Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Approximation Algorithms Part One: Constant factor approximations
1. Advanced Algorithms – COMS31900
Approximation algorithms part one
Constant factor approximations
Benjamin Sach
2. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
(this second part is the definition of NP-hard)
3. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
(this second part is the definition of NP-hard)
‘yes/no’ problems
4. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
(this second part is the definition of NP-hard)
5. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
(this second part is the definition of NP-hard)
we can solve B
using A
6. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
(this second part is the definition of NP-hard)
7. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
(this second part is the definition of NP-hard)
8. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
They are the ‘hardest’ problems in NP
(this second part is the definition of NP-hard)
9. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
They are the ‘hardest’ problems in NP
(this second part is the definition of NP-hard)
that you can’t solve them in polynomial time (i.e. that P = NP)
Most computer scientists (I’ve met) believe
10. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
They are the ‘hardest’ problems in NP
(this second part is the definition of NP-hard)
A polynomial time algorithm for an
NP-complete problem is worth
(a lot more than) a million dollars
11. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
They are the ‘hardest’ problems in NP
(this second part is the definition of NP-hard)
12. NP-completeness recap
A problem A is NP-complete if
NP is the class of decision problems we can
check the answer to in polynomial time
A is in NP
Every B in NP has a polynomial time reduction to A
If we could solve A quickly we could solve every problem in NP quickly
They are the ‘hardest’ problems in NP
So if a problem is NP-complete, we give up right?
(this second part is the definition of NP-hard)
25. Bin packing
1
1
Problem pack all items into the fewest possible bins
4/8
4/8
7/8
2/8
2/8
3/8
The BINPACKING problem is known to be NP-hard
26. Bin packing
1
1
Problem pack all items into the fewest possible bins
4/8
4/8
7/8
2/8
2/8
3/8
The BINPACKING problem is known to be NP-hard
and the decision version. . . “Can you pack the items into at most k bins?”
is NP-complete
27. Bin packing
1
1
Problem pack all items into the fewest possible bins
4/8
4/8
7/8
2/8
2/8
3/8
The BINPACKING problem is known to be NP-hard
and the decision version. . . “Can you pack the items into at most k bins?”
is NP-complete
In the decision version,
k is part of the input
28. Bin packing
1
1
Problem pack all items into the fewest possible bins
4/8
4/8
7/8
2/8
2/8
3/8
The BINPACKING problem is known to be NP-hard
29. Bin packing
1
1
Problem pack all items into the fewest possible bins
4/8
4/8
7/8
2/8
2/8
3/8
The BINPACKING problem is known to be NP-hard
but fortunately we can approximate
50. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
and s be the number of non-empty bins (using Next fit)
51. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
52. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i)
53. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I
54. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I
the sum of the
item weights
55. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
the sum of the
item weights
56. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
the sum of the
item weights
the optimal number of bins
57. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
58. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
therefore s 2 · Opt
59. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
therefore s 2 · Opt in other words the Next Fit is never worse than twice the optimal
60. Next fit
1
4/8
2/8
4/8
7/8
2/8
3/8
Next fit runs in O(n) time but how good is it?
Let fill(i) be the sum of item sizes in bin i
Observe that fill(2i − 1) + fill(2i) > 1 (for 1 2i s)
and s be the number of non-empty bins (using Next fit)
so s/2 <
1 2i s
fill(2i − 1) + fill(2i) I Opt
therefore s 2 · Opt
61. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
within an α factor of Opt
62. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
within an α factor of Opt
63. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
• If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
64. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
• If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
• If P is a minimisation problem (like BINPACKING), Opt s α · Opt
65. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
• If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
• If P is a minimisation problem (like BINPACKING), Opt s α · Opt
We have seen a 2-approximation algorithm for BINPACKING
66. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
• If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
• If P is a minimisation problem (like BINPACKING), Opt s α · Opt
We have seen a 2-approximation algorithm for BINPACKING
the number of bins used, s is always between Opt and 2 · Opt
67. Approximation Algorithms
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
• If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
• If P is a minimisation problem (like BINPACKING), Opt s α · Opt
We have seen a 2-approximation algorithm for BINPACKING
the number of bins used, s is always between Opt and 2 · Opt
In the examples we consider, α will be a constant but it could depend on n (the input size)
70. First fit decreasing (FFD)
1
1
Step 1: Sort the items into non-increasing order
4/8 4/8
7/8
3/82/82/8
71. First fit decreasing (FFD)
1
1
4/8 4/8
7/8
3/8 2/8 2/8
Step 1: Sort the items into non-increasing order
72. First fit decreasing (FFD)
1
1
4/8 4/8
7/8
3/8 2/8 2/8
Step 2: Put each item in the first (left-most) bin it fits in
73. First fit decreasing (FFD)
1
1
4/8 4/8 3/8 2/8 2/8
Step 2: Put each item in the first (left-most) bin it fits in
7/8
74. First fit decreasing (FFD)
1
1
4/8 3/8 2/8 2/8
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
75. First fit decreasing (FFD)
1
1
3/8 2/8 2/8
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
76. First fit decreasing (FFD)
1
1
2/8 2/8
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
3/8
77. First fit decreasing (FFD)
1
1
2/8
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
3/8
2/8
78. First fit decreasing (FFD)
1
1
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
3/8
2/8
2/8
79. First fit decreasing (FFD)
1
1
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
3/8
2/8
2/8
this will be important
for the proof
80. First fit decreasing (FFD)
1
1
Step 2: Put each item in the first (left-most) bin it fits in
7/8
4/8
4/8
3/8
2/8
2/8
82. First fit decreasing (FFD)
1
1
7/8
4/8
4/8
3/8
2/8
2/8
FFD runs in O(n2) time but how good is it?
83. First fit decreasing (FFD)
1
7/8
4/8
4/8
3/8
2/8
2/8
FFD runs in O(n2) time but how good is it?
84. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
ss
85. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
j
ss
86. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
j
ss
87. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
j
ss
88. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
j
ss
89. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
j
ss
90. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
because we packed big things first and each thing was
packed in the lowest numbered bin
j
ss
91. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
because we packed big things first and each thing was
packed in the lowest numbered bin
j
ss
92. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
j
ss
93. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
each of these items has to be in a different bin (even in Opt)
j
ss
94. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
each of these items has to be in a different bin (even in Opt)
So Opt uses at least 2s
3 bins
j
ss
95. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 1: Bin j contains an item of size > 1/2
Every bin j j contains an item of size > 1/2
each of these items has to be in a different bin (even in Opt)
So Opt uses at least 2s
3 bins
or. . . s
3Opt
2
j
ss
96. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
j
ss
97. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
when FFD packed the first item into bin j,
j
ss
98. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
when FFD packed the first item into bin j,
j
ss
99. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
when FFD packed the first item into bin j,
j
ss
100. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
101. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
102. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
103. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
2+ 2+ 2+ 2+
104. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
2+ 2+ 2+ 2+
(we only use a new bin when the item won’t fit in any previous bin)
105. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
2+ 2+ 2+ 2+
106. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
2+ 2+ 2+ 2+
107. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
when FFD packed the first item into bin j,
j
1. all bins j, (j + 1), . . . , (s − 2), (s − 1) were empty
ss
2. and all unpacked items had size 1/2
(because we pack in non-increasing order)
2+ 2+ 2+ 2+ 1+
108. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
j
ss
2+ 2+ 2+ 2+ 1+
109. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
j
ss
2+ 2+ 2+ 2+ 1+
110. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
111. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
112. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
otherwise we would have packed them there
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
113. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
114. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
115. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
recall I is the total weight of all items
j
ss
2+ 2+ 2+ 2+ 1+
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
116. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
recall I is the total weight of all items
pairing these with these
considerconsider
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
117. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
recall I is the total weight of all items
pairing these with these
considerconsider
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
this + this
> 1
118. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
recall I is the total weight of all items
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
119. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1}
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
120. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1} 2s/3 − 1
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
121. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
This gives a total of 2(s − j) + 1 items, none of which fits into bins 1, 2, 3, . . . , (j − 1)
so I > min{j − 1, 2(s − j) + 1} 2s/3 − 1
by plugging in j = 2s/3
j
ss
so Bins j, (j + 1), . . . , (s − 2), (s − 1) each contain at least two items
and bin s contains at least one item
122. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
As 2s/3 − 1 < I
j
ss
123. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
As 2s/3 − 1 < I and I Opt
j
ss
124. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
As 2s/3 − 1 < I
we have that 2s/3 − 1 < Opt
and I Opt
j
ss
125. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
As 2s/3 − 1 < I
we have that 2s/3 − 1 < Opt
and I Opt
. . . but both sides are integers. . .
so 2s/3 Opt
finally . . . 2s/3 2s/3 Opt
or s (3/2)Opt
j
ss
126. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
j
ss
127. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
Case 2: Bin j contains only items of size 1/2
Case 1: Bin j contains an item of size > 1/2
in both cases. . . s
3Opt
2
j
ss
128. First fit decreasing (FFD)
Consider bin j = 2s
3 (s is the number of bins FFD uses on this input)
So FFD is a 3/2-approximation algorithm for BINPACKING
Case 2: Bin j contains only items of size 1/2
Case 1: Bin j contains an item of size > 1/2
in both cases. . . s
3Opt
2
j
ss
129. Approximation Algorithms Summary
An algorithm A is an α-approximation algorithm for problem P if,
◦ A runs in polynomial time
◦ A always outputs a solution with value s
Here P is an optimisation problem with optimal solution of value Opt
If P is a maximisation problem,
Opt
α s Opt
within an α factor of Opt
If P is a minimisation problem (like BINPACKING), Opt s α · Opt
We have seen Next Fit which is a 2-approximation algorithm for BINPACKING
which runs in O(n) time
and First Fit Decreasing which is a 3/2-approximation algorithm for BINPACKING
which runs in O(n2) time
Bin Packing is NP-hard so solving it exactly in polynomial time would prove that P = NP