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Core AIPEP Frameworks | Examples of Good Practice | Accreditation Tools
In response to its research findings, the first iteration of AIPEP produced three key outcome and framework documents. These Frameworks provide guidance and direction for future action on curriculum transformation, student support, and workforce capabilities. Read the core Frameworks and Guidelines by clicking below.
My Indigenous Psychology Journey
Associate Professor Monica Thielking, Chair, Psychological Sciences at Swinburne University of Technology
Following the inaugural Heads of Departments and Schools of Psychology Association in Australia – Indigenous Psychology Education Project (HoDSPA-IAPEP) meeting in Sydney in April 2018, my colleague, Dr Ben Bullock, and I were inspired to develop the Swinburne Indigenous Psychology Committee (SIPC), which meets once a month to discuss how we can make important changes in our Department to transform staff Indigenous Psychology awareness and teaching and to increase Indigenous student enrolments, particularly in our Honours and postgraduate levels.
We were (and continue to be) very fortunate to have the support and guidance of the Moondani Toombadool Centre at Swinburne, which is responsible for all Aboriginal and Torres Strait Islander matters at Swinburne, including governance, student services, teaching and learning, research, staff, culture, engagement and governance. We started with limited knowledge, but a whole lot of purpose, and so SIPC is a true demonstration of 'from little things big things grow'. Since 2018 we have created a Terms of Reference for SIPC which mirrors the goals of AIPEP and the Swinburne RAP, established Indigenous psychology student recruitment pathways, and this year was successful in obtaining university recognition for our efforts through workload support for SIPC leadership. Our department is now closely integrated with other initiatives across the university in improving Indigenous health training, and regularly engages in important events to improve Indigenous cultural knowledge, such as regular yarning meetings, whole of staff training at the Koorie Heritage Trust, and an Indigenous literature book and movie club.
The role of leadership that seeks to cultivate best practice and innovation is crucial. As Department Chair, my continued and dedicated leadership of this initiative, of making time to listen to staff and to be present, of supporting staff to 'give it a go', guiding staff to be involved in professional development op at Indigenous psychology initiatives, connecting staff with Indigenous academics and celebrating staff success (however small) is key to building a culture of innovation and change in Indigenous Psychology. Being open about my own journey, and actively demonstrating a commitment to Indigenous Psychology, like always including an Acknowledgement of Country in staff meetings and guest lectures, or advocating for workload, for example, is a message to all staff and students that Indigenous Psychology is valued and matters. I am proud of the many staff members in my department who are now also leading the way in different areas of Indigenous Psychology teaching and learning.
To reiterate my earlier point, none of our achievements would be possible without the support and leadership of Indigenous staff at Swinburne, the Moondani Toombadool Centre and groups like AIPEP. I strongly encourage all psychology schools and departments who have an intention to make a difference in Indigenous psychology higher education teaching and learning to join with their university's Indigenous Centre and staff, to turn to AIPEP for resources and support, to honour the need for Indigenous self-determination, and to go gently, go bravely and go purposefully towards Indigenous psychological teaching and learning.
Decolonisation Journey
Associate Professor Paul Rhodes, Convener Psychologists for Social Justice, The University of Sydney
The Clinical Psychology Program at The University of Sydney is a typically white course for mostly privileged students. It is one of the countries most "elite" programs, a word that is obviously problematic when it comes to issues of culture and colonialism. In the past five years we have been on a journey trying to understand how we might leverage our privilege and engage in a process of reform regarding Indigenous knowledge and cultural humility. This has been both an academic and personal journey involving a reckoning with our own colourblindness and ignorance regarding these issues.
With the support of many, including Pat Dudgeon, Rob Brockman, our own National Centre for Cultural Competence, black scholars overseas we are seeing changes.
We have enrolled our first two Cadigal applicants
We have begun to include Indigenous Psychology in the program through standalone lectures and also integrated into our foundation Unit of Study
We have conducted two international studies on Decolonising Clinical Psychology
We have a student, Ash Wright, studying the Indigenous decolonisation of attachment, a project with The Children's Court aimed at challenging child removals
We are taking advantage of the Cadigal program at Usyd and providing Indigenous candidates with a fast track to interview
We are currently working on systemic change at the undergraduate level
These things occur very slowly but they add up.
These initiatives are also supported by Psychologists for Social Justice, a group of over 1000 psychologists who have been formed during this time, centred on Activist Practice (find us on Facebook). We aim to challenge the hegemony of traditional psychology and hold regular seminars and publish The Activist Practitioner.
The additional benefit is that Indigenous knowledge and participation challenges the field in important epistemological ways, opening up an awareness that our traditions are not to be taken for granted and that there are other ways of knowing and practicing. There are now research projects on eco-anxiety, place-based distress, social determinants on mental health, topics that we wouldn't have typically been embraced.
Decolonising Psychology (InPsych)
Professor Pat Dudgeon, University of Western Australia; Dr Dawn Darlaston-Jones, University of Notre Dame; Associate Professor Jeneva Ohan, University of Western Australia; Danielle Amiet, Monash University; Dr Meegan Kilcullen, James Cook University; & Dr Kate Derry, University of Western Australia
Read the full text here.

redpajama_set_name: RedPajamaCommonCrawl


[USPRwire, Wed Dec 05 2018] According to the World Bank, waste generation rate is rapidly increasing and the amount of waste generated per person per day is approximately 0.74 Kg. Bin liners are one of the efficient solutions, which are used to control the amount of waste produced. Bin liners are the most convenient way of handling waste and are widely used worldwide. Bin liners are manufactured in various colors, such as black, white, pink, green, etc. which easily helps to segregate different types of waste stored differently.
The rising demand for eco- friendly bin liners due to the increasing environmental awareness among the consumers is the primary driver for the bin liners market. The rising trend of online shopping reflects an increase in the bin liners market. Bin bags are enhanced with odor control benefits which is another factor driving the growth of the bin liners market. Bin liners are available in a variety of closures including tie top, straight top, etc. The tie-top type of bin liners is trending in the market due to its easy transportation and handling. Implementation of stringent regulations and taxation by the government of most of the countries over the use of plastic bags to diminish the harmful effects on the environment are expected to restrain the bin liners market growth. The plastic bin liners are now prominently replaced by sustainable options in the hotels, house, offices and any other place where a bin liner is used. Owing to its value-added applications of bin liners, right from the handling of waste to transportation and distribution of garbage, the market for bin liners is expected to create significant growth and investment opportunities for players operating in the bin liner market, especially in the developing countries.


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E3 2018: Where and when to watch all the major press conferences
Amanda Farough
If you're not going to be braving the crowds at the LACC and LA Live for this year's E3, here are the details on where to watch the major press conferences, including Bethesda, Sony, Xbox, EA, and Nintendo.
Remember when E3 started a day before the showroom floor opened and it still managed to be fairly chill, even though it was about a bajillion degrees outside? Yeah, me neither. These days, E3 is a week-long event that includes press conferences from all of the major players, including some new faces on the presser scene — Square Enix and Devolver Digital. If you're staying home this year and catching the action from the comfort of your office (or your couch), here are the details on when and where to watch the streams.
Electronic Arts E3 2018: Saturday, June 9 @ 11AM PT
EA is kicking off this year's press conferences on Saturday with a showcase of what they're bringing to EA Play — their EA oasis in LA Live, just a stone's throw from the convention center. Tune in on EA Play, Twitch, and YouTube at 11AM Pacific.
Xbox E3 2018: Sunday, June 10 @ 1PM PT
With Crackdown 3 delayed until 2019, I'm not entirely certain what Microsoft will be showcasing at their press conference. Will Xbox fill out that first-party lineup with more than an annual Forza title? Tune in on Mixer, Twitch, and YouTube at 1PM Pacific.
Bethesda E3 2018: Sunday, June 10 @ 6:30PM PT
Bethesda's #BE3 conference is always a good time, regardless of what they end up revealing during the conference. We'll get more details on Fallout 76 and Prey, more than likely. Tune in on Bethesda's dashboard, Twitch, and YouTube at 6:30PM PT.
Devolver Digital E3 2018: Sunday, June 10 @ 8PM PT
Devolver's "press conference" during E3 2017 was a poke at the traditional E3 pressers. We might see more of that fun-loving, satirical goodness during their conference this year. If nothing else, it'll be good (probably NSFW) fun all the way around. Tune inTwitch and YouTube at 8PM PT.
Square Enix E3 2018: Monday, June 11 @ 11AM PT
It's been a few years since Square Enix has held a dedicated press conference, but with so many releases on the horizon for the Japanese publisher, it's going to be quite the year. Tune in onTwitch andYouTube at 11AM PT.
Ubisoft E3 2018: Monday, June 11 @ 1PM PT
Another big year for Ubisoft, with a new Assassin's Creed on the way, hopefully more info on Beyond Good And Evil 2, and The Division 2. Last year's Ubisoft press conference was the highlight of E3 for me. Tune in onTwitch andYouTube at 1PM PT.
PC Gaming Show E3 2018: Monday, June 11 @ 3PM PT
For all ye PC hardware junkies out there (yes, I'm including myself on that list), PC Gaming Show is back for another E3 with a showcase of what's hot in PC gaming. Tune in on the PC Gaming Show website, Twich, andYouTube at 3PM PT.
PlayStation E3 2018: Monday, June 11 @ 6PM PT
PlayStation's E3 press conferences can yield a considerable amount of magic — 2016's reveal of the now critically-acclaimed God of War was one of the best reveals I've ever seen. They'll be running a tight show this year, likely with a lot of video, so it'll be trailers, trailers, trailers. (And hopefully some info on The Last Of Us 2.) Tune in on PlayStation, Twitch, andYouTube at 6PM PT.
Nintendo E3 2018: Tuesday, June 12 @ 9AM PT
Nintendo always closes out the press conferences at E3. Once upon a time, they used to do it live. Now they've managed to stitch together their press conference as a longer Nintendo Direct. We'll get more info on Smash and Pokemon, for sure. Tune in on Nintendo's website, Twitch, andYouTube at 9AM PT.
Follow GameDaily.biz's E3 Coverage


Hypergammon is a variant of backgammon.
Each player begins with three pieces, rather than fifteen. At the start of the game, each player begins with one piece on each of the three most distant points of the board (the 22-point, the 23-point and the 24-point).
Hypergammon is often played using the Jacoby Rule, which is also used in backgammon money games. Under this rule, gammons and backgammons will not count extra until the cube has been doubled.
As a consequence of the reduced number of pieces, hypergammon relies more heavily on luck than does backgammon. The lead in a game can swing very quickly. The difficulty of getting two pieces on the same point means that both players nearly always have blots on the board. The combination of the fact that bearing off can be completed with a single lucky roll of doubles and that it is nearly impossible to protect blots from being sent to the bar results in gammons being scored much more often than in backgammon.


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Home Human Body Bones Humerus Bone Anatomy
Humerus Bone Anatomy
Are you interested in the human body and want to know about humerus bone anatomy? The humerus is a long bone of the upper appendage, which reaches out from the shoulder to the elbow.
The proximal area of the humerus explains with the glenoid fossa of the scapula, framing the glenohumeral joint. Distally, at the elbow joint, the humerus expresses with the leader of the range and the trochlear score of the ulna.
In this article, we will discuss humerus bone anatomy. Its hard tourist spots and clinical relationships.
Proximal Landmarks
The proximal humerus is set apart by a head, anatomical neck, careful neck, more prominent and lesser tubercles and intertubercular sulcus (a tubercle is a round knob, and implies the connection site of a muscle or tendon).
The upper end of the humerus comprises of the head. These appearances medially, upwards and in reverse and is isolated from the more noteworthy and lesser tubercle by the anatomical neck.
The more noteworthy tubercle is found along the side on the humerus. It has a foremost and back face. The more prominent tubercle fills in as connection site for three of the rotator sleeve muscles – supraspinatus, infraspinatus and teres minor – they join predominant, center and mediocre aspects individually.
The lesser tubercle is considerably littler and all the more medially situated on the bone. It just has a front face. It gives a connection to the last rotator sleeve muscle; subscapularis.
Isolating the two tubercles is a profound sorrow, called the intertubercular sulcus, or furrow. The ligament of the long head of biceps brachii rises up out of the shoulder joint and goes through this section.
The pole of the humerus is the site of connection for different muscles. Cross-segment sees uncover it to be round proximally and leveled distally.
On the horizontal side of the humeral shaft is a roughened surface where the deltoid muscle connects. This is referred to is as the deltoid tuberosity.
The outspread (or winding) groove is a shallow despondency that runs slantingly down the back surface of the humerus, parallel to the deltoid tuberosity. The spiral nerve and profundabrachii conduit lie in this depression.
The accompanying muscles join to the humerus along its pole:
Anteriorly – coracobrachialis, deltoid, brachialis, brachioradialis.
Posteriorly – average and sidelong leaders of the triceps (the winding score differentiates their separate starting points).
Distal Region
The horizontal and average outskirts of the distal humerus shape average and parallel supraepicondylar edges. The horizontal supraepicondylar edge is more roughened, giving the site of the basic root of the lower arm extensor muscles.
Promptly distal to the supraepicondylar edges are extracapsular projections of bone, the sidelong and average epicondyles. Both can be palpated at the elbow. The average is the bigger of the two and expands all the more distally. The ulnar nerve goes in a section on the back part of the average epicondyle where it is discernable.
Distally, the trochlea is found medially and expands onto the back part of the bone. Horizontal to the trochlea is the capitulum, which explains with the range.
Additionally situated on the distal bit of the humerus are three miseries, known as the coronoid, outspread and olecranon fossae. They oblige the lower arm bones amid development at the elbow.
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Home » News » Media Releases » Student Safeguards Must be in Place by New School Year
Student Safeguards Must be in Place by New School Year
Cassy O'Connor MP - Wednesday, 10 November 2021
Tags: Child Safety, Children and Young People, Commission of Inquiry
Cassy O'Connor MP | Greens Leader and Education spokesperson
Tasmanian parents need greater reassurance that their children are safe in the state's public school system.
The lack of transparency demonstrated by the Gutwein Government over the Inquiry in to the Department of Education's responses to child sexual abuse has done nothing to foster trust.
Professors Stephen Smallbone and Tim McCormack's full report of the inquiry in to sexual abuse in the State's public school system was received by government in June - nearly five months ago. It was released through Right to Information, on the same day the Education Minister tabled a summary document.
The report details a culture of cover up within education in Tasmania. The Government does little to reassure Tasmanians that cultural change is underway by attempting to hide details of the report.
When scrutinised by opposition parties in Parliament this morning, Minister Courtney claimed the government has been transparent. That would be more believable had they not sat on the report for five months before suddenly tabling a summary document in Parliament earlier on the same day that the full report was released through the statutory Right to Information process.
The report revealed "some record of concern" about more than 40 current employees, including at least three principals, with half potentially requiring further investigation. Perhaps of even greater concern, the Department has no way of knowing whether or not relief teachers have red flags against their name.
This would have been distressing news to wake up to for any parent sending their kids off to school this morning.
In the government's response to the report's recommendations, we are told many will take more than a year to implement. Those recommendations were made by Professors Smallbone and McCormack to fix serious issues identified with the safety of children and young people in schools. Their implementation is urgent to keep children safe.
Minister Courtney was unable to explain why their implementing safeguards will take so long.
Tasmanian parents need to know students are as safe as possible in public schools. Minister Courtney must explain to them why all the systems and safeguards for students won't be in place by the start of the 2022 school year.
Recognising National Survivors' Da...
Today we recognise the pain and honour the courage of all survivors of sexual assault and institutional abuse.
Cassy O'Connor MP | Media Release | 15 Nov 2022
National Child Protection Week
With the backdrop of the Commission of Inquiry, this National Child Protection Week is a particularly sombre one in Tasmania
Cassy O'Connor MP | Media Release | 5 Sep 2022
ROGS Confirm Child Safety Record Going...
Today's Productivity Commission Report on Government Services (RoGS) again highlights the Rockliff Government's appalling child safety record.
Cassy O'Connor MP | Media Release | 24 Jan 2023
Draft Youth Justice Blueprint a Good Pl...
While the draft plan represents a major shift for youth justice and lays out the steps forward, it doesn't solve Tasmania's current youth justice failings.
Apology to Victim and Survivors of Inst...
Joining us in Parliament, out on the lawns, over at Elizabeth St Pier & on screens across the country, are Tasmanians - victim survivors & people who love them.
Cassy O'Connor MP | Parliamentary Activity | 8 Nov 2022


Cannabissearch.com Buzz Rating: Apollo 99 gives an incredibly medicinal effect, providing feelings of relaxation, while reducing inflammation, wherever it may occur in the body. Some cerebral effects are noted for being slightly psychedelic and upbeat, with this strain.
Dutch Flowers created a marvelous hybrid, with their crossing of Apollo 13 x Cinderella 99. These two famous strains are now combined to make a variety with tight, large buds that are covered in potent resin. They are noted for having excellent flavors of pepper, hash and freshly upturned soil. Aromas are of earth and pine. Apollo 99 makes a great smoke for any time of the day or night, with a very well balanced effect. Not too up and not too down. Just right.
This variety should only double in height, by the end of flowering. With that in mind, growers can feel comfortable vegetating this one as long as six weeks, before flipping the switch to flowering. When grown outdoors, these ladies will be ready to come off the vine by the middle of October. Yields have shown to be average to slightly above average, producing buds with incredible quality that any grower would be pleased with.
Apollo 99 0 out of 5 based on 0 ratings. 0 user reviews.


New Bug Seat includes: seat back, seat pan, tilt, recline, angle adjustable flip up footplate and tool, left/right hip guide and arm rest. (Base Options available below) Please see the "Specs" tab for specific sizes. Select color below.
Select One (or multiple!) Base for your Seat, or Select "None" if you already have a Base.
Spunky Italian design isn't just for cars and clothes...the Ormesa New Bug Special Needspushchair offers Moms & kids awesome features! Tilt in space technology, 2 interchangeable bases: 4-wheel base & Hi-Low Feeder...and, it comes in green! New Modern Colors and updated sizing allows for more growth. The Ormesa New Bug is a versatile seating system. It is ideal for children with moderate to severe positioning needs. Size one/two for children up to 45 inches tall/ 77 lbs. Size three/four for children/teens up to 69 inches tall and 99 lbs.
Questions about the Ormesa New Bug Seating System (2-Base Options) others have asked!
Can the seat unit without the wheel base be used in the car as a car seat?
No, the New Bug seat cannot be interchanged as a car seat.
This is a frequent question but when the logistics are broken down, it becomes clear that a combo system never really works.
To secure a seat into a vehicle, significant effort is involved. It would not really be possible to unhook a car seat, lift a child and seat together, and place them on a stroller base. The alternative... what do you do with the child after you take them out of the car seat while you are removing the car seat from the vehicle and securing it onto the stroller.
Separate car seats that are dedicated for use are the most popular. The ones (Columbia and Special Tomato) that are removable and "strollerable" are rarely family favorites due to the logistics discussed above.
The New Bug can be placed in a forward or parent facing position by changing the configuration of the mounting plate.
ORMESA warrants the product for Two Years from the date of the purchase. Always use original replacement parts or you will risk voiding the warranty.
THE GUARANTEE DOES NOT COVER WEAR PARTS, which are subject to wear and tear, such as the upholsteries and the wheels.


text:
require File.dirname(__FILE__) + "/../../spec_helper"

describe OpenStruct, "to_hash" do
  it "should return the hash it works with" do
    OpenStruct.new({:foo => :bar}).to_hash.should == {:foo => :bar}
  end
end

redpajama_set_name: RedPajamaGithub


Tee – The Newest Bead from Preciosa!
Excited to announce the latest bead! We'll be posting more samples and finished projects soon!
I have a question . How do you do a knot that will stay on a Stretch Bead Cord ? AN ANSWER WILL BE APPRECIATED .
Here is one good step by step I used.
Previous Previous post: Jade, the Beaded Mannequin!


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This page contains a more detailed guide to significant events concerning the Stephen Joseph Theatre In The Round, Scarborough, in 1986.
February: Robin Herford is appointed Director of Productions / joint Artistic Director of the Stephen Joseph Theatre In The Round to run the company on a day-to-day basis whilst Alan Ayckbourn (who remains Artistic Director) take his two year sabbatical at the National Theatre.
April: Russ Allen becomes the theatre's publicity officer. He is notorious for both publicity stunts and a very public presence during his two year tenure.
4 April: It is reported in the press that the theatre is £22,000 (later revealed to be £17,000) in debt to the Midland Bank following a publication of a letter from the bank to the theatre; it is not clear why the theatre's press office chose to leak a letter suggesting it had financial difficulties.
Late April: The theatre's press officer, Russ Allen, claims the theatre could be the target of terrorists following the US bombing of Libya.
May: Scarborough Building Society begins sponsorship at the theatre. The company provides funding towards a new pay, Stephen Mallatratt's Touch Wood & Whistle.
Summer: Alan Ayckbourn leaves the theatre to begin his two year sabbatical as a Company Director at the National Theatre.
29 May: The Spring / Summer season opens with Time & Time Again by Alan Ayckbourn.
1 June: The Sunday Night concerts are renamed Scarborough Rock and the summer season includes famed folk singer Martin Carthy.
16 July: World premiere of Stephen Mallatratt's Touch Wood & Whistle; the first new play written for the company by the actor, playwright & screenwriter since his return in 1985 following a six year absence from the company.
July: A short-lived partnership between the theatre's restaurant The Square Cat and the vegetarian chef Sarah Brown - who ran a restaurant in the town - comes to a close after two months with publicity officer Russ Allen saying: "most of our customers are not vegetarians and there is a demand for meals containing meat."
31 July: Relations between the theatre and BBC Radio York seriously deteriorate leading the radio station to abandon plans to broadcast the theatre's Scarborough Rock concerts and threatening to stop covering the theatre if its staff were 'verbally abused by your staff, as happened on a recent occasion.'
August: A new initiative ostensibly aimed at streamlining the theatre's amateur programme into a single season leads to several societies complaining to the Scarborough Evening News about the publicity officer Russ Allen's comments and a perceived acrimonious attitude towards the amateurs as well as sharply increased costs.
20 August: It is reported a thief broke into the building, only to steal the theatre's alarm bell!
September: The theatre launches its first regular drama workshops for adults led by Alison Keyes called the Open Door theatre project. Aimed at 16 - 25 year olds, it offers a chance to learn writing, acting and directing skills.
17 September: The Autumn / Winter season opens with Benefactors by Michael Frayn.
November: The Financial Times critic Michael Coveney reports Alan Ayckbourn is considering not returning to Scarborough from the National Theatre. As Alan has always stated he would return to Scarborough, never actually stood down as Artistic Director and made plans for the two years until his return, the report is dismissed as gossip and based on the theatre's rocky relationship with Scarborough Council. No credible source is ever been given for the report.
In response to Michael Coveney's story above, it emerges Alan Ayckbourn has never drawn his wage as Artistic Director of the theatre (since 1972), has invested more than £70,000 into the theatre and gives 1% of his London royalties to the theatre.
13 November: BBC Radio 2's Around Midnight is broadcast live from the theatre with guests including Alan Ayckbourn and Robin Herford.
23 December: Alan Ayckbourn is appointed a Freeman of the Borough of Scarborough.
Click here to go to 1987.


text:
require File.expand_path(File.join(File.dirname(__FILE__),'..','..','test_helper'))
require 'new_relic/agent/samplers/cpu_sampler'

class NewRelic::Agent::Samplers::CpuSamplerTest < Test::Unit::TestCase

  def setup
    @original_jruby_version = JRUBY_VERSION if defined?(JRuby)
  end

  def teardown
    set_jruby_version_constant(@original_jruby_version) if defined?(JRuby)
  end

  def test_correcly_detecting_jruby_support_for_correct_cpu_sampling
    if defined?(JRuby)
      set_jruby_version_constant '1.6.8'
      refute_supported_on_platform

      set_jruby_version_constant '1.7.0'
      assert_supported_on_platform

      set_jruby_version_constant '1.7.4'
      assert_supported_on_platform
    else
      assert_supported_on_platform
    end
  end

  #
  # Helpers
  #

  def assert_supported_on_platform
    assert_equal NewRelic::Agent::Samplers::CpuSampler.supported_on_this_platform?, true, "should be supported on this platform"
  end

  def refute_supported_on_platform
    assert_equal NewRelic::Agent::Samplers::CpuSampler.supported_on_this_platform?, false, "should not be supported on this platform"
  end

  def set_jruby_version_constant(string)
    Object.const_set('JRUBY_VERSION', string)
  end

end


text:
\section{Introduction}
\label{sec:intro}

Comparison-based algorithms have been thoroughly studied in computer
science. This includes algorithms for problems such as
\textsc{Minimum}, \textsc{Median}, \textsc{Sorting},
\textsc{Searching}, \textsc{Dictionaries}, \textsc{Priority Queues},
and many others. The cost measure analyzed is almost always the total
number of comparisons performed by the algorithm, either in the worst
case or the expected case. Recently, another type of cost measure has
been introduced~\cite{afshani:fragile-ESA19} which instead considers
how many comparisons each individual element is subjected during the
course of the algorithm. In~\cite{afshani:fragile-ESA19}, a
comparison-based algorithm is defined to have {\em fragile complexity\xspace} $\f{n}$ if
each individual input element participates in at most $\f{n}$
comparisons. The fragile complexity\xspace of a computational problem is the best possible
fragile complexity\xspace of any comparison-based algorithm solving the problem.
\stefan{Define out-fragile complexity\xspace complexity}

This cost measure has both theoretical and practical motivations. On
the theoretical side, it raises the question of to what extent the
comparisons necessary to solve a given problem can be spread evenly
across the input elements. On the practical side, this question is
relevant in any real world situation where comparisons involve some
amount of destructive impact on the elements being compared (hence the
name of the cost measure). As argued in~\cite{afshani:fragile-ESA19},
one example of such a situation is ranking of any type of consumable
objects (wine, beer, food, produce), where each comparison reduces the
available amount of the objects compared. Here, an algorithm like
\textsc{QuickSort}, which takes a single object and partitions the
whole set with it, may use up this pivot element long before the
algorithm completes. Another example is sports, where each comparison
constitutes a match and takes a physical toll on the athletes
involved. If a comparison scheme subjects one contestant to many more
matches than others, both fairness to contestants and quality of
result are impacted---finding a winner may not be very useful if this
winner has a high risk of being injured in the process. The negative
impact of comparisons may also be of non-physical nature, for instance
when there is a privacy risk for the elements compared, or when bias
grows if few elements are used extensively in comparisons.

\subsection{Previous work}
In~\cite{afshani:fragile-ESA19}, the study of algorithms' fragile complexity\xspace was
initiated and a number of upper and lower bounds on the fragile complexity\xspace for
fundamental problems was given. The problems studied included
\textsc{Minimum}, the \textsc{Selection}, \textsc{Sorting}, and
\textsc{Heap Construction}, and both deterministic and randomized
settings were considered.
In the deterministic setting, \textsc{Minimum} was shown to have fragile complexity\xspace
$\Omega(\log n)$ and \textsc{Sorting} to have fragile complexity\xspace $O(\log n)$. Since
\textsc{Sorting} can solve \textsc{Selection}, which can solve
\textsc{Minimum}, the fragile complexity\xspace of all three problems is $\Theta(\log
n)$. The authors then consider randomized algorithms, as well as a
more fine-grained notion of fragile complexity\xspace, where the objective is to protect
selected elements such as the minimum or median (i.e., the element to
be returned by the algorithm), possibly at the expense of the
remaining elements.
Among other results, it is shown in~\cite{afshani:fragile-ESA19} that
\textsc{Minimum} can be solved incurring expected $O(1)$ comparisons
on the minimum element itself, at a price of incurring expected
$O(n^{\varepsilon})$ on each of the rest. Also a more general
trade-off between the two costs is shown, as well as a close to
matching lower bound. For \textsc{Selection}, similar results are
given, including an algorithm incurring expected $O(\log \log n)$
comparisons on the returned element itself, at a price of incurring
expected $O(\sqrt{n})$ on each of the rest.

An earlier body of work relevant for the concept of fragile complexity\xspace is the study
of sorting networks, started in 1968 by Batcher~\cite{Batc68}. In
sorting networks, and more generally comparator networks, the notion
of depth (the number of layers, where each layer consists of
non-overlapping comparators) and size (the total number of
comparators) correspond to fragile complexity and standard worst case
complexity, respectively, in the sense that a network with
depth~$f(n)$ and size~$s(n)$ can be converted into a comparison-based
algorithm with fragile complexity~$f(n)$ and standard
complexity~$s(n)$ by simply simulating the network.

Batcher, as well as a number of later
authors~\cite{Dowd-Perl-Rudolph-Saks/89,Parberry/92,IPL::ParkerP1989,books/garland/Pratt72}, gave sorting networks with
$\ensuremath{\mathcal{O}}(\log^2 n)$ depth and $\ensuremath{\mathcal{O}}(n\log^2 n)$ size.
For a long time it was an open question whether better results were
possible. In 1983, Ajtai, Koml{\'o}s, and
Szemer{\'e}di~\cite{aks-halvers,aks} answered this in the affirmative
by constructing a sorting network of $\ensuremath{\mathcal{O}}(\log n)$ depth and
$\ensuremath{\mathcal{O}}(n\log n)$ size. This construction is quite complex and involves
expander graphs~\cite{HooLinWig06,journals/fttcs/Vadhan12}. It was
later modified by
others~\cite{Chvatal/92,conf/stoc/Goodrich14,Paterson/90,journals/algorithmica/Seiferas09},
but finding a simple, optimal sorting network, in particular one not
based on expander graphs, remains an open problem.  Comparator
networks for other problems, such as selection and heap construction
have also been studied
~\cite{alekseev:selection-69,brodal:heap-98,Jimbo,Pippenger91,Yao.merging}.

While comparator networks are related to fragile complexity\xspace in the sense that
results for comparator networks can be transfered to the fragile complexity\xspace setting by
simple simulation, it is demonstrated in~\cite{afshani:fragile-ESA19}
that the two models are not equivalent: there are problems where one
can construct fragile algorithms with the same fragile complexity\xspace, but with
strictly lower standard complexity (i.e., total number of comparisons)
than what is possible by simulation of comparison networks. These
problems include \textsc{Selection} and \textsc{Heap Construction}.

\subsection{Our Contribution}
In many settings, the classical worst case complexity of
comparison-based algorithms can be lowered if additional information
on the input is known. For instance, sorting becomes easier than
$\Theta(n \log n)$ if the input is known to be close to
sorted. Another example is searching in a sorted set of elements,
which becomes easier than $O(\log n)$ if we know an element of rank
close to the element searched for. Such algorithms may be described as
\emph{adaptive} to input restrictions (using the terminology from the
sorting setting~\cite{DBLP:journals/csur/Estivill-CastroW92}). Given
that the total number of comparisons can be lowered in such
situations, the question arises whether also reductions in the fragile complexity\xspace
are possible under these types of input restrictions.

In this paper, we expand the study of the fragile complexity\xspace of comparison-based
algorithms to consider the impact of a number of classic input
restrictions. We show that searching for the predecessor in a sorted
array has fragile complexity\xspace $\Theta(\log k)$, where $k$ is the rank of the query
element, both in a randomized and a deterministic setting. For
predecessor searches, we also show how to optimally reduce the
amortized fragile complexity\xspace of the elements in the array. We also prove the
following results: Selecting the $k$th smallest element has expected
fragile complexity\xspace $O(\log\log k)$ for the element selected.  Deterministically
finding the minimum element has fragile complexity\xspace $\Theta(\log(\ensuremath{\mathrm{Inv}}))$ and
$\Theta(\log(\ensuremath{\mathrm{Runs}}))$, where $\ensuremath{\mathrm{Inv}}$ is the number of inversions in a
sequence and $\ensuremath{\mathrm{Runs}}$ is the number of increasing runs in a sequence.
Deterministically finding the median has fragile complexity\xspace
$O(\log(\ensuremath{\mathrm{Runs}}) + \log\log n)$ and $\Theta(\log (\ensuremath{\mathrm{Inv}}))$.
Deterministic sorting has fragile complexity\xspace $\Theta(\log (\ensuremath{\mathrm{Inv}}))$ but it has fragile complexity\xspace
$\Theta(\log n)$ regardless of the number of runs.

\section{Searching}
\label{sec:searching}

The problem of predecessor searching is, given a sorted array~$A$ with
$n$ elements, $A[0]..A[n-1]$, answer queries of the form ``What is the
index of the largest element in $A$ smaller than $x$?''  Binary search
is the classic solution to the predecessor search problem. It achieves
$\log n$ fragile complexity\xspace for~$x$, and fragile complexity\xspace at most one for each element of~$A$.
We can improve on this in two ways.  The first is where we try to keep
the fragile complexity\xspace of~$x$ small, which is possible if we know something about
the rank of~$x$. We show that the optimal dependency on
the rank of~$x$ is $\Theta(\log k)$ where $k$ is its rank, both for deterministic and
randomized algorithms.\footnote{For simplicity of exposition, we assume the
  rank is close to one, but the result clearly holds for rank distance
  to other positions in~$A$.}
The second setting is where we are concerned with the
fragile complexity\xspace of the other elements.  While there is no way to improve a
single search, classical deterministic binary search will always do
the first comparison with the same element (typically the median).
Hence we consider deterministic algorithms that improve the amortized
fragile complexity\xspace of any element of the array~$A$ over a sequence of searches.

\subsection{Single search}

\begin{theorem}\label{thm:rank}
  Let $A$ be a sorted array.
  Determining the predecessor of an element $x$ within~$A$ has fragile complexity\xspace $\Theta(\log k)$ for deterministic and randomized algorithms, where $k$ is the rank of $x$ in~$A$.
\end{theorem}
\begin{proof}
The upper bound follows from standard exponential
search~\cite{conf/stoc/Fredman75}: We compare~$x$ to
$A[2],A[4],A[8],\ldots$ until we find the smallest~$i$ such that
$x<A[2^i]$.  We perform a binary search with the initial interval
$[2^{i-1},2^i]$.  If $x$ has the predecessor $A[k]$, this requires
$O(\log k)$ comparisons.

For the lower bound assume we have a deterministic algorithm to
determine the rank of an element~$x$.  If the answer of the algorithm
is~$k$, let $B_k$ be the bit-string resulting from concatenating the
sequence of the outcomes of the comparisons performed by the
algorithm, the $i$-th bit $B_k[i]=0$ for $x<A[k]$, otherwise it is~1.
Because the algorithm is deterministic and correct, all these
bit-strings are different and they are a code for the numbers
$1,\ldots,n$.  Now, for any~$k$, consider the uniform distribution on
the numbers~$0,\ldots,k-1$, a distribution with entropy~$\log k$.  By
Shannon's source coding theorem, the average code length must be at
least $\log k$, i.e., $\sum_{i=0}^{k-1} |B_i| \ge k\log k$.

For a contradiction, assume there would be an algorithm with $|B_i|\le \log i$ (the binary logarithm itself).
Then for $k> 1$, $\sum_{i=0}^{k-1} |B_i| < k\log k$, in contrast to Shannon's theorem.

The bound $\sum_{i=0}^{k-1} |B_i| \ge k\log k$ also holds for randomized algorithms if the queries are drawn uniformly from~$[1,\ldots,k]$, following Yao's principle:
Any randomized algorithm can be understood as a collection of deterministic algorithms from which the 'real' algorithm is drawn according to some distribution.
Now each deterministic algorithm has the lower bound, and the average number of comparisons of the randomized algorithm is a weighted average of these.
Hence the lower bound also holds for randomized algorithms.\qed
\end{proof}
\subsection{Sequence of searches}

As mentioned, in binary search, the median element of the array will be compared with every query element. Our goal here is to develop a search strategy so as to ensure that data far away from the query will only infrequently be involved in a comparison.
Data close to the query must be queried more frequently. While we prove this formally in Theorem~\ref{t:lb}, it is easy to see that predecessor and successor of a query must be involved in comparisons with the query in order to answer the query correctly.

\begin{theorem} \label{t:search}
There is a search algorithm that
for any sequence of predecessor searches $x_1, x_2, \ldots, x_m$ in a sorted array $A$ of size $n$
the number of comparisons with any $y\in A$ is $O\lrp{ \log n+ \sum_{i=1}^m \frac{1}{d(x_i,y)}}$
 where $d(x,y)$ is the number of elements between $x$ and $y$ in $A$, inclusive.
The runtime is $O(\log n)$ per search and the structure uses $O(n)$ bits of additional space.
\end{theorem}

\def\text{\sl offset}{\text{\sl offset}}
\begin{proof}
We use the word interval to refer to a contiguous range of $A$; when we index an interval, we are indexing $A$ relative to the start of the interval.
Call an aligned interval $I$ of $A$ of rank $i$ to be $( A[k\cdot 2^i]\ldots A[(k+1) \cdot 2^i])$ for some integer $k$, i.e., the aligned intervals of $A$ are the dyadic intervals of $A$. There are $O(n)$ aligned intervals of $A$, and for each aligned interval $I$ of rank $i$ we store an offset $I.\text{\sl offset}$ which is in the range $[0,2^i)$, and it is initialized to 0.

The predecessor search algorithm with query $x$ is a variant of recursive binary search, where at each step an interval $I_q$ of $A$ is under consideration, and the initial recursive call considers the whole array $A$. Each recursive call proceeds as follows: Find the largest $i$ such that there are at least three rank-$i$ aligned intervals in $I_q$, use $I_m$ to denote  the middle such interval (or an arbitrary non-extreme one if there are more than three), and we henceforth refer to this recursive call as a rank-$i$ recursion. Compare $I_m[I_m.\text{\sl offset}]$ with $x$, and then increment $I_m.\text{\sl offset}$ modulo $2^i$. Based on the result of the comparison, proceed recursively as in binary search.
The intuition is by moving the offset with every comparison, this prevents a single element far from the search from being accessed too frequently.
We note that the total space used by the offsets is $O(n)$ words, which can be reduced to $O(n)$ bits if the offsets are stored in a compact representation.

\pagebreak[3]
\noindent First, several observations:\nopagebreak
\begin{compactenum}
\item \label{p1} In a rank-$i$ recursion, $I_q$ has size at least $3 \cdot 2^i$ (since there must be at least three rank-$i$, size $2^i$ aligned intervals in $I_q$) and at most $8 \cdot 2^i$, the latter being true as if it was this size there would be three rank-$i+1$ intervals in $I_q$, which would contradict $I_m$ having rank $i$.

\item If $I_q$ has size $k$ then if there is a recursive call, it is called with an interval of size at most~$\frac{7}{8}k$. This is true by virtue of $I_m$ being rank-$i$ aligned with at least one rank-$i$ aligned interval on either side of $I_m$ in $i$. Since $I_q$ has size at most $8\cdot 2^i$, this guarantees an eighth of the elements of $I_q$ will be removed from consideration as a result of the comparison in any recursive call.

\item \label{p3} From the previous two points, one can conclude that for a given rank $i$, during any search there are at most 7 recursions with of rank $i$. This is because after eight recursions any rank-$i$ search will be reduced below the minimum for rank $i$:  $8 \cdot 2^i \cdot \lrp{\frac{7}{8}}^8 <3 \cdot 2^i$.
\end{compactenum}

For the analysis, we fix an arbitrary element $y$ in $A$ and use the potential method to analyse the comparisons involving $y$.
Let $\mathcal{I}_y=\{I_y^1,I_y^2\ldots\}$ be the $O(\log n)$ aligned intervals that contain $y$, numbered such that $I_y^i$ has rank $i$.
Element $y$ will be assigned a potential relative to each aligned interval $I_y^i\in \mathcal{I}_y$ which we will denote as $\varphi_y(I_y^i)$.
Let $t_y(I_y^i)$ be number of times $I_y^i.\text{\sl offset}$ needs to be incremented before $I_y^i[I_y^i.\text{\sl offset}]=y$, which is in the range $[0,2^i)$.
The potential relative to $I^i_y$ is then defined as $\varphi_y(I_y^i)\coloneqq \frac{2^i-t_y(I^i_y)}{2^i}$, and the potential relative to $y$ is defined to be the sum of the potentials relative to the intervals in $\mathcal{I}_y$: $\varphi_y \coloneqq \sum_{I^i_y\in\mathcal{I}_y} \varphi_y(I_y^i)$.

How does $\varphi_y(I_y^i)$ change during a search?
First, if there is no rank-$i$ recursive call during the search to an interval containing $y$, it does not change as $I_y^i.\text{\sl offset}$ is unchanged.
Second, observe from point \ref{p3} that a search can increase $\varphi_y (I_y^i)$ by only $\frac{7}{2^i}$.
Furthermore if $y$ was involved in a comparison during a rank-$i$ recursion, there will be a loss of $1-\frac{1}{2^i}$ units of potential in $\varphi_y(I_y^i)$ as the offset of $I_y^i$ changes from 0 to $2^i-1$.

Following standard potential-based amortized analysis, the amortized number of comparisons involving $y$ during a search is the actual number of comparisons (zero or one) plus the change in the potential $\varphi_y $.
Let $i_{\min}$ be the smallest value of $i$ for which there was a rank-$i$ recursion that included $y$. As the maximum gain telescopes, the potential gain is at most $\frac{14}{2^{i_{\min}}}$, minus 1 if $y$ was involved in a comparison.
Thus the amortized number of comparisons with $y$ in the search is at most $\frac{14}{2^{i_{\min}}}$.

Observe that if there was a rank-$i$ recursion that included $y$, that $d(x,y)$ is at most $8\cdot 2^i$ by point~\ref{p1}.
This gives $d(x,y)\leq 8\cdot 2^i \leq 8\cdot 2^{i_{\min}}$.
Thus the amortized cost  can be restated as being at most
$ \frac{14}{2^{i_{\min}}}\leq \frac{112}{d(x,y)}$.

To complete the proof, the total number of comparisons involving $y$ over a sequence of searches is the sum of the amortized costs plus any potential loss. As the potential $\varphi_y$ is always nonnegative and at most $\lrc{\log n}$ (1 for each $\varphi_y(I^i_y)$), this gives the total cost as $O\lrp{ \log n+ \sum_{i=1}^m \frac{1}{d(x_i,y)}}$. \qed
\end{proof}

Note that the above proof was designed for easy presentation and not an optimal constant. Also note that this theorem implies that if the sequence of searches is uniformly random, the expected fragility of all elements is $O(\frac{\log n}{n})$, which is asymptotically the best possible since random searches require $\Omega(\log n)$ comparisons in expectation.

\subsection{Lower Bounds.} It is well-known that comparison-based searching requires $\Omega(\log n)$ comparisons per search. In our method, taking a single search $x_i$ summing over the upper bound on amortized cost of the number of comparisons with $y$, $\frac{42}{d(x_i,y)}$, for all $y$ yields a harmonic series which sums to $O(\log n)$. But we can prove something stronger:

\begin{theorem} \label{t:lb}
There is a constant $c$ such that if a predecessor search algorithm has an amortized number of comparisons of $f(d(x_i,y))$ for an arbitrary $y$ for every sequence of predecessor searches $x_1, x_2, \ldots x_m$, then $\sum_{k=1}^p f(k) \geq c \log p$ for all $p \leq n$.
\end{theorem}

\begin{proof}
This can be seen by looking at a random sequence of predecessor searches for which the answers are uniform among $A[0]\ldots A[p-1]$, if the theorem was false, similarly to the proof of Theorem~\ref{thm:rank}, this would imply the ability to execute such a sequence in $o(\log p)$ amortized time per operation. \qed
\end{proof}
\riko{perhaps we could do this even nicer}

This shows that a flatter asymptotic tradeoff between $d(x_i,y)$ and the amortized comparison cost is impossible; more comparisons are needed in the vicinity of the search than farther away.
For example, a flat amortized number of comparisons of $\frac{\log n}{n}$ for all elements would sum up to $O(\log n)$ amortized comparisons over all elements, but yet would violate this theorem.

\subsection{Extensions.}

Here we discuss extensions to the search method above. We omit the proofs as they are simply more tedious variants of the above.

One can save the additional space used by the offsets of the intervals through the use of randomization. The offsets force each item in the interval to take its turn as the one to be compared with, instead one can pick an item at random from the interval. This can be further simplified into a binary search where at each step one simply picks a random element for the comparison amongst those (in the middle half) of the part of the array under consideration.

To allow for insertions and deletions, two approaches are possible. The first is to keep the same array-centric view and simply use the packed-memory array \cite{DBLP:conf/icalp/ItaiKR81,DBLP:conf/sigmod/Willard86,DBLP:journals/iandc/Willard92} to maintain the items in sorted order in the array. This will give rise to a cost of $O(\log^2 n)$ time which is inherent in maintaining a dynamic collection of items ordered in an array \cite{DBLP:journals/siamcomp/BulanekKS15} (but no additional fragility beyond searching for the item to insert or delete as these are structural changes).
The second approach would be to use a balanced search tree such as a red-black tree \cite{DBLP:conf/focs/GuibasS78}. This will reduce the insertion/deletion cost to $O(\log n)$ but will cause the search cost to increase to $O(\log^2 n)$ as it will take $O(\log n)$ time to move to the item in each interval indicated by the offset, or to randomly choose an item in an interval. The intervals themselves would need to allow insertions and deletions, and would, in effect be defined by the subtrees of the red-back tree.
It remains open whether there is a dynamic structure with the fragility results of Theorem~\ref{t:search} where insertions and deletions can be done in $O(\log n)$ time.

\section{Selection}
\label{sec:selection}

In this section we consider the problem of finding the $k$-th smallest element of an unsorted array.
There is a randomized algorithm that selects the $k$-th smallest element with expected fragile complexity of $O(\log \log n)$ for the selected element~\cite{afshani:fragile-ESA19}.
We consider the question if this complexity can be improved for small $k$.
In this section we define a sampling method that, combined with the algorithm given in~\cite{afshani:fragile-ESA19}, selects the $k$-th smallest element with expected $O(\log \log k)$ comparisons.

Next, we define the filtering method \textsc{ReSet} in a tail-recursive fashion. The idea of this procedure is the following: First, we build a random half size sample $A_1$ from the input set $X$. Later, we continue recursively constructing a random half sample $A_i$ from the previous sample $A_{i-1}$ until we get a random sample $A_{\ell}$ of size $O(k+1)$. Once $A_{\ell}$ is given, then a set $A'_{\ell}$ of size $O(k)$ is given for the previous recursive call. Using such set, a new subset $A'_{\ell-1}$ is given from the previous sample $A_{\ell-1}$ where its expected size is $O(k)$. This process continuous until a final subset $\mathcal{C}$ is given from the input set $X$ such that its expected size $O(k)$ and it contains the $k$-th smallest element of $X$.

\noindent\fbox{\scalebox{0.95}{\begin{minipage}{\textwidth}
\begin{small}
\begin{algorithmic}[1]
	 \Procedure{ReSet}{$X, k$}
	 	\Comment{Returns a small subset $\mathcal{C}$ of $X$ that contains the $k$-th element}
			\State Let $n=|X|$ and $\mathcal{C} = \emptyset$
			\If{$k \geq \frac{n}{2}-1$}
			    \Comment{The set has size $O(k+1)$}
			    \State Let $A'=X$
			\Else
			    \Comment{Recursively construct a sample of expected size $O(k+1)$}
			    \State Sample $A$ uniformly at random from $X$, $|A|= \frac{n}{2}$
			    \State Let $A'=$ \textsc{ReSet($A, k$)}
			\EndIf
			\State Choose the ($k+1$)-th smallest element $z$ from $A'$ (by standard linear time selection)
			\State Let $\mathcal{C}=\{x \in X : x\leq z\}$
			\State \Return $\mathcal{C}$
	\EndProcedure
\end{algorithmic}
\end{small}
\end{minipage}}
}

In the following theorem we show that the combination of the \textsc{ReSet} procedure and the randomized selection algorithm in~\cite{afshani:fragile-ESA19}, results in expected $O(\log \log k)$ comparisons for the selected element.

\begin{theorem}
Randomized selection is possible in expected fragile complexity\\ $O(\log \log k)$ in the selected element.
\end{theorem}
\begin{proof}
Let us show that the following procedure for selecting the $k$-th element in a set $X$ with $|X|=n$, gives an expected fragile complexity $O(\log \log k)$ in the $k$-th element:

\emph{If $k > n^{\frac{1}{100}}$, then let $S'=X$. If $k \leq n^{\frac{1}{100}}$, then sample uniformly at random $S$ from $X$, where $|S|=\frac{n}{k}$. Let $C=$ \textsc{ReSet}($S, k$) and select the $k+1$-th smallest element $z$ from $\mathcal{C}$ by standard linear time selection. Let $S'=\{x \in X: x\leq z\}$. Finally, apply to $S'$ the randomized selection algorithm of~\cite{afshani:fragile-ESA19}.}

Let $x_k$ denote the $k$-th smallest element in $X$ and let $f_k$ denote the fragile complexity\xspace of $x_k$.
Note that if $x_k \in S$, then, before constructing $S'$, $f_k$ is given by the fragile complexity\xspace of $x_k$ in $\textsc{ReSet}(S,k)$ plus $O(|\mathcal{C}|)$ when finding the ($k+1$)-th smallest element in $\mathcal{C}$. Otherwise, $x_k$ is not compared until $S'$ is constructed.
On the other hand, recall that the expected $f_k$ in the algorithm in~\cite{afshani:fragile-ESA19} is $O(\log \log m)$ where $m$ is the size of the input set. Hence, the expected $f_k$ after selecting the $k+1$-th element in  $\mathcal{C}$ is 1 when creating $S'$ plus the expected $f_k$ in the randomized selection algorithm in~\cite{afshani:fragile-ESA19} that is $\sum_{|S'|}O(\log \log |S'|)\mathbb{P}[|S'|]=\mathbb{E}[O(\log \log |S'|)]$. Thus, $\mathbb{E}[f_k]=(\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S]+\mathbb{E}[|\mathcal{C}|])\mathbb{P}[x_k \in S] + 0\mathbb{P}[x_k \notin S] + 1 + \mathbb{E}[O(\log \log |S'|)]$. Since the logarithm is a concave function,  $\mathbb{E}[O(\log \log |S'|)]\leq O(\log \log (\mathbb{E}[|S'|]))$.
Therefore, if we prove that: (i) the expected fragile complexity\xspace of $x_k$ before creating $S'$ is $O(1)$ and (ii) $\mathbb{E}[|S'|]=c'k^{c}$ for some constants $c$ and $c'$. Then, we obtain that $\mathbb{E}[f_k]\leq O(1)+1+O(c\log\log k+\log c') = O(\log \log k)$, as desired. In order to prove (i) and (ii) we consider 2 cases: (1) $k > n^{\frac{1}{100}}$, (2) $k \leq n^{\frac{1}{100}}$.
\\
\emph{Case 1)} $S' = X$ and it makes no previous comparisons in any element, proving (i). In addition, $S'$ has size less than $k^{100}$. Thus, (ii) holds.
\\
\emph{Case 2)} $S$ is a sample of $X$ with size $\frac{n}{k}$ and $S'=\textsc{ReSet}(S, k)$. \\
First, let us show (i). If $x_k \notin S$, then there are no previous comparisons. Hence, the expected fragile complexity\xspace of $x_k$ before constructing $S'$ is given by $(\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S'] + \mathbb{E}[|\mathcal{C}|])\mathbb{P}[x_k \in S]+ 0$. Since $S$ is an uniform random sample with size $\frac{n}{k}$, $\mathbb{P}[x_k \in S] = \frac{1}{k}$, it suffices to show that $\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S'] + \mathbb{E}[|\mathcal{C}|]=O(k)$, which gives an expectation of $O(k)\frac{1}{k}=O(1)$, proving (i). So, let us show that $\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S'] + \mathbb{E}[|\mathcal{C}|]=O(k)$.
Let $A_0=S$ and let $A_1$ be the sample of $A_0$ when passing through line 6 in \textsc{ReSet}. Similarly, denote by $A_i$ to the sample of $A_{i-1}$ in the $i$-th recursive call of \textsc{ReSet} and let $A'_i=\textsc{ReSet}(A_{i}, k)$. Note that by definition $A'_0=\mathcal{C}$.
Let $\ell+1$ be the number of recursive calls in $\textsc{ReSet}(S, k)$.

Since $A_i$ is a uniform random sample of size $\tfrac{|A_{i-1}|}{2}$ for all $i\geq 1$, $\mathbb{P}[x \in A_i| x \in A_{i-1}]=2^{-1}$ and $\mathbb{P}[x \in A_i| x \notin A_{i-1}]=0$. Hence, $\mathbb{P}[x_{k} \in A_i]=\mathbb{P}[x_k \in \cap_{i=0}^{i} A_i]=2^{-i}$. Note that the number of comparisons of $x_k$ in \textsc{ReSet} is given by the number of times $x_k$ is compared in lines 8 and 9. Thus, for each $i$-th recursive call: if $x_k \in A_i$, then $x_k$ is compared once in line 9; and if $x_k \in A_i\cap A'_i$, then $x_k$ is compared at most $|A'_i|$ times in line 8. Otherwise, $x_k$ is not compared in that and the next iterations.
Thus, $\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S'] + \mathbb{E}[|\mathcal{C}|] \leq \sum_{i=0}^{\ell} (1+\mathbb{E}[|A'_i|])\mathbb{P}[x_k \in A_i]= \sum_{i=0}^{\ell} 2^{-i}(1+\mathbb{E}[|A'_i|])\leq 2(1+ \mathbb{E}[|A'_i|])$.
Let us compute $\mathbb{E}[|A'_i|]$.
Since the ($\ell$$+$$1$)-th iteration $\textsc{ReSet}(A_{\ell},k)$ passes through the if in line 3, there is no new sample from $A_{\ell}$.
Thus, $A'_{\ell}$ is given by the $k+1$ smallest elements of $A_{\ell}$. Therefore, $\mathbb{E}[|A'_{\ell}|]=k+1$ Denote by $a'^{i}_j$ to the $j$-th smallest element of $A'_i$. For the case of $0\leq i<\ell$, we have $A'_i=\{x\in A_{i+1} : x\leq a'^{i+1}_{k+1}\}$. Hence, $\mathbb{E}[|A'_i|]=\mathbb{E}[|\{x\in A_{i+1} : x\leq a'^{i+1}_{k+1}\}|] =\mathbb{E}[|\{x \in A_{i+1} : x \leq a'^{i+1}_1]\}|] + \sum_{j=1}^{k} \mathbb{E}[|\{x \in A_{i+1} :  a'^{i+1}_{j-1}< x \leq a'^{i+1}_{j}]\}|]\leq \sum_{j=1}^{k+1}\sum_{t=1}^{\infty} t2^{-1}(2^{-(t-1)})=2(k+1)$.
Therefore, $\mathbb{E}[f_k\text{ in \textsc{ReSet}}| x_k \in S'] + \mathbb{E}[|\mathcal{C}|]=\sum_{i=0}^{\ell} 2^{-i}(1+\mathbb{E}[|A'_i|])\leq 2+ 2\mathbb{E}[|A'_i|]=O(k)$ proving (i).
Finally, let us show (ii): For simplicity, let $c_j$ denote the $j$-th smallest element of $\mathcal{C}$. Then, $\mathbb{E}[|S'|]=\mathbb{E}[|\{x\in X: x\leq c_1\}|]+\sum_{j=1}^{k}\mathbb{E}[|\{x\in X: c_j \leq x\leq c_{j+1}\}|]\leq \sum_{j=1}^{k+1}\sum_{j=0}^{\infty} j k^{-1}(1-k^{-1})^{j-1}=k(k+1)=O(k^2)$, proving (ii). \qed
\end{proof}
\section{Sorting}
\label{sec:sorting}

When the input is known to have some amount of existing order, sorting
can be done faster than $\Theta(n \log n)$. Quantifying the amount of
existing order is traditionally done using measures of
disorder~\cite{DBLP:journals/csur/Estivill-CastroW92}, of which \ensuremath{\mathrm{Inv}}{}
and \ensuremath{\mathrm{Runs}}{} are two classic examples.\footnote{The measure \ensuremath{\mathrm{Inv}}{} is
  defined as the total number of inversions in the input, where each
  of the ${n\choose 2}$ pairs of elements constitute an inversion if
  the elements of the pair appear in the wrong order. The measure
  \ensuremath{\mathrm{Runs}}{} is defined as the number of runs in the input, where a run is a maximal consecutive ascending subsequence.}
A sorting algorithm is adaptive to a measure of disorder if it
is faster for inputs with a smaller value of the measure. For the
above measures, run times of $O(n \log (\ensuremath{\mathrm{Inv}}/n))$ and
$O(n \log (\ensuremath{\mathrm{Runs}}))$ can be achieved. These results are best possible
for comparison-based sorting, by standard information-theoretic
arguments based on the number of different inputs having a given maximal
value of the measure.

The fact~\cite{aks,afshani:fragile-ESA19} that we can sort all inputs
in $\Theta(n \log n)$ time and $\Theta(\log n)$ fragile complexity\xspace can be
viewed as being able to distribute the necessary comparisons evenly
among the elements such that each element takes part in at most
$\Theta(\log n)$ comparisons.
Given the running times for adaptive sorting stated above, it is
natural to ask if for an input with a given value of \ensuremath{\mathrm{Inv}}{} or
\ensuremath{\mathrm{Runs}}{} we are able to sort in a way that distributes the necessary
comparisons evenly among the elements, i.e., in a way such that each
element takes part in at most $O(\log (\ensuremath{\mathrm{Inv}}))$ or $O(\log (\ensuremath{\mathrm{Runs}}))$
comparisons, respectively.
In short, can we sort in fragile complexity\xspace $O(\log (\ensuremath{\mathrm{Inv}}))$ and
$O(\log (\ensuremath{\mathrm{Runs}}))$?  Or more generally, what problems can we solve with
fragile complexity\xspace adaptive to \ensuremath{\mathrm{Inv}}{} and \ensuremath{\mathrm{Runs}}{}? In this section, we study
the fragile complexity\xspace of deterministic algorithms for \textsc{Minimum},
\textsc{Median}, and \textsc{Sorting} and essentially resolve their
adaptivity to \ensuremath{\mathrm{Inv}}{} and \ensuremath{\mathrm{Runs}}{}.

\begin{theorem}
\textsc{Minimum} has fragile complexity\xspace $\Theta(\log (\ensuremath{\mathrm{Runs}}))$.
\end{theorem}
\begin{proof}
For the upper bound: identify the runs in $O(1)$ fragile complexity\xspace by a scan of the input. Then, use a
tournament on the heads of the runs since the minimum is the minimum of the heads of the runs.
For the lower bound: apply the logarithmic lower
bound for \textsc{Minimum}~\cite{afshani:fragile-ESA19} on the heads
of the runs. \qed
\end{proof}

\begin{theorem}
\textsc{Sorting} has fragile complexity\xspace $\Theta(\log n)$, no matter what value of
\ensuremath{\mathrm{Runs}}{} is assumed for the input.
\end{theorem}
\begin{proof}
The upper bound follows from general sorting. For the lower bound: the input
consisting of a run~$R$ of length $n-1$ and one more element~$x$ has
$\ensuremath{\mathrm{Runs}} = 2$, but $\log n$ comparisons on $x$ can be forced by an
adversary before the position of $x$ in $R$ is determined.\qed
\end{proof}

\begin{theorem}
\textsc{Median} has fragile complexity\xspace $O(\log (\ensuremath{\mathrm{Runs}}) + \log\log n)$.
\end{theorem}
\begin{proof}
Assume that $4 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n < n/2$, since otherwise the
claimed fragile complexity\xspace is $O(\log n)$ for which we already have a median
algorithm~\cite{afshani:fragile-ESA19}. Consider the rank space
$[1,n]$ (i.e., the indices of the input elements in the total sorted
order) of the input elements and consider the rank interval $[a,b]$
around the median defined by $a = n/2 - 4 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$
and $b = n/2 + 4 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$. In each step of the
algorithm, elements are removed in two ways: type~A removals and
type~B removals. A removal of type~A is a balanced removal, where a
number of elements with ranks in $[1,a-1]$ are removed and the same
number of elements with ranks in $[b+1,n]$ are removed. The key
behind the type~A removal is that the median element of the set
prior to the removal is the same as the median of the set after the
removal, if the median prior to the removal has a rank in $[a,b]$.

A removal of type~B is a removal of elements with arbitrary
rank. However, the total number of elements removed by type~B removals
is at most $7 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$ during the\ entire run of the
algorithm. Hence, repeated use of type~A and type~B removals will
maintain the invariant that the median of the remaining elements has a
rank in $[a,b]$.

We now outline the details of the algorithm. The first step is to
identify all the runs in $O(1)$ fragile complexity\xspace by a scan. A run will be
considered {\em short} if the run consists of fewer than
$7 \cdot \log n$ elements and it will be considered {\em long} otherwise. A
step of the algorithm proceeds by first performing a type~B removal
followed by a type~A removal.  A type~B removal consists of removing
all short runs that are present. The short runs that are removed will
be reconsidered again at the end once the number of elements under
consideration by the algorithm is less than
$64 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$.

Once a type~B removal step is completed, only long runs remain under
consideration.  We now describe a type~A removal step. Note that a
long run may become short after a type~A removal step, in which case
it will be removed as part of the next type~B removal step. Each run
can become short (and be removed by a type~B removal) only once, hence
the total number of elements removed by type~B removals will be at
most $7 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$, as claimed.

In the following, let $\mathcal{N}$ denote the elements under
consideration just before a type~A removal (i.e., the elements of the
remaining long runs), and let $N = |\mathcal{N}|$. The algorithm stops
when $N \le 64 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$.

To execute the type~A removal step, the algorithm divides each long
run $R$ into blocks of length $\log n$. The blocks of a run are
partitioned by a {\em partitioning} block. The partitioning block has
the property that there are at least $|R|/7$ elements of $R$ whose
values are less than the values in the partitioning block and at least
$5|R|/7$ elements of $R$ whose value are greater than the elements in
the partitioning block. One element~$x_R$ is selected from the
partitioning block. We will refer to this element as a {\em
partitioning} element.
These partitioning elements are then sorted
into increasing order, which incurs a cost of $O(\log (\ensuremath{\mathrm{Runs}}))$ fragile complexity\xspace
on each of the partitioning elements.  The runs are then arranged in
the same order as their partitioning elements.  Label this sequence of
runs as $R_1, R_2, \dots, R_k$, and let $t$ be the largest index such
that $\sum_{i=1}^{t-1} |R_i| < N/8.$

Since the partitioning element $x_{R_t}$ is smaller than all the elements in the blocks with values greater than their
respective partitioning blocks
in $R_t, R_{t+1},$ $\dots, R_k$, we have that $x_{R_t}$ is smaller than $(7/8)(5N/7) = 5N/8$ of the
remaining elements. Hence in rank it is at least $N/8$ below the median
of the remaining elements. By the invariant on the position in rank
space of this median and the fact that $N > 64 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$, we note
that $x_{R_t}$ has a rank below~$a$. We also note that all the elements
below the partitioning blocks in $R_1, R_{2}, \dots, R_t$ have value less than
$x_{R_t}$. This constitutes at least $(1/8)(N/7) = N/56$
elements in~$\mathcal{N}$ with rank below~$a$. Therefore,
we can remove $N/56$ elements with rank below~$a$. In a similar manner, we can find
at least $N/56$ elements in~$\mathcal{N}$ with rank
above~$b$. Removal of these $2N/56 = N/28$ elements
in~$\mathcal{N}$ constitutes a type~A removal step.

Since the number of elements under consideration, i.e. $N$,
decreases by a constant factor at each step, the algorithm
performs $O(\log n)$ type~A and type~B removal steps before we have
$N \le 64 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$.  Since each block under
consideration in a type~A removal step has size $\log n$, we can guarantee that each element in a
partitioning block
only needs to be selected as a partitioning element $O(1)$ times. This implies
that a total cost of $O(\log (\ensuremath{\mathrm{Runs}}))$ fragile complexity\xspace is incurred on each
element once we have that $N \le 64 \cdot \ensuremath{\mathrm{Runs}} \cdot \log n$.

We now describe the final step of the algorithm.
At this point, the algorithm combines the last $\mathcal{N}$ elements
with all the short runs removed during its execution up to this point, forming the
set~$\mathcal{S}$. This set is the original elements
subjected to a series of type~A removals, each of which are balanced
and outside the rank interval $[a,b]$. Hence, the median
of~$\mathcal{S}$ is the global median. As
$|\mathcal{S}| = O(\ensuremath{\mathrm{Runs}} \cdot \log n)$, we can find this median in
$O(\log(\ensuremath{\mathrm{Runs}} \cdot \log n)) = O(\log (\ensuremath{\mathrm{Runs}}) + \log\log n)$
fragile complexity\xspace~\cite{afshani:fragile-ESA19}, which dominates the total
fragile complexity\xspace of the algorithm.

We note that for $\ensuremath{\mathrm{Runs}} = 2$, we can improve the above result to
$O(1)$ fragile complexity\xspace as follows. Let the two runs be $R_1$ and $R_2$, with
$|R_1| \le |R_2|$.  Compare their middle elements $x$ and $y$ and
assume \ $x \le y$. Then the elements in the first half of $R_1$ are
below $n/2$ other elements, and hence are below the median. Similarly,
the elements in the last half of $R_2$ are above the median. Hence, we
can remove $|R_1|/2$ elements on each side of the median by removing
that many elements from one end of each run. The median of the
remaining elements is equal to the global median. By recursion, we in
$\log |R_1|$ steps end up with $R_1$ reduced to constant length. Then
$O(1)$ comparisons with the center area of $R_2$ will find the median.
Because both runs lose elements in each recursive step, both $x$ and
$y$ will be new elements each time. The total fragile complexity\xspace of the
algorithm is therefore $O(1)$. \qed
\end{proof}

\begin{theorem}
\textsc{Minimum} has fragile complexity\xspace $\Theta(\log (\ensuremath{\mathrm{Inv}}))$.
\end{theorem}
\begin{proof}
Lower bound: For any $k$, consider the instances composed of
$\sqrt{k}$ elements in random order followed by $n - \sqrt{k}$
larger elements in sorted order. These instances have $\ensuremath{\mathrm{Inv}} \le
k$. Finding the minimum is equal to finding the minimum on the first
$\sqrt{k}$ elements, which has a lower
bound~\cite{afshani:fragile-ESA19} of
$\log \sqrt{k} = \Omega(\log k)$ on its fragile complexity\xspace.

For the upper bound, we will remove a subset~$I$ of size $O(\ensuremath{\mathrm{Inv}})$
which leaves a single sorted run~$R$. We can find the mininum in~$I$
in $O(\log(\ensuremath{\mathrm{Inv}}))$ fragile complexity\xspace by a tournament tree, which can then be
compared to the head of~$R$ for the final answer.

We find~$I$ and~$R$ in $O(1)$ fragile complexity\xspace during a scan of the input as
follows, using~$R$ as a stack. For each new element~$e$ scanned, we
compare it to the current top element~$f$ of~$R$. If $e$ is larger, we
push~$e$ to the top of~$R$. If~$e$ is smaller, it forms an inversion
with~$f$, and we include $e$ in $I$. We also put a mark on~$f$. If an
element on the stack gets two marks, we pop it, include it in~$I$,
remove one of its marks (which will account for its inclusion in~$I$)
and move the second mark to the new top element~$f'$ of~$R$. If $f'$
now has two marks, this process continues until an element with only a
single mark is created (or the stack gets empty). An element is
compared with exactly one element residing earlier in the input (when
the element is scanned). To count comparisons with elements residing
later in the input, call such a comparison large or small, depending
on whether the other element is larger or smaller. It is easy to see
that elements on the stack always have between zero and one marks,
that an element with zero marks has participated in one large
comparison, and that an element with one mark has either participated
in at most two larger comparisons or one smaller comparison. Hence,
the fragile complexity\xspace of the process is~$O(1)$. By the accounting scheme, $I$
is no larger than $\ensuremath{\mathrm{Inv}}$ plus the number of marks, which is also at
most~$\ensuremath{\mathrm{Inv}}$. \qed
\end{proof}

\begin{theorem}
\textsc{Median} has fragile complexity\xspace $\Theta(\log (\ensuremath{\mathrm{Inv}}))$.
\end{theorem}
\begin{proof}
As \textsc{Median} solves \textsc{Minimum} via padding with $n$
elements of value~$-\infty$, the lower bound follows from the lower
bound on~\textsc{Minimum}. For the upper bound, find~$R$ and $I$ as in
the upper bound for \textsc{Minimum}, sort~$I$ in
fragile complexity\xspace~$O(\log(\ensuremath{\mathrm{Inv}}))$ and use the algorithm for \textsc{Median}
for $\ensuremath{\mathrm{Runs}} = 2$. \qed
\end{proof}

\begin{theorem}
\textsc{Sorting} has fragile complexity\xspace $\Theta(\log (\ensuremath{\mathrm{Inv}}))$.
\end{theorem}
\begin{proof}
The lower bound follows from the lower bound on~\textsc{Minimum}.
For the upper bound, find~$R$ and $I$ as in the upper bound for
\textsc{Minimum} and let each element recall its position in the
input. Divide the sorted sequence~$R$ into contiguous blocks of size
$|I|$ and let $R_i$ be the set of $i$'th elements of all blocks.
With the $i$'th element of~$I$ we perform an exponential search on
$R_i$, starting from the block where the element's position in the
input is. If the search moves a distance~$k$, the element from~$I$
participated in at least~$k$ inversions in the input, so
$k \le \ensuremath{\mathrm{Inv}}$ and hence the incurred fragile complexity\xspace for the element
is~$O(\log k) = O(\log(\ensuremath{\mathrm{Inv}}))$. A fragile complexity\xspace of~$O(1)$ is incurred on the
elements of~$R$, as each~$R_i$ is used once. After this, each
element from~$I$ knows its position in~$R$ within a window of
size~$|I|$. If the window of an element and a block overlaps, we
call the element and the block associated. Each block of~$R$ is
associated with at most~$|I| = O(\ensuremath{\mathrm{Inv}})$ elements, and each element
is associated with at most two blocks. For each block in turn, we
now sort its associated elements and merge them into the block
(except for tail elements overlapping the next block). The sorting
incurs~$O(\log(\ensuremath{\mathrm{Inv}}))$ fragile complexity\xspace, as does the merging if we use
exponential merging~\cite{afshani:fragile-ESA19}. We remove all
inserted elements from any association with the next block. Then we
continue with the next block. \qed
\end{proof}

redpajama_set_name: RedPajamaArXiv


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text:
Far Cry New Dawn Review
To think that Far Cry 5 is less than a year old can be quite a shocking revelation especially given that Far Cry New Dawn is here and it is a brand new fully-fledged Far Cry experience. While this is not an expansion, it is a smaller, and shorter Far Cry game but has a large emphasis on an end-game grind, more than its predecessor.
Check out our Far Cry New Dawn video review below:
Far Cry New Dawn is everything you would expect from the Far Cry recipe. It has a large open world to loot and explore, a vast array of awesome guns to craft and kill things with, lots of side activities to keep you busy and of course, all the explosions that light up your screen. Far Cry New Dawn also brings a couple of new mechanics into the mix that adds a few new gameplay elements.
Some of these don't work as well as I hoped but they do enough to make the game stand out from last year's release. The new RPG system sees you go up against higher rank enemies that are tougher to kill especially if you don't have the higher gear to fight them with, and the ability to scavenge each outpost and reset it in order to up the difficulty and gain Ethanol and rare cool-looking outfits is a great way to enable players to keep grinding these outposts well after they complete the game.
Far Cry New Dawn takes place 17 years after the events of Far Cry 5's "end of the world" ending. Society is no longer. Instead, small groups of people have formed refuge together in an attempt to create a sanctuary. You take on the role of a survivor sent to a place called Prosperity in order to help them grow and evolve. A deadly group of rebels known as the Highwaymen led by crazy twins Lou and Mickey who intercept your train and things go south.
After a rather epic intro, you are introduced to New Dawn's take on Hope County, a cast of great characters and the beautiful open world to explore. The aim of the game is to build up Prosperity by helping "Specialists" across the map, bringing them back to the home camp, levelling up their department and take down the Highwaymen once and for all.
The Far Cry series has been known for its strong antagonist leads and the twins Mickey and Lou sort of keep it together. I wished their backstories were a little more fleshed out but they are two crazy ladies that are just enjoying the end of the world and running a gang of ruthless Highwaymen and killing people. After Joseph Seed, it is hard to follow his presence and while Ubisoft tried to portray these two as badass as possible, their characters are not deep out enough to make the player care about them.
The lands of Hope County have shifted, lakes have been formed and mountains have sprung up due to the damage of the nuclear fallout. This means that the overall map is quite different. Sure, you will come across some memorable landmarks but it is hard to nail down just where you are standing in contrast to the location in Far Cry 5.
It is the age of the "Superbloom" meaning most of the land and its landmarks are now covered in gorgeous vegetation. The animals have also mutated slightly with large horns, the bears have rock-hard skin and the bison are as deadly as ever with massive stone-like skulls that can bash you into next week.
The same can be said for the locations as they are now very different but still recognizable to a certain degree. The Highwaymen have taken over landmarks across the map and turned them into their own base all in an attempt to gather the rarest of resource in the lands, Ethanol. There is no currency in the game, instead, you take over these outposts, scavenge them, race to supply drops from the sky, and collect everything you find in order to trade, craft weapons and items and upgrade Prosperity.
This new economy gives the player guidance but it is not as scarce as I hoped it to be. I never once did not have enough circuit boards to craft an elite weapon nor did I have to go and grind Ethanol to level up a specific department. The resources are plenty which is a good and bad thing. In one way, it lets you focus on the world, gameplay and story but in another way, it completely throws the survival aspect out of the window.
With that being said, the things you can craft are fantastic. Weapons in Far Cry New Dawn all have this great "post-apocalyptic" look and feel to them and most of them all bare a makeshift and homemade appearance. The saw launcher shoots out, well, saw blades that ricochet off enemies and cause havoc and the overall weapons in the game all have this clunky and unpolished feeling to them. Well, most of them anyway. Some of them are extremely over the top and don't fit into the game's theme and setting at all. I mean why would you have a unicorn flamethrower during the end of the world? It makes no sense.
These weapons then add to the combat which is as familiar as ever. Jumping, sliding, taking down enemies from above, and of course, the satisfying stealth is all present and accounted for. With the new ranking system, you need to have higher-ranked skills unlocked to silently take down the level 3 and elite enemies but it makes no difference once you have a double-barrel shotgun equipped as you can just pump led into them and they die pretty quickly.
I think that is my biggest issue with the game and how it changes difficulty so drastically. At first, you struggle and take on every challenge as thoughtfully as possible but soon after you have unlocked decent weapons, be it in-game or through the Ubisoft Club, things get way too easy and the scavenging for ammo, pumping bullets into enemies and dying really fast all get thrown out the window. I would have loved to see the game keep up the challenge throughout the experience and not just for the first act.
Even the ranked enemies get way too easy and taking on the "scavenged" outposts at the highest rank just needs a bow, disabling all the alarms from the distance and you are set. My co-op partner and I farmed the same outpost four to five times in 15 minutes to unlock the full Knight Set from For Honor without any challenge at all.
Character progression all depends on the level and state of Prosperity and unlocking and upgrading specialists gives you access to new skills, weapons and throwables. Again, you can just farm an outpost on the highest level and unlock everything mid-game. With that being said, the progression system is a great feature that keeps you going and striving to save all the specialists and unlock all the perks.
One of the game's best features is no doubt the new Expeditions mode. Throughout the game, as you upgrade this portion of Prosperity, you will gain access to more Expeditions. These missions take place outside of Hope County and take you to great-looking smaller maps across the US. The nuclear fallout hit them too so the look and feel tie in well with the overall game.
It is here where you and a partner, need to head into the map, avoid or kill everything, find a backpack and get to the extraction point all while surviving the onslaught of enemies. When you pick up the bag, a GPS alarm triggers after 30 seconds. All the enemies start to swarm you as you try to make your way to the extraction point and survive while the chopper arrives.
These missions where a great break from the usual Hope County exploration and all offered a unique place to visit, all with unique approaches and combat opportunities. My favourite one is still the H.M.S MacCoubrey that had me enter a giant ship, get the bag and drive off the edge in a passenger motorcycle. It was fun and rewarding as all the loot you get from these missions are used to craft the highest gear.
Everything then changes in Far Cry New Dawn after a certain point where you obtain special powers called Eden's Touch. With these powers, you can double jump, take down enemies while healing yourself and even go into a rage and bash everyone back to the 1800s. These expedition missions and outposts all then become a little easier but your approach changes which is variety great too.
An average session with Far Cry New Dawn took me to familiar places to hunt treasure, kill a bunch of Highwaymen, recruit new fangs for hire, gather some ethanol and upgrade my home base. It is a cycle that works and you never feel bored as the world has enough to keep you busy and distracted. Throw in co-op with friendly fire and you have yourself one of the best ways to spend your time.
The game lasts for around 25 hours while doing the main quests. There is also the opportunity to go back and complete all the outposts on the highest difficulty for that set gear and complete the great Expeditions on various difficulties too. These are fun and rewarding so there is a reason to go back.
Unfortunately, Far Cry New Dawn's biggest downfall is in its rank system. It could have been something really great that offered new challenges and layers of gameplay but in the end, it fell completely flat as it is simply not fleshed out enough to warrant any of your attention. The first time you see an elite guard is intimidating but as soon as you realize you can cheese him with a mounted gun or dynamite things take a turn for the worst.
But the magic is in the game's world and activities and we all know Ubisoft are masters at open world titles. This is no exception. It offers an explosive take on the apocalypse with memorable characters, great action and a decent amount of unique activities to keep the experience fresh. Far Cry New Dawn is a Far Cry game, and that should be enough to for anyone.
This review was based on a review code sent to us by Ubisoft
Available On: PS4, Xbox One, PC | Reviewed On: PS4 Pro | Release Date: 15 February 2019 | Price: R599
Marco Cocomello
Managing Editor at Glitched Africa
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text:
honey mushroom
Friedrich Bergius
Wellcome Library, London
(1884–1949). In 1931 German chemist Friedrich Bergius was a corecipient, with Carl Bosch, of the Nobel Prize for Chemistry. Bergius and Bosch were instrumental in developing a method to convert coal dust and hydrogen directly into gasoline and lubricating oils without isolating intermediate products.
Bergius was born on October 11, 1884, in Goldschmieden, near Breslau, Germany (now Wroclaw, Poland). He was educated at the Universities of Breslau, Leipzig, and Berlin and at technical schools in Karlsruhe and Hannover. He described his research in The Use of High Pressure in Chemical Actions (1913). These studies led to his work on converting coal into liquid hydrocarbons. He also researched the conversion of wood into sugar and of sugar into other food products. This work helped to provide Germany with food during World War II.
After the war, Bergius left Germany and moved to Argentina. He died on March 30, 1949, in Buenos Aires.


During a recent weekend, I had the opportunity to hear nearly 90 tuba players audition for my orchestra, the National Symphony Orchestra. Having served on several audition committees over the past seven years, I have heard more than 25 separate auditions for NSO vacancies. Most vacancies have occurred in the string section, but in the recent past I have heard auditions for flute, clarinet, English horn, French horn and timpani. While each of these instruments has a distinct nature and varying inherent challenges the audition procedure remains the same. This most recent round of tuba auditions prompted me to make some observations on this subject with an eye toward the preparation and training necessary to assemble a successful audition.
It has been said that traffic police can pull over just about any vehicle at any time for a citation, because the odds are very good that most drivers are breaking motor vehicle laws. Be it excessive speed, an incomplete stop, faulty equipment, failure to wear seat-belts, tailgating, cell phone use, littering, uninspected/uninsured/unregistered vehicles—the list goes on and on. As an audition committee member, I frequently feel like a traffic cop with the ability to write up a ticket for an audition candidate at almost any point in any given excerpt; playing out of tune, incorrect tempo, wrong style, playing too fast/slow, playing too inaccurately and so on. The thing is, as a committee member, I am not seeking to penalize the musical scofflaws of the world. I am looking for the good drivers! I want you to play well. Believe me, committees are pulling for you not against you.
When so many candidates "break the law" over the course of an audition, it does become disheartening for the committee. Remember every member of that committee was an audition winner at one point, so they are uniquely qualified for their post. The recent tuba audition announcement garnered 225 resumes from applicants around the globe. Approximately 80 were encouraged to attend, and all of those candidates had either graduated with a degree from a college/conservatory of music or were currently enrolled in studies. Many were also currently employed as full time professional musicians.
The three "must haves" of auditioning are what I call the three t's: (time, tuning and tone). As an audition committee member, I wonder why so many candidates are unable to master these seemingly simple musical tasks. Am I wrong to assume that four years (or more) of training at a good college/conservatory of music is adequate enough time to master these three basic objectives?
Too often, I encounter young brass musicians in training who are technically gifted yet musically bereft. Most need a bit more exposure, experience and time spent living life to become seasoned players. The typical music education in the U.S. places an emphasis on developing technical mastery of ones' instrument. Nurturing the artistic and creative elements of a student are usually secondary to technique. That said, should an audition committee expect even more from this music school graduates' grasp of the three t's? Do the audition candidates realize their faults? Was it a lack of talent, aptitude or poor instruction that failed them in their quest? Do they believe that they are capable of winning a major audition even with their faults? In my experience, a player with the "total package" usually wins and a good committee is seldom fooled by anything short of that.
Perhaps even more disturbing for me as I listen to auditions is when a candidate simply fails to make music. I would say that making music involves the transference of thoughts, ideas, moods, styles and/or emotions from the performer to the listener. If the performer has failed to transmit these things successfully to the listener there is "transmission failure." Better call AAMCO. It could be that the candidates are so bogged down in the muck of technical mastery that they have forgotten to simply put forth their musical ideas in a coherent manner. In other words, they cannot see the forest for the trees.
I have never sat on an audition committee that found a candidate to be excessively musical. Audiences and audition committees want to hear what is inside a performer's musical soul. Much of the selected orchestral audition repertoire for brass instruments lies within a low to medium degree of difficulty. Yes, the rest is usually very difficult. The easy stuff is often selected so that a committee can lob a musical softball at a candidate and see how far they can hit it. Can they take something simple, shape it nicely and present it as a polished product? Too often, candidates will make the wrong musical decisions or, worse yet, make NO decision. The successful ones will take the softball into the upper deck because they made their musical decisions with conviction and communicate them effectively to the committee.
My main objective in offering all of these rantings and criticisms is to point out that all of the basics of music making must be firmly established in order to put together a winning audition. Playing in time, in tune with a good tone should all be givens. The ability to convey your musical personality is also very important. When you step onto that concert stage, be it behind a screen or not, remember that you are there to make music for an audience, seen or unseen. Granted, the audition experience is an artificial one; your audience is actually a committee assembled to critique you, and your music consists of brief snippets of pieces intended to fit into a much larger symphonic puzzle. Even with these limitations, the audition candidate must make music somehow. Believe me, it can be done. I have done it a few times and have been lucky enough to hear many others do it as well.


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