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Date: Fri, 27 Mar 2015 06:24:01 -0700
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Grill or toast 4 slices of Italian bread til crispy. Arrange on plates or a small platter the arugula,2 thin sliced fennels, pears 2 thin sliced and 200 g. of gorgonzola crumbles. To serve drizzle over a little balsamic white or red and evoo then hit it with salt and coarse cracked black pepper. this will serve 4 as a starter or side dish.
--
This must be my interpretation of Kade's recipe since I wrote it in my notes when I saved her recipe
o 1/2 cup walnut or pecan halves, toasted
o about 4 cups arugula or torn salad greens, trimmed washed & dried
o 1/2 cup balsamic or red wine vinaigrette
o 2 to 3 pears, cored & sliced Don't Peel PEARS
o Radish slices
o cuke peeled, seeded and cut in half moons
o peapods sliced on diagonal
o small red onion sliced thin
o 1/4 lb gorgonzola, crumbled This message was sent to [email protected]


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Long story short, I stepped on a nail Monday night while feeding the woodstove. So, yesterday I spent the day on the couch reading cookbooks. I have a copy of Professional Cooking. Since it is baking season, I thought I'd check out Chapter 24. In this chapter, baker's percentages are explained. Wondering if anyone has converted "home" recipes to baker's percentages. I have all the ingredients to make my grandma's brown sugar shortbread cookies. I thought I'd weigh all the ingredients and make her recipe according to baker's percentages and another batch the conventional way and see which I like better. I'd love to have a discussion on baker's percentages and how to use those for home baking. (Baking is a much more interesting pursuit than painting--I


Many algorithms require to compute (-1)^n (both integer), usually as a factor in a series. That is, a factor that is -1 for odd n and 1 for even n. In a C or C++ environment, one often sees:

#include<iostream>
#include<cmath>
int main(){
   int n = 13;
   std::cout << std::pow(-1, n) << std::endl;
}
What is better or the usual convention? (or something else),

std::pow(-1, n)
std::pow(-1, n%2)
(n%2?-1:1)
(1-2*(n%2))  // (gives incorrect value for negative n)
EDIT: In addition, user @SeverinPappadeux proposed another alternative based on array lookups. My version of it is:

const int res[] {-1, 1, -1}; // three elements are needed for negative modulo results
const int* const m1pow = res + 1;
..
m1pow[n%2]
c++ c algorithm cmath
shareimprove this question
edited 2 days ago

asked Mar 17 at 22:14

alfC
1,4831225
44
pow is a floating-point function so I'd avoid that. n%2 ? -1 : 1 seems the most straightforward.   Matt McNabb Mar 17 at 22:17
7
Another variant would be (n & 1) ? -1 : 1. Some people find it less readable, but a comment next to it clears that up and bit-masking is fast.   pjs Mar 17 at 22:22
4
(n&1)? -1: 1 should be faster -- it's equivalent in this case but not in general, so it may not be automatically optimized.   Ben Voigt Mar 17 at 22:23
3
@BenVoigt Since n%2 is immediately converted to bool here, I would expect identical codegen.   T.C. Mar 17 at 22:23
4
@T.C.: Probably. But note that 1 - 2*(n&1) is correct, while (1-2*(n%2)) from the question can give -1, 1, or 3   Ben Voigt Mar 17 at 22:24
show 8 more comments
5 Answers
activeoldestvotes
up vote
25
down vote
You can use (n & 1) instead of n % 2 and << 1 instead of * 2 if you want to be super-pedantic, er I mean optimized.
So the fastest way to compute in an 8086 processor is:

1 - ((n & 1) << 1)

I just want to clarify where this answer is coming from. The original poster alfC did an excellent job of posting a lot of different ways to compute (-1)^n some being faster than others.
Nowadays with processors being as fast as they are and optimizing compilers being as good as they are we usually value readability over the slight (even negligible) improvements from shaving a few CPU cycles from an operation.
There was a time when one pass compilers ruled the earth and MUL operations were new and decadent; in those days a power of 2 operation was an invitation for gratuitous optimization.

shareimprove this answer
edited Mar 18 at 1:23

answered Mar 17 at 22:23

Eli Algranti
3,94311122
8
n<<1 is defined as n * 2 by the C standard (for non-negative n). , so this is about readability rather than optimization.   Matt McNabb Mar 17 at 22:27
5
Are you going to show the benchmarks that show that this the fastest way?   juanchopanza Mar 17 at 22:28
3
A good optimizing compiler might not compile the branch in ((n&1)?-1:1), but if you can do it yourself, why not?   Lee Daniel Crocker Mar 17 at 22:37
2
@juanchopanza if I had an 8086 processor and an 8086 era C compiler I would but there is no need just look at the number of cycles it takes to do a multiplication vs shift and integer remainder vs logical and. It is much trickier with current compilers and modern 1 cycle multiplication processors.   Eli Algranti Mar 17 at 23:07
8
I wonder how many people have an 8086 processor and an 8086 era C compiler. It seems strange to advise using a bit shift in lieu of an integer multiplication by 2 without a strong caveat.   juanchopanza Mar 18 at 6:33
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up vote
23
down vote
Usually you don't actually calculate (-1)^n, instead you track the current sign (as a number being either -1 or 1) and flip it every operation (sign = -sign), do this as you handle your n in order and you will get the same result.

EDIT: Note that part of the reason I recommend this is because there is rarely actually semantic value is the representation (-1)^n it is merely a convenient method of flipping the sign between iterations.

shareimprove this answer
edited Mar 18 at 16:33

answered Mar 17 at 22:17

Guvante
11.6k1745
8
There's no need for tracking anything. The result is a very simple function of n.   juanchopanza Mar 17 at 22:27

Yes, you see all sort of things, including this. That is in par the origin of the question.   alfC Mar 17 at 22:33

@alfC: I will try and revise, part of my point is that the actual definition is more akin to "flip the sign every other member" but it is convenient to represent that as (-1)^n, as that is mathematical notation and correct. Thus this methodology matches the original intention, rather than being a workaround like such tricks usually are.   Guvante Mar 17 at 23:23
7
Why multiply by -1 if you can just flip the sign: sign=-sign;?   Ruslan Mar 18 at 4:49
6
+1 This answer raise an excellent point! There is "rarely semantic value" in (-1)^n, that is perfectly true and the most important observation to make :)   Ant Mar 18 at 11:56
add a comment
up vote
3
down vote
Many algorithms require to compute (-1)^n (both integer), usually as a factor in a series. That is, a factor that is -1 for odd n and 1 for even n.
Consider evaluating the series as a function of -x instead.

shareimprove this answer
answered Mar 18 at 20:22

Ian Ollmann
49316
add a comment
up vote
2
down vote
What about

(1 - (n%2)) - (n%2)
n%2 most likely will be computed only once

UPDATE

Actually, simplest and most correct way would be using table

const int res[] {-1, 1, -1};

return res[n%2 + 1];
shareimprove this answer
edited Mar 18 at 20:37

answered Mar 18 at 1:32

Severin Pappadeux
1,06215
1
For n = -1 it gives 3 (just like 1-(1-2*(n%2)))   alfC Mar 18 at 4:42

A modulo is still (much) slower than a mask   Flavien Volken Mar 18 at 14:01
1
@Alnitak: The conversion is only valid for non-negative values of n (e.g. unsigned type).   R.. Mar 18 at 18:05
3
Neither the question nor OP's code int n preclude n from having a negative value - thus making this solution unusable. Even if not practicable that n should even have a negative value, a compiler will likely not conclude that and thus limit optimizations. Suggest simply preface with unsigned n;   chux Mar 18 at 20:36
1
@chux Right, it should be n%2u, which should produce exactly the same code as n&1.   David Schwartz Mar 18 at 20:45
show 6 more comments
up vote
1
down vote
First of all, the fastest isOdd test I do know (in an inline method)

/**
* Return true if the value is odd
* @value the value to check
*/
inline bool isOdd(int value)
{
return (value & 1);
}
Then make use of this test to return -1 if odd, 1 otherwise (which is the actual output of (-1)^N )

/**
* Return the computation of (-1)^N
* @n the N factor
*/
inline int minusOneToN(int n)
{
return isOdd(n)?-1:1;
}


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<td align="left" valign="middle" class="mobile-block" width="33%" style="font-family:Georgia; font-size:14px; padding:0 20px; background-color:#333333; color:#b8b8b8; text-align:left; line-height:1.2em; font-size:1px;">Grill or toast 4 slices of Italian bread til crispy. Arrange on plates or a small platter the arugula,2 thin sliced fennels, pears 2 thin sliced and 200 g. of gorgonzola crumbles. To serve drizzle over a little balsamic white or red and evoo then hit it with salt and coarse cracked black pepper. this will serve 4 as a starter or side dish.<br />
  --<br />
  This must be my interpretation of Kade's recipe since I wrote it in my notes when I saved her recipe<br />
  o<strong>1/2</strong>cup walnut or pecan halves, toasted<br />
  oabout<strong>4</strong>cups arugula or torn salad greens, trimmed washed &amp; dried<br />
  o<strong>1/2</strong>cup balsamic or red wine vinaigrette<br />
  o<strong>2 to 3</strong>pears, cored &amp; sliced Don't Peel PEARS<br />
  oRadish slices<br />
  ocuke peeled, seeded and cut in half moons<br />
  opeapods sliced on diagonal<br />
  osmall red onion sliced thin<br />
  o<strong>1/4</strong>lb gorgonzola, crumbled</td>
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<td align="center" style='text-align: right;background-color:#333333;'>Long story short, I stepped on a nail Monday night while feeding the woodstove. So, yesterday I spent the day on the couch reading cookbooks. I have a copy of Professional Cooking. Since it is baking season, I thought I'd check out Chapter 24. In this chapter, baker's percentages are explained. Wondering if anyone has converted "home" recipes to baker's percentages.

I have all the ingredients to make my grandma's brown sugar shortbread cookies. I thought I'd weigh all the ingredients and make her recipe according to baker's percentages and another batch the conventional way and see which I like better. I'd love to have a discussion on baker's percentages and how to use those for home baking.

(Baking is a much more interesting pursuit than painting--I</td>
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