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What's an elegant way to unify X,Y with (1,2), (1,-2), (-1,2), (-1,-2), (2,1), (2,-1) , (-2,1), (-2,-1)?
Doing it this way seems error prone and tedious:
foo(1,2). foo(1,-2). foo(-1,-2). ... ... ...
And this way seems too expensive:
foo(X,Y) :- L = [1...
Started by Absolute0 on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
Foo(X, Y) :- foo(-....
Foo(2, 1).
Foo(1, 2).
Foo(X,Y):- foo0(X,Y); foo0(Y,X).
A further development on what was commented:
generate_pairs_foo(X,Y) :- L = [1,-1,2,-2], member(X,L]), member(Y,[2,-2]).
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What's the most most ruby-like way of converting from an Array like [:one, 1, :two, 2] to a Hash like {:one => 1, :two => 2} ?
Started by Bob Aman on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Here's how I would do it:
Hash[*array]
Hash[*[:one, 1, :two, 2]] #=> {:one => 1, :two => 2}.
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If x^2+y^2=1, is x+y=1 ?
(1) xy=0
(2) y=0 [Reveal] Spoiler: OA E
Last edited by Bunuel on Thu May 17, 2012 1:19 am, edited 1 time in total. Edited the OA
Started by Bunuel on
, 15 posts
by 8 people.
Answer Snippets (Read the full thread at gmatclub):
(1) xy=0 --> either x=0 or y=0 :
if x=0 , then x^2+y^2=y^2=1 and y=1 or ....
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Ask your Facebook Friends
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In mathematics the identity (1 + sqrt(2))^2 = 3 + 2*sqrt(2) holds true. But in floating point (IEEE 754, using single precision i.e. 32 bits) calculations it's not the case, as sqrt(2) doesn't have an exact representation in binary.
So does using a approximated...
Started by Naximus on
, 18 posts
by 18 people.
Answer Snippets (Read the full thread at stackoverflow):
Generally I use [(1 + sqrt(2))^2] - [3 + 2*sqrt(2)] < 0.00001 to test equality
In double....
When comparing floating point purposes.
Multiplying by 2, etc.
To that same number on both sides: Adding 1 vs.
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(1=1) simple right? But something has been bothering me, can that be proven?
Take the old one apple plus another apple equals two apples. This seems simple enough but it is quite frankly the second i realized math was not my strong point. Its my greatest...
Started by Samuel Smith on
, 15 posts
by 5 people.
Answer Snippets (Read the full thread at mymathforum):
That one prolly deserves its own rant.
On you guys..
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Portal 2: $6.79
2 pack: $11.89
Portal 1 + 2: $8.49
http://store.steampowered.com/app/620/
Started by 4LC4PON3 on
, 20 posts
by 19 people.
Answer Snippets (Read the full thread at hardforum):
- The latter half of 1 just felt! I really wanted the Portal 1 & 2
-GONE/GIFTED- Keep meaning to try this, good dealGreat deal on some great titles....
I enjoyed 2 a lot more than 1.
Is cheap enough for me to buy now...
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How do I detect programmatically in which ring (-1, 0, 1, 2, 3) I am running?
Started by Andrew J. Brehm on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
(SEH, Windows, kernel mode)
bool ring_lower_0 = false....
The easiest way is, to just run the (x86) command and catch the corresponding error .
Unless you're a device driver, you'll always be running in Ring 3 (for systems that have "rings", per se) .
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1) long to wide question:
I have a dataset with 3 columns: person, event, frequency. If the frequency is zero, the row is not in the table. Is there a simple way using basic R functions or libraries to convert this table to wide format, with one row per...
Started by Patrick McCann on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Data[filteredRows,]
For the first part of your question, you can either use... .
Re: the first part, you can filter out the rows with Frequency == 0 by
filteredRows <- data$Frequency != 0 ## restrict ourselves to looking at the data where Frequency != 0 .
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Hi,
I got the following regex that almost does the work but does not exclude zero ...How to do that?
^(\d|\d{1,9}|1\d{1,9}|20\d{8}|213\d{7}|2146\d{6}|21473\d{5}|214747\d{4}|2147482\d{3}|21474835\d{2}|214748364[0-7])$
Also can anybody explain a bit how...
Started by Tanmoy on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
In essence....
The regex you pasted works:
The most important thing to remember is that 2**31 - 1 = 2147483647 (a number;= 2**31 - 1
In addition to the other answers, your regex doesn't work (besides allowing 0 with 10 digits).
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Prove (1/ 1+ cos x) + (1/ 1 - cos x) = 2/ sin² x? using identities
tan x = sin x / cos x
OR
sin²x + cos²x = 1
Started by Biladi on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at yahoo):
1 / (1 + cos x)] + [1 / (1 - cos x)] = 2 / sin² x
the LCD of the two fractions on the left hand + cos x) / (1 - cos² x)]
= [(1 - cos x) + (1 + cos x)] / (1 - cos² x)
= 2 / (1....
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