinventing-the-wheel, and doesn't state they have to use recursion. It is also tagged performance, and this is almost definitely more performant. – Peilonrayz May 27 at 16:14
• @Peilonrayz That's a fair point. I read the title to mean that the program should involve recursion, but it could just as well mean simply that his version does. – Sedsarq May 27 at 16:17

A basic recursive idea is to repeatedly reduce the problem to a simpler version of the same kind, until it reaches a base case which can no longer be reduced. Recursive solutions are often "simple" in the sense of not using many lines of code, so that's something to look for.

Let's see how this can be applied to this particular problem. Take the word "cow", with the permutations

cow, cwo, ocw, owc, wco, woc

Notice how the first character stays the same, until all permutations of the "tail" has been covered. That's the reduction step you need: For each character in the string, make that the first character, and then call the function again with the rest of the characters. Keep track of the character selection order as that will be the word. Then remember to catch the base case: If there's no characters left in the string, we're done removing characters and the word is complete.

def lexperm(s, word=''):
if not s:
print(word)
for i, char in enumerate(s):
lexperm(s[:i] + s[i+1:], word + char)

Notice how this doesn't require swapping or reversing anything, because we're removing all characters, in every order.

I've read over a couple of your past questions, and I can't really describe how unsightly the code is without violating SE's rules.

Skip over the section "A Foolish Consistency is the Hobgoblin of Little Minds" as you don't seem to know what good code looks like and you arn't interacting with legacy code. It's better to be a Hobgoblin for now, as you'll produce better code than you are now.

Get a couple of linter wrappers, like Flake8, Prospector or Coala. Make sure to turn all warnings on, as they don't normally by default.

Fixing all the errors these tools raise helps keep your code clean and make it so when you produce larger programs it's easier to understand what your code is doing. You may protest and say, "but people on Code Review understand my code!" And you would be right, but that is because you're developing solutions to coding challenges, not real life problems that are more intricate and require larger amounts of code.

• I understand. I will keep this in mind. Thank you :) – Justin May 27 at 16:59

I would recommend looking at itertools.permutations

from itertools import permutations
for p in permutations("cow"):
joined = "".join(p)
print(joined)

Output:

cow
cwo
ocw
owc
wco
woc

You can use it multiple ways:

1. Test your algorithm regarding output
2. Compare to your algorithm regarding speed
3. Check the implementation to get an idea how to it is done

Regarding 3, there is even an equivalent version (optimized but still readable) in the docs.


text:
ISSN 0439-755X
CN 11-1911/B

Acta Psychologica Sinica ›› 2013, Vol. 45 ›› Issue (4): 391-405.

### Effect of Semantic Decomposability on Understanding of Chinese Three-Character Idioms

MA Lijun;ZHANG Jijia;DU Kai

1. (1 School of Economy and Management, Guangzhou University of Chinese Medicine, Guangzhou 510006, China) (2 Departmemt of Psychology, Renmin University of China, Beijing 100873, China) (3 Center for Psychological Application, South China Normal University, Guangzhou 510631, China)
• Received:2012-03-29 Published:2013-04-25 Online:2013-04-25
• Contact: ZHANG Jijia

Abstract: Idiomatic expressions are widely used as one of figurative languages when people sensed that idiom contained more than its surface meaning. For a long time, researchers have done a lot of work studying the mechanism of processing idioms, and two models were proposed: noncompositional model and compositional model. Those two models, however, can only explain one aspect of processing idioms. Therefore, hybrid model (Cutting & Bock, 1997) came into being in explaning how people process idioms: top-down process and bottom-up process co-exist. Compositionality was introduced in psycholinguistics by Gibbs and his colleagues (Gibbs, Nayak, & Cutting, 1989), and later developed into the so-called idiom semantic decomposition hypothesis, which made great contribution to generating hybrid model. According to this view, idioms are at least partly decomposable, and speakers have shared intuitions about how the meanings of the parts contribute to the idiomatic meaning. The degree of analyzability determines the access of idioms’ figarative meaning. During recent years, semantic analyzability has become very popular among psycholinguists, and the claims of the idiom semantic decomposition hypothesis are now widely accepted (Libben & Titone, 2008). Some questions remain unsettled, however, including: (1) whether semantic decomposability affects understanding idioms independently; (2) if so, how semantic decomposability affects it; (3) whether semantic decomposition hypothesis is suitable in processing Chinese modifier-noun structure; (4) which model can be applied in accessing Chinese three-character idioms. The present study tried to answer those questions by exploring semantic priming task to investigate how Chinese three-character idioms with verb-object structure and with modifier-noun structure were processed. We intended to reveal the role semantic decomposability played in processing idioms in early (SOA=300ms in Experiment 1) and later (SOA=800ms in Experiment 2) period. The results showed that semantic decomposability affected semantic activation independently. In early period, as to high decomposable idioms with verb-object structure and with modifier-noun structure, we found significant activation in their literal meanings; on the other hand, in later period, as to low decomposable idioms with modifier-noun structure, we found significant accuracy in activating literal and figurative meanings. The results also showed that the degree of semantic decomposability resulted in different effects of semantic priming. In early period, as to low decomposable idioms with verb-object structure, we found significant activation in both literal and figurative meanings, and we found activation in literal meanings when idioms with verb-object structure and with modifier-noun structure were high decomposable. In later period, significant activation in both literal and figurative meanings was found in low decomposable idioms with verb-object structure. But significant activation in figurative meanings was only found in high decomposable idioms with modifier-noun structure. These results supported the notion that semantic decomposability plays a role in processing idioms in early period; while in later period, as to high decomposable idioms with verb-object structure, the simultaneous activation of literal and figuratvie meanings weakens the advantage of semantic priming. So we made conclusions that (1) semantic decomposability independently affects understanding idioms; (2) figurative meanings are activated more easily when idioms are low semantic decomposable; whereas literal meanings tend to be easily activated when idioms are high semantic decomposable; (3) the privilege of activation in literal meaning exists in processing idioms. All those results partly support the Graded Salience Hypothesis and hybrid hypothesis.

format: plain


text:
# Tag Info

4

The given matrix can not represent a covariance matrix since it would imply that asset 1 is negatively correlated to asset 2 and asset 3. But asset 2 is negatively correlated to asset 3 which contradicts the first statement. In general a covariance matrix has to be positive semi-definite and symmetric, and conversely every positive semi-definite symmetric ...

3

This however, goes against the conventional wisdom that variance becomes smaller as you hold the portfolio longer. Which conventional wisdom says this? If the variance decreases with time, then the likelyhood of getting a return close to the expected return increases (Cecbycev's inequality). So you are telling me, I know more about the long-time ...

3

This pdf says on page two that the paper was never published. I don't know the reason but you could try to mail the authors of the papers were the article is mentioned. Since it was never published it might be less encumbered by copyright than usual.

2

There is a simple reason to use prefer $CE$ to pure utility: $CE$ is independent of utility units. Thus it allows direct comparison. The cash equivalent of a risky portfolio is the certain amount of cash that provides the same utility that portfolio. So for portfolio $w$ we can define $CE$ via $U(CE)=E[U(w)]$ or $CE=U^{-1}(E[U(w)])$. Note that for ...

2

I do not see any advantage in this approach whatsoever, nor would I believe, as you suggested, that "many" use this kind of approach. In fact I find it horribly wrong. Using a single variable (CE in this case) to represent a non-trivial risk-return construct implies the ability to map such relationship to one variable representations. Everybody values risk ...

2

See "Some hypothesis tests for the covariance matrix when the dimension is large compared to the sample size" by Ledoit and Wolf. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aos/1031689018

2

The unconstrained mean-variance problem $$w_{mv,unc}\equiv argmax\left\{ w'\mu-\frac{1}{2}\lambda w'\Sigma w\right\}$$ can easily be found by taking the derivative $$\frac{\partial}{\partial w}\left(w'\mu-\frac{1}{2}\lambda w'\Sigma w\right)=\mu-\lambda\Sigma w$$ setting it to zero, and solving for $w$. This gives ...

2

The answer to the original question is simple: the Chopra-Ziemba paper is highly flawed and unreliable. Note that the framework is in-sample and based on a utility function. It has nothing to do with out-of-sample behavior of the mean vs. the covariance in an optimization. Estimation error grows linearly in the mean but quadratically in the covariance. At ...

1

In mean-variance analysis, you combine different assets to minimize variance and maximize expected return. The hyperbola is not a function of the number of assets, but of their mean and variance. If the efficient frontier where a tangent to the y-axis (which can't be) or nearly a tangent, that would mean you would have almost zero portfolio-variance, which ...

1

Generally I would annualize risk and returns even when an asset's returns/general time series (ts) does not span over the full year So, both, FB and G present risk and return over the past year. For risk and return that is calculated over longer periods I would not include an asset in the portfolio of which you have no ts available to measure risk and ...

1

You still want to perform portfolio optimization. Put everything into one bucket, run 'global' portfolio optimization, build the portfolio. Even if you prefer Sharpe ratios, you should do that on the overall portfolio - not just on individual ones. Be careful of sharpe ratios for low risk, low return assets. Dividing one small number by another small number ...

1

The term in sample and out of sample are commonly used in any kind of optimization or fitting methods (MVO is just a particular case). When you make the optimization, you compute optimal parameters (usually the weights of the optimal portfolio in asset allocation) over a given data sample, for example, the returns of the securities of the portfolio for the ...

1

I do not think they are directly applicable to MVO because inherently you always model the efficient frontier or asset selection on in-sample data and the result is measured out-of-sample. You can't say, "hey I model it in-sample over 2005 data and then I measure the performance of the portfolio over 2006 data and compare that with results derived from 2010 ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# 习题3-11换抵挡转置

Kickdown

Description

A research laboratory of a world-leading automobile company has received an order to create a special transmission mechanism, which allows for incredibly efficient kickdown – an operation of switching to lower gear. After several months of research engineers found that the most efficient solution requires special gears with teeth and cavities placed non-uniformly. They calculated the optimal flanks of the gears. Now they want to perform some experiments to prove their findings.

The first phase of the experiment is done with planar toothed sections, not round-shaped gears. A section of length n consists of n units. The unit is either a cavity of height h or a tooth of height 2h . Two sections are required for the experiment: one to emulate master gear (with teeth at the bottom) and one for the driven gear (with teeth at the top).

There is a long stripe of width 3h in the laboratory and its length is enough for cutting two engaged sections together. The sections are irregular but they may still be put together if shifted along each other.

The stripe is made of an expensive alloy, so the engineers want to use as little of it as possible. You need to find the minimal length of the stripe which is enough for cutting both sections simultaneously.

Input

The input file contains several test cases, each of them as described below.There are two lines in the input, each contains a string to describe a section. The first line describes master section (teeth at the bottom) and the second line describes driven section (teeth at the top). Each character in a string represents one section unit – 1 for a cavity and 2 for a tooth. The sections can not be flipped or rotated.Each string is non-empty and its length does not exceed 100.

Output

For each test case, write to the output a line containing a single integer number – the minimal length of the stripe required to cut off given sections.

Sample Input

2112112112

2212112

12121212

21212121

2211221122

21212

Sample Output

10

8

15

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char s[105], t[105];
int len1, len2, len;
while(cin>>s>>t)
{
int i, j, x, y;
int len1 = strlen(s), len2 = strlen(t);
for( i=0; i<len1; i++ )
{
bool flag = true;
for( j=0; j<len2 && j+i<len1 ; j++ )
if(s[i+j]=='2'&&t[j]=='2')
{flag = false; break;}
if(flag)
break;

}
x = max(len1, len2+i);
i = j = 0;
for( i=0; i<len2; i++ )
{
bool flag = true;
for( j=0; j<len1 && j+i<len2; j++ )
if(t[i+j]=='2'&&s[j]=='2')
{flag = false; break ;}
if(flag)
break;
}
y = max(len2, len1+i);
cout<<min(x,y)<<endl;
}
return 0;
}

————————————————

©2019 CSDN 皮肤主题: 大白 设计师: CSDN官方博客


text:
# QT CLI Build

• Hello,

I am trying to build a project with batch.
Problem is I have installed another version of MinGW and the build fails due to some conflict.
In Qt Creator I was able to set the environment variable "LIBRARY_PATH" to the QT mingw lib folder.
This setting works, so that the project can be build from the IDE.

Unfortunately it doesnt work for my batch script
the scipt looks like the following:

``````set LIBRARY_PATH="C:\Qt\5.13.0\mingw73_32\lib"
qmake
mingw32-make
``````

the makefile still has includes from my other mingw installation.
How can I solve this problem?

• @oliver123 Use absolute path to qmake to make sure you're using correct one.
Same for mingw32-make.

Did you add anything from the other MinGW installation to PATH?

• It works now, I added the path to the QT mingw32-make to my system path and rebooted.
Not sure if this was the cause, but not it works

• @oliver123 said in QT CLI Build:

It works now,

Great!, so please don't forget to mark your post as solved!. Thanks.

• Hi,

@oliver123 said in QT CLI Build:

It works now, I added the path to the QT mingw32-make to my system path and rebooted.
Not sure if this was the cause, but not it works

You shouldn't do that at the system level. This will come back and bite you in the future.

Only configure the command line you are using with that path. Its also how you work with e.g. Visual Studio. They provide a set of shortcuts that do some initialisation of the terminal and you can then use it for building what you want.


text:
Chebyshevskii Sbornik
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 Chebyshevskii Sb.: Year: Volume: Issue: Page: Find

 Personal entry: Login: Password: Save password Enter Forgotten password? Register

 Chebyshevskii Sb., 2014, Volume 15, Issue 1, Pages 7–18 (Mi cheb321)

This article is cited in 4 scientific papers (total in 4 papers)

Some residual properties of soluble groups of finite rank

D. N. Azarov

Ivanovo State University

Abstract: The generalization of one classical Smel'kin's theorem for polycyclic groups is obtained. A. L. Smelkin proved that if $G$ is a polycyclic group, then it is a virtually residually finite $p$-group for any prime $p$. Recall that a group $G$ is said to be a residually finite $p$-group if for every nonidentity element $a$ of $G$ there exists a homomorphism of the group $G$ onto some finite $p$-group such that the image of the element $a$ differs from 1. A group $G$ will be said to be a virtually residually finite $p$-group if it contains a finite index subgroup which is a residually finite $p$-group.
One of the generalizations of the notation of polycyclic group is a notation of soluble finite rank group. Recall that a group $G$ is said to be a group of finite rank if there exists a positive integer $r$ such that every finitely generated subgroup in $G$ is generated by at most $r$ elements. For soluble groups of finite rank the following necessary and sufficient condition to be a residually finite $\pi$-group for some finite set $\pi$ of primes is obtained.
If $G$ is a group of finite rank, then the group $G$ is a residually finite $\pi$-group for some finite set $\pi$ of primes if and only if $G$ is a reduced poly-(cyclic, quasicyclic, or rational) group. Recall that a group $G$ is said to be a reduced group if it has no nonidentity radicable subgroups. A group $H$ is said to be a radicable group if every element $h$ in $H$ is an $m$th power of an element of $H$ for every positive number $m$.
It is proved that if a soluble group of finite rank is a residually finite $\pi$-group for some finite set $\pi$ of primes, then it is a virtually residually finite nilpotent $\pi$-group. We prove also the following generalization of Smel'kin's theorem.
Let $\pi$ be a finite set of primes. If $G$ is a soluble group of finite rank, then the group $G$ is a virtually residually finite $\pi$-group if and only if $G$ is a reduced poly-(cyclic, quasicyclic, or rational) group and $G$ has no $\pi$-radicable elements of infinite order. Recall that an element $g$ in $G$ is said to be $\pi$-radicable if $g$ is an $m$th power of an element of $G$ for every positive $\pi$-number $m$.

Keywords: finite rank group, soluble group, polycyclic group, nilpotent group, residually finite $p$-group.

Full text: PDF file (280 kB)
References: PDF file   HTML file
UDC: 512.543
Received: 31.01.2014

Citation: D. N. Azarov, “Some residual properties of soluble groups of finite rank”, Chebyshevskii Sb., 15:1 (2014), 7–18

Citation in format AMSBIB
\Bibitem{Aza14}
\by D.~N.~Azarov
\paper Some residual properties of soluble groups of finite rank
\jour Chebyshevskii Sb.
\yr 2014
\vol 15
\issue 1
\pages 7--18
\mathnet{http://mi.mathnet.ru/cheb321}

Linking options:
• http://mi.mathnet.ru/eng/cheb321
• http://mi.mathnet.ru/eng/cheb/v15/i1/p7

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Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. D. N. Azarov, “Approksimatsionnye svoistva nilpotentnykh grupp”, Model. i analiz inform. sistem, 22:2 (2015), 149–157
2. D. N. Azarov, “Approksimatsionnye svoistva abelevykh grupp”, Vestn. Tomsk. gos. un-ta. Matem. i mekh., 2015, no. 3(35), 5–11
3. D. V. Goltsov, “Approksimiruemost fundamentalnoi gruppy konechnogo grafa grupp kornevym klassom grupp”, Chebyshevskii sb., 17:3 (2016), 64–71
4. D. N. Azarov, N. S. Romanovskii, “Finite homomorphic images of groups of finite rank”, Siberian Math. J., 60:3 (2019), 373–376
•  Number of views: This page: 188 Full text: 64 References: 32

 Contact us: math-net2021_10 [at] mi-ras ru Terms of Use Registration to the website Logotypes © Steklov Mathematical Institute RAS, 2021


text:
# EM 12c, EM 13c : Enterprise Manager Cloud Control Agent Blocked Due to Plugin Mismatch and emoms_pbs.trc File Reports Error code : NOT_FOUND_IN_INPUT (Doc ID 1919191.1)

Last updated on DECEMBER 06, 2021

## Applies to:

Enterprise Manager Base Platform - Version 12.1.0.1.0 and later
Information in this document applies to any platform.

## Symptoms

Enterprise Manager Cloud Control Agent status shows that it is blocked:

Example:

$/AGENT_INST/bin/emctl status agent Blocked Reason shows Plugin mismatch found between agent and repository. Please perform an agent resync to unblock the agent. Mismatched plugins: [ The plug-in is oracle.sysman.db In this example, issue is with the DB Plugin: oracle.sysman.db. Same issue may happen due to other plugins too. SQL query output from repository DB as SYSMAN shows 'NOT_FOUND_IN_INPUT' SQL> select * from MGMT_BLOCKED_AGENTS where target_guid=(Select target_guid from mgmt_targets where target_name like '%<AGENT_HOST_NAME>%‘); The plug-in is oracle.sysman.db Version in pla inventory : xxxxx Version on agent : null Content type : AGENT Error code : TARGETS_BEING_MONITORED Error Msg : Targets are being actively monitored by this plug-in and hence upgrade/downgrade/delete is not possible, The plug-in is oracle.sysman.db Version in pla inventory : xxxxx Version on agent : null Content type : AGENT Error code : NOT_FOUND_IN_INPUT Error Msg : This plug-in is present in the inventory for the given agent but has not been provided in the input] from agent https://<myagent.domain>:3872/emd/main/ does not match with repository. Agent was blocked due to this.$..\gc_inst\em\EMGC_OMS\sysman\emoms_pbs.trc reports the following message

Error Msg : This plug-in is present in the inventory for the given agent but has not been provided in the input] from agent https://<myagent.domain>:3872/emd/main/ does not match with repository. Agent was blocked due to this match
java.lang.Exception: The plugin, [ The plug-in is oracle.sysman.db
Version in pla inventory : xxxxxxxxx
Version on agent : null
Content type : AGENT
................
Version on agent : null
Content type : AGENT
Error code : NOT_FOUND_IN_INPUT
Error Msg : This plug-in is present in the inventory for the given agent but has not been provided in the input,  The plug-in is oracle.sysman.db
Version in pla inventory : xxxxxxxx
Version on agent : null
Content type : DISCOVERY
Error code : NOT_FOUND_IN_INPUT
Error Msg : This plug-in is present in the inventory for the given agent but has not been provided in the input,  The plug-in is oracle.sysman.oh
Version in pla inventory : xxxxxxx
Version on agent : null
Content type : DISCOVERY
Error code : NOT_FOUND_IN_INPUT


text:
# Math Help - Implicit differentiation

1. ## Implicit differentiation

Hello people,

This question just doesnt help me at all!

4x^4 + 8xy - 16y = 14

using implicit differentiation,

Help me out please with steps and explain.

Thank you so much!!!!!!

2. If a function is defined in implicit form...

$f(x,y)=0$ (1)

... then is...

$y^{'} = -\frac{f_{x}^{'} (x,y)}{f_{y}^{'} (x,y)}$ (2)

$f(x,y)= 4x^{4} + 8xy -16y -14$

$f_{x}^{'} (x,y) = 16x^{3} + 8y$

$f_{y}^{'} (x,y) = 8x - 16$ (3)

... so that...

$y^{'} = - \frac{2x^{3} + y}{x-2}$ (4)

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
If a function is defined in implicit form...

$f(x,y)=0$ (1)

... then is...

$y^{'} = -\frac{f_{x}^{'} (x,y)}{f_{y}^{'} (x,y)}$ (2)

$f(x,y)= 4x^{4} + 8xy -16y -14$

$f_{x}^{'} (x,y) = 16x^{3} + 8y$

$f_{y}^{'} (x,y) = 8x - 16$ (3)

... so that...

$y^{'} = - \frac{2x^{3} + y}{x-2}$ (4)

Kind regards

$\chi$ $\sigma$
I completely understand that until you get to step (4)

Wait I get it, your reducing with 8. I thought you weren't allowed to do that though.


text:
max planck institut
informatik

# MPI-I-97-1-016

## Faster deterministic sorting and priority queues in linear space

### Thorup, Mikkel

MPI-I-97-1-016. July 1997, 9 pages. | Status: available - back from printing | Next --> Entry | Previous <-- Entry

Abstract in LaTeX format:
The RAM complexity of deterministic linear space sorting of
integers in words is improved from $O(n\sqrt{\log n})$ to
$O(n(\log\log n)^2)$. No better
bounds are known for polynomial space. In fact, the techniques give a
deterministic linear space priority queue supporting insert and delete in
$O((\log\log n)^2)$ amortized time and find-min in constant time. The priority
queue can be implemented using addition, shift, and
bit-wise boolean operations.
Acknowledgement:
References to related material:

123 KBytes
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URL to this document: http://domino.mpi-inf.mpg.de/internet/reports.nsf/NumberView/1997-1-016
BibTeX
@TECHREPORT{Mikkel97,
AUTHOR = {Thorup, Mikkel},
TITLE = {Faster deterministic sorting and priority queues in linear space},
TYPE = {Research Report},
INSTITUTION = {Max-Planck-Institut f{\"u}r Informatik},
NUMBER = {MPI-I-97-1-016},
MONTH = {July},
YEAR = {1997},
ISSN = {0946-011X},
}