Contains a lot of different flag masks.
Note that this is not the conventional starting point for gene numbering: Section "E. Coli genome starting point".
As of 2021, last updated 2016, and python 3.5 appears to be mandatory or else:which apparently broke in 3.6: stackoverflow.com/questions/41343263/provide-classcell-example-for-python-3-6-metaclass and
RuntimeError: __class__ not set defining 'AbstractBaseUser' as <class 'django.contrib.auth.base_user.AbstractBaseUser'>. Was __classcell__ propagated to type.__new__?pyenv install fails on Ubuntu 20.10, so... fuck. Workarounds at:but am I in the mood considering that the ancient Django version would require an immediate port anyways? Repo is at Django 1.0, while newest is now already Django 3. The Rails one is broken for the same reason. Fuck 2.
Often just called collimated light due to the collimator being the main procedure to obtain it.
However, you move very far away from the source, e.g. the Sun, you also get essentially parallel light.
Exactly 131 commit apoptosis in the hermaphrodite.
www.wormatlas.org/celllineages.html contains the full lineage as some huge and impossible to view images. This image was taken directly from The embryonic cell lineage of the nematode Caenorhabditis elegans where it is split across many pages, it is a thing of beauty on the PDF.
www.wormatlas.org/celllistsulston.htm contains a non-hierarchical table with the cells and their names.
One of the most beautiful molecular machines known!
The first one with such complexity that was uncovered.
The ATP synthase complex is so large that Power, Sex, Suicide by Nick Lane (2006) page 123 contains a cryoEM image of several ATP synthases on small membrane vesicles, this is the paper: pubs.acs.org/doi/abs/10.1021/bi00437a031# under a fucking paywall.
Advanced quantum field theory lecture by Tobias Osborne (2017) Lecture 2 by 
Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16Using the TLB makes translation faster, because the initial translation takes one access per TLB level, which means 2 on a simple 32 bit scheme, but 3 or 4 on 64 bit architectures.
The TLB is usually implemented as an expensive type of RAM called content-addressable memory (CAM). CAM implements an associative map on hardware, that is, a structure that given a key (linear address), retrieves a value.
Mappings could also be implemented on RAM addresses, but CAM mappings may required much less entries than a RAM mapping.
linear physical
------ --------
00000 00001
00001 00010
00010 00011
FFFFF 00000For each process, the virtual address space looks like this:
------------------ 2^32 - 1
Stack (grows down)
v v v v v v v v v
------------------
(unmapped)
------------------ Maximum stack size.
(unmapped)
-------------------
mmap
-------------------
(unmapped)
-------------------
^^^^^^^^^^^^^^^^^^^
brk (grows up)
-------------------
BSS
-------------------
Data
-------------------
Text
-------------------
------------------- 0The kernel maintains a list of pages that belong to each process, and synchronizes that with the paging.
If the program accesses memory that does not belong to it, the kernel handles a page-fault, and decides what to do:
When an ELF file is loaded by the kernel to start a program with the
exec system call, the kernel automatically registers text, data, BSS and stack for the program.The
brk and mmap areas can be modified by request of the program through the brk and mmap system calls. But the kernel can also deny the program those areas if there is not enough memory.brk and mmap can be used to implement malloc, or the so called "heap".mmap is also used to load dynamically loaded libraries into the program's memory so that it can access and run it.Stack allocation: stackoverflow.com/questions/17671423/stack-allocation-for-process
Calculating exact addresses Things are complicated by:
- Address Space Layout Randomization.
- the fact that environment variables, CLI arguments, and some ELF header data take up initial stack space: unix.stackexchange.com/questions/145557/how-does-stack-allocation-work-in-linux/239323#239323
Why the text does not start at 0: stackoverflow.com/questions/14795164/why-do-linux-program-text-sections-start-at-0x0804800-and-stack-tops-start-at-0
x86 Paging Tutorial Single level paging scheme numerical translation example by Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16Suppose that the OS has setup the following page tables for process 1:and for process 2:
entry index entry address page address present
----------- ------------------ ------------ -------
0 CR3_1 + 0 * 4 0x00001 1
1 CR3_1 + 1 * 4 0x00000 1
2 CR3_1 + 2 * 4 0x00003 1
3 CR3_1 + 3 * 4 0
...
2^20-1 CR3_1 + 2^20-1 * 4 0x00005 1entry index entry address page address present
----------- ----------------- ------------ -------
0 CR3_2 + 0 * 4 0x0000A 1
1 CR3_2 + 1 * 4 0x12345 1
2 CR3_2 + 2 * 4 0
3 CR3_2 + 3 * 4 0x00003 1
...
2^20-1 CR3_2 + 2^20-1 * 4 0xFFFFF 1When process 1 tries to access a linear address, this is the physical addresses that will be actually accessed:
linear physical
--------- ---------
00000 001 00001 001
00000 002 00001 002
00000 003 00001 003
00000 FFF 00001 FFF
00001 000 00000 000
00001 001 00000 001
00001 FFF 00000 FFF
00002 000 00003 000
FFFFF 000 00005 000To switch to process 2, the OS simply sets
cr3 to CR3_2, and now the following translations would happen:linear physical
--------- ---------
00000 002 0000A 002
00000 003 0000A 003
00000 FFF 0000A FFF
00001 000 12345 000
00001 001 12345 001
00001 FFF 12345 FFF
00004 000 00003 000
FFFFF 000 FFFFF 000Step-by-step translation for process 1 of logical address
0x00000001 to physical address 0x00001001:- split the linear address into two parts:
| page (20 bits) | offset (12 bits) | - look into Page table 1 because
cr3points to it. - The hardware knows that this entry is located at RAM address
CR3 + 0x00000 * 4 = CR3:
*0x00000because the page part of the logical address is0x00000
*4because that is the fixed size in bytes of every page table entry - since it is present, the access is valid
- by the page table, the location of page number
0x00000is at0x00001 * 4K = 0x00001000. - to find the final physical address we just need to add the offset:
00001 000 + 00000 001 --------- 00001 001because00001is the physical address of the page looked up on the table and001is the offset.The offset is always simply added the physical address of the page. - the hardware then gets the memory at that physical location and puts it in a register.
Another example: for logical address
0x00001001:- the page part is
00001, and the offset part is001 - the hardware knows that its page table entry is located at RAM address:
CR3 + 1 * 4(1because of the page part), and that is where it will look for it - it finds the page address
0x00000there - so the final address is
0x00000 * 4k + 0x001 = 0x00000001
The problem with a single-level paging scheme is that it would take up too much RAM: 4G / 4K = 1M entries per process.
If each entry is 4 bytes long, that would make 4M per process, which is too much even for a desktop computer:
ps -A | wc -l says that I am running 244 processes right now, so that would take around 1GB of my RAM!The Quantum Story by Jim Baggott (2011) page 2 mentions how newton's support for the corpuscular theory of light led it to be held for a very long time, even when evidence of the wave theory of light was becoming overwhelming.
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