Hi...
I have a sequence of jobs and its predecessors..
Input
Job_Name Predecessor
A NULL
B1 A
B2 A
B3 B1
C B3
C B2
So based on these i have to generate the level Number
What i mean is
Let A be level 1
for B1 to happen it should have done A
so B1 level is A+1 = 1+1 = 2 (12 Replies)
as you can see there is a delimiter after c8 "::". Awk sees the rest as fields because it doesn't recognize spaces and tabs as delimiters. So i am basically looking to isolate 20030003ba13f6cc. Can anyone help?
c8::20030003ba13f6cc disk connected configured unknown (2 Replies)
Hi,
Is there a way to create sequence numbers in unix
i have a set of batches(which contain records) and i want to assign a number to every batch. how can i do that? (1 Reply)
Hi All,
I have a requirement where in I have an input as follows:-
input=1-4,6,8-10,12-15
I need to explode this range into an output file as follows:-
1
2
3
4
6
8
9
10
12
13
14
15
My input may vary like 1,5-9,11-13,15-17....... (3 Replies)
Hi Guys,
I have a file with numbers in sequence.
The sequence have been broken somewhere.. I need to find out at which number the sequence has been broken...
For an example, consider this sequence, it needs to give me output as 4 (as 5 is missing) and 6(as 7 is missing)
Thanks for... (3 Replies)
how can i generate following sequence
for a given input 1,2,3,4,5
1->2
2->3
3->4
4->5
1->2,3
1,2->3
2->3,4
2,3->4
3->4,5
3,4->5
1->2,3,4
1,2->3,4
1,2,3->4 (4 Replies)
i want to generate a random number through a script, and even if anyone reads the script, they wont be able to figure out what the random number is. only the person who setup the script would know it.
something like this could work: random
the full thread is here:
... (13 Replies)
How do I generate line numbers in Vi?
I have this:
,'04-90020-039N','61423','2GDV00039-0002', SYSDATE);
,'04-90020-040D','61423','2GDV00046-0001', SYSDATE);
,'04-90020-041N','61423','2GDV00038-0002', SYSDATE);
,'04-90020-043D','61423','2GDV00047-0001', SYSDATE);... (3 Replies)
I am setting this thread to this bsd forum, though it may fit into bash. But as using bsd and the terminal, I would like to generate a random sequence of alphanumerical digits, such as I use to do so on linux by typing just
mcookiethis one gives me a pretty random password, but it does not on bsd... (0 Replies)
Discussion started by: 1in10
0 Replies
LEARN ABOUT DEBIAN
bup-margin
bup-margin(1) General Commands Manual bup-margin(1)NAME
bup-margin - figure out your deduplication safety margin
SYNOPSIS
bup margin [options...]
DESCRIPTION
bup margin iterates through all objects in your bup repository, calculating the largest number of prefix bits shared between any two
entries. This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.
For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45. That means a 46-bit
hash would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
its first 46 bits.
The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects. Since SHA-1 hashes have 160 bits,
that leaves 115 bits of margin. Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits
with far fewer objects.
If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see if
you're getting dangerously close to 160 bits.
OPTIONS --predict
Guess the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
from the guess. This is potentially useful for tuning an interpolation search algorithm.
--ignore-midx
don't use .midx files, use only .idx files. This is only really useful when used with --predict.
EXAMPLE
$ bup margin
Reading indexes: 100.00% (1612581/1612581), done.
40
40 matching prefix bits
1.94 bits per doubling
120 bits (61.86 doublings) remaining
4.19338e+18 times larger is possible
Everyone on earth could have 625878182 data sets
like yours, all in one repository, and we would
expect 1 object collision.
$ bup margin --predict
PackIdxList: using 1 index.
Reading indexes: 100.00% (1612581/1612581), done.
915 of 1612581 (0.057%)
SEE ALSO bup-midx(1), bup-save(1)BUP
Part of the bup(1) suite.
AUTHORS
Avery Pennarun <apenwarr@gmail.com>.
Bup unknown-bup-margin(1)