Related
I have a problen in scilab
How can I plot functions containing if and < like
function y = alpha(t)
if (t < 227.8) then
y = 0.75;
elseif (t < 300) then
y = 2.8 - 0.009 .* t;
else
y = 0.1;
end
endfunction
and
function [r]=minus_alpha(t)
r = 1 - alpha(t)
endfunction
When I use
x = linspace(0,300)
plot(x, alpha(x))
I got the error message
WARNING: Transposing row vector X to get compatible dimensions
plot2d: falsche Größe für Eingangsargument: inkompatible Größen.
Error 999 : in plot2d called by plot
Sorry for german mix. Thank you.
You can avoid explicit loop and be more efficient using the followin code
function y = alpha(t)
y=0.1*ones(t);
y(t<227.8)=0.75;
i=t>=227.8&t<300;
y(i)=2.8 - 0.009 .* t(i);
endfunction
It is really sad to see a great majority of Scilab community is not aware of vectorized operations. You can change your function to:
function y = alpha(t)
y = 0.1;
if t < 227.8 then
y = 0.75;
elseif t < 300 then
y = 2.8 - 0.009 * t;
end
y = 1 - y;
endfunction
and then use feval to broadcast the function over the sequence:
x = linspace(0, 300);
plot2d(x, feval(x, alpha));
which results:
rule of thumb if you are using for loop you need to revise your code and if someone is offering you a code where there is unnecessary for loop you shouldn't probably use it.
All the proposed answers are overcomplicated considering that the function alpha in the original demand is piecewise-affine. In Scilab in can be coded that way:
x = linspace(0,400,1000);
plot(x,linear_interpn(x,[227.8 300],[0.75 0.1]))
i.e. you just have to know the nodes coordinates (here abscissae) and value of the function at nodes. The function linear_interpn does also multilinear interpolation, it is worth knowing it guys...
If you check the output of your alpha(x), you will see that it is just a scalar (not a vector). I guess you wanted something like this, so it's necessary to iterate through t to compute each value of y based on the value of t:
clc;
clear;
function y = alpha(t)
for i=1:size(t,"*")
if t(i) < 227.8 then
y(i) = 0.75;
elseif t(i) < 300 then
y(i) = 2.8 - 0.009 * t(i);
else
y(i) = 0.1;
end
end
endfunction
x = linspace(0,300);
plot2d(x,alpha(x));
If you find the answer useful, please do not forget to accept it, so others will see that your problem is solved.
Before your answers (thank you) my workaround was a combination of indicator functions composed with floor and exp( -t^2):
function y = alpha(t)
y = floor(exp(-(t .* (t-T1)) / (T1*T1))) * 0.75
+ floor(exp(-((t-T2) .* (t- T1) / (2000)))) .* (2.8-0.009 .* t)
+ floor(exp(-((t-T2) .* (t-1000) / (200000))))*0.1
endfunction
I'me trying to understand some Action Script 3 features in order to port some code.
Code 1
How does the "++" influences the index part mean? If idx_val=0 then what xvaluer index will be modified?
xvaluer(++idx_val) = "zero";
Code 2
Then I have this: what is the meaning of this part of code?
What is being assigned to bUnicode in the last 3 lines?
(can you explain me the "<<"s and ">>"s)
bUnicode = new Array(2);
i = (i + 1);
i = (i + 1);
bUnicode[0] = aData[(i + 1)] << 2 | aData[(i + 1)] >> 4;
i = (i + 1);
bUnicode[1] = aData[i] << 4 | aData[(i + 1)] >> 2;
Code 3
I haven't the faintest idea of what is happening here.
What is "as" ? What is the "?" ?
bL = c > BASELENGTH ? (INVALID) : (s_bReverseLPad[c]);
Code 4
What is "&&" ?
if ((i + 1) < aData.length && s_bReverseUPad(aData((i + 1))) != INVALID)
Code 5
What is "as" ? What is the "?" ?
n2 = c < 0 ? (c + 256) as (c)
bOut.push(n1 >> 2 & 63)
bOut.push((n1 << 4 | n2 >> 4) & 63)//What is the single "&" ?
bOut.push(n2 << 2 & 63)
Finally, what are the differences between "||" and "|", and between "=" and "==" ?
Code 1: ++i is almost the same thing as i++ or i += 1; The only real difference is that it's modified before it is evaluated. Read more here.
Code 2: << and >> are bitwise shifts, they literally shift bits by one place. You really need to understand Binary before you can mess about with these operators. I would recommend reading this tutorial all the way through.
Code 3: This one is called Ternary Operator and it's actually quite simple. It's a one line if / else statement. bL = c > BASELENGTH ? (INVALID) : (s_bReverseLPad[c]); is equivalent to:
if(c > BASELENGTH) {
bL = INVALID;
} else {
bL = s_bReverseLPad[c];
}
Read more about it here.
Code 4: "The conditional-AND operator (&&) performs a logical-AND of its bool operands, but only evaluates its second operand if necessary." There is also the conditional-OR operator to keep in mind (||).
As an example of the AND operator here is some code:
if(car.fuel && car.wheels) car.move();
Read more about it here.
Code 5: From AS3 Reference: as "Evaluates whether an expression specified by the first operand is a member of the data type specified by the second operand." So basically you're casting one type to another, but only if it's possible otherwise you will get null.
& is Bitwise AND operator and | is Bitwise OR operator, again refer to this article.
= and == are two different operators. The former(=) is called Basic Assignment meaning it is used when you do any kind of assignment like: i = 3;. The later(==) is called Equal to and it is used to check if a value is equal to something else. if(i == 3) // DO STUFF;. Pretty straight forward.
The only part that doesn't make sense to me is the single question mark. Ternary Operator needs to have both ? and :. Does this code actually run for you? Perhaps a bit more context would help. What type is c?
n2 = c < 0 ? (c + 256) as (c)
Lets assume core1 and core2 try writing their variables a and b to same memory location.
How can UB be explained here?
We dont know if a or b is written to that memory location(as a last action).
We dont even know what is written there (a garbage)
Even the target memory address can be miscalculated(segfault?).
Some logical gates make wrong currents and CPU disables itself
CPU's frequency information becomes corrupt and goes high overclock(and break itself)
Can I assume only the first option is valid for all vendors of CPU( and GPU)?
I just converted below code into a parallel GPU code and it seems to be working fine.
Generic code:
for (j=0; j<YRES/CELL; j++) // this is parallelized
for (i=0; i<XRES/CELL; i++) // this is parallelized
{
r = fire_r[j][i];
g = fire_g[j][i];
b = fire_b[j][i];
if (r || g || b)
for (y=-CELL; y<2*CELL; y++)
for (x=-CELL; x<2*CELL; x++)
addpixel(i*CELL+x, j*CELL+y, r, g, b, fire_alpha[y+CELL][x+CELL]);
//addpixel accesses neighbour cells' informations and writes on them
//and makes UB
r *= 8;
g *= 8;
b *= 8;
for (y=-1; y<2; y++)
for (x=-1; x<2; x++)
if ((x || y) && i+x>=0 && j+y>=0 && i+x<XRES/CELL && j+y<YRES/CELL)
{
r += fire_r[j+y][i+x];
g += fire_g[j+y][i+x];
b += fire_b[j+y][i+x];
}
r /= 16;
g /= 16;
b /= 16;
fire_r[j][i] = r>4 ? r-4 : 0; // UB
fire_g[j][i] = g>4 ? g-4 : 0; // UB
fire_b[j][i] = b>4 ? b-4 : 0;
}
Opencl:
" int i=get_global_id(0); int j=get_global_id(1);"
" int VIDXRES="+std::to_string(kkVIDXRES)+";"
" int VIDYRES="+std::to_string(kkVIDYRES)+";"
" int XRES="+std::to_string(kkXRES)+";"
" int CELL="+std::to_string(kkCELL)+";"
" int YRES="+std::to_string(kkYRES)+";"
" int x=0,y=0,r=0,g=0,b=0,nx=0,ny=0;"
" r = fire_r[j*(XRES/CELL)+i];"
" g = fire_g[j*(XRES/CELL)+i];"
" b = fire_b[j*(XRES/CELL)+i];"
" int counterx=0;"
" if (r || g || b)"
" for (y=-CELL; y<2*CELL; y++){"
" for (x=-CELL; x<2*CELL; x++){"
" addpixel(i*CELL+x, j*CELL+y, r, g, b, fire_alpha[(y+CELL)*(3*CELL)+(x+CELL)],vid,vido);"
" }}"
" r *= 8;"
" g *= 8;"
" b *= 8;"
" for (y=-1; y<2; y++){"
" for (x=-1; x<2; x++){"
" if ((x || y) && i+x>=0 && j+y>=0 && i+x<XRES/CELL && j+y<YRES/CELL)"
" {"
" r += fire_r[(j+y)*(XRES/CELL)+(i+x)];"
" g += fire_g[(j+y)*(XRES/CELL)+(i+x)];"
" b += fire_b[(j+y)*(XRES/CELL)+(i+x)];"
" }}}"
" r /= 16;"
" g /= 16;"
" b /= 16;"
" fire_r[j*(XRES/CELL)+i] = (r>4 ? r-4 : 0);"
" fire_g[j*(XRES/CELL)+i] = (g>4 ? g-4 : 0);"
" fire_b[j*(XRES/CELL)+i] = (b>4 ? b-4 : 0);"
Here is picture of some rare artifacts of a 2D NDrangeKernel 's local boundary UB. Can these kill my GPU?
On xf86 and xf86_64 architectures it means We dont know if a or b is written to that memory location(as a last action), because load/store operations of 32 (for both) or 64 bit (xf86_64 only) memory aligned datatypes are atomic.
On other architectures usually We dont even know what is written there (a garbage) is a valid answer - for sure on RISC architectures, I currently don't know on GPU's.
Note that The fact the code works doesn't imply that it is correct and in the 99% of the times it's the source of sentences like "there's a compiler bug, the code was working until the previous version" or "the code works on the development machine. The server selected for production is broken" :)
EDIT:
On NVidia GPUs we have weakly-ordered memory model. In the description on the Cuda C Programming guide it's not explicitly stated that store operations are atomic. The write operations come from the same thread, so it does not mean that load/store operations are atomic.
For the code above, IMHO the first option is the only possible one. Basically, if you assume that you have enough threads/processors to execute all the loops in parallel, the inner nested loops (the x and y ones) will have undetermined values.
For example, if we consider only the
r += fire_r[j+y][i+x];
section, the value at fire_r[j+y][i+x] can be the original one just as well as the result of another instance of the same loop being finished in another thread.
Given a 2D circle with 2 angles in the range -PI -> PI around a coordinate, what is the value of the smallest angle between them?
Taking into account that the difference between PI and -PI is not 2 PI but zero.
An Example:
Imagine a circle, with 2 lines coming out from the center, there are 2 angles between those lines, the angle they make on the inside aka the smaller angle, and the angle they make on the outside, aka the bigger angle.
Both angles when added up make a full circle. Given that each angle can fit within a certain range, what is the smaller angles value, taking into account the rollover
This gives a signed angle for any angles:
a = targetA - sourceA
a = (a + 180) % 360 - 180
Beware in many languages the modulo operation returns a value with the same sign as the dividend (like C, C++, C#, JavaScript, full list here). This requires a custom mod function like so:
mod = (a, n) -> a - floor(a/n) * n
Or so:
mod = (a, n) -> (a % n + n) % n
If angles are within [-180, 180] this also works:
a = targetA - sourceA
a += (a>180) ? -360 : (a<-180) ? 360 : 0
In a more verbose way:
a = targetA - sourceA
a -= 360 if a > 180
a += 360 if a < -180
x is the target angle. y is the source or starting angle:
atan2(sin(x-y), cos(x-y))
It returns the signed delta angle. Note that depending on your API the order of the parameters for the atan2() function might be different.
If your two angles are x and y, then one of the angles between them is abs(x - y). The other angle is (2 * PI) - abs(x - y). So the value of the smallest of the 2 angles is:
min((2 * PI) - abs(x - y), abs(x - y))
This gives you the absolute value of the angle, and it assumes the inputs are normalized (ie: within the range [0, 2π)).
If you want to preserve the sign (ie: direction) of the angle and also accept angles outside the range [0, 2π) you can generalize the above. Here's Python code for the generalized version:
PI = math.pi
TAU = 2*PI
def smallestSignedAngleBetween(x, y):
a = (x - y) % TAU
b = (y - x) % TAU
return -a if a < b else b
Note that the % operator does not behave the same in all languages, particularly when negative values are involved, so if porting some sign adjustments may be necessary.
An efficient code in C++ that works for any angle and in both: radians and degrees is:
inline double getAbsoluteDiff2Angles(const double x, const double y, const double c)
{
// c can be PI (for radians) or 180.0 (for degrees);
return c - fabs(fmod(fabs(x - y), 2*c) - c);
}
See it working here:
https://www.desmos.com/calculator/sbgxyfchjr
For signed angle:
return fmod(fabs(x - y) + c, 2*c) - c;
In some other programming languages where mod of negative numbers are positive, the inner abs can be eliminated.
I rise to the challenge of providing the signed answer:
def f(x,y):
import math
return min(y-x, y-x+2*math.pi, y-x-2*math.pi, key=abs)
For UnityEngine users, the easy way is just to use Mathf.DeltaAngle.
Arithmetical (as opposed to algorithmic) solution:
angle = Pi - abs(abs(a1 - a2) - Pi);
I absolutely love Peter B's answer above, but if you need a dead simple approach that produces the same results, here it is:
function absAngle(a) {
// this yields correct counter-clock-wise numbers, like 350deg for -370
return (360 + (a % 360)) % 360;
}
function angleDelta(a, b) {
// https://gamedev.stackexchange.com/a/4472
let delta = Math.abs(absAngle(a) - absAngle(b));
let sign = absAngle(a) > absAngle(b) || delta >= 180 ? -1 : 1;
return (180 - Math.abs(delta - 180)) * sign;
}
// sample output
for (let angle = -370; angle <= 370; angle+=20) {
let testAngle = 10;
console.log(testAngle, "->", angle, "=", angleDelta(testAngle, angle));
}
One thing to note is that I deliberately flipped the sign: counter-clockwise deltas are negative, and clockwise ones are positive
There is no need to compute trigonometric functions. The simple code in C language is:
#include <math.h>
#define PIV2 M_PI+M_PI
#define C360 360.0000000000000000000
double difangrad(double x, double y)
{
double arg;
arg = fmod(y-x, PIV2);
if (arg < 0 ) arg = arg + PIV2;
if (arg > M_PI) arg = arg - PIV2;
return (-arg);
}
double difangdeg(double x, double y)
{
double arg;
arg = fmod(y-x, C360);
if (arg < 0 ) arg = arg + C360;
if (arg > 180) arg = arg - C360;
return (-arg);
}
let dif = a - b , in radians
dif = difangrad(a,b);
let dif = a - b , in degrees
dif = difangdeg(a,b);
difangdeg(180.000000 , -180.000000) = 0.000000
difangdeg(-180.000000 , 180.000000) = -0.000000
difangdeg(359.000000 , 1.000000) = -2.000000
difangdeg(1.000000 , 359.000000) = 2.000000
No sin, no cos, no tan,.... only geometry!!!!
A simple method, which I use in C++ is:
double deltaOrientation = angle1 - angle2;
double delta = remainder(deltaOrientation, 2*M_PI);
I was reading Joel's book where he was suggesting as interview question:
Write a program to reverse the "ON" bits in a given byte.
I only can think of a solution using C.
Asking here so you can show me how to do in a Non C way (if possible)
I claim trick question. :) Reversing all bits means a flip-flop, but only the bits that are on clearly means:
return 0;
What specifically does that question mean?
Good question. If reversing the "ON" bits means reversing only the bits that are "ON", then you will always get 0, no matter what the input is. If it means reversing all the bits, i.e. changing all 1s to 0s and all 0s to 1s, which is how I initially read it, then that's just a bitwise NOT, or complement. C-based languages have a complement operator, ~, that does this. For example:
unsigned char b = 102; /* 0x66, 01100110 */
unsigned char reverse = ~b; /* 0x99, 10011001 */
What specifically does that question mean?
Does reverse mean setting 1's to 0's and vice versa?
Or does it mean 00001100 --> 00110000 where you reverse their order in the byte? Or perhaps just reversing the part that is from the first 1 to the last 1? ie. 00110101 --> 00101011?
Assuming it means reversing the bit order in the whole byte, here's an x86 assembler version:
; al is input register
; bl is output register
xor bl, bl ; clear output
; first bit
rcl al, 1 ; rotate al through carry
rcr bl, 1 ; rotate carry into bl
; duplicate above 2-line statements 7 more times for the other bits
not the most optimal solution, a table lookup is faster.
Reversing the order of bits in C#:
byte ReverseByte(byte b)
{
byte r = 0;
for(int i=0; i<8; i++)
{
int mask = 1 << i;
int bit = (b & mask) >> i;
int reversedMask = bit << (7 - i);
r |= (byte)reversedMask;
}
return r;
}
I'm sure there are more clever ways of doing it but in that precise case, the interview question is meant to determine if you know bitwise operations so I guess this solution would work.
In an interview, the interviewer usually wants to know how you find a solution, what are you problem solving skills, if it's clean or if it's a hack. So don't come up with too much of a clever solution because that will probably mean you found it somewhere on the Internet beforehand. Don't try to fake that you don't know it neither and that you just come up with the answer because you are a genius, this is will be even worst if she figures out since you are basically lying.
If you're talking about switching 1's to 0's and 0's to 1's, using Ruby:
n = 0b11001100
~n
If you mean reverse the order:
n = 0b11001100
eval("0b" + n.to_s(2).reverse)
If you mean counting the on bits, as mentioned by another user:
n = 123
count = 0
0.upto(8) { |i| count = count + n[i] }
♥ Ruby
I'm probably misremembering, but I
thought that Joel's question was about
counting the "on" bits rather than
reversing them.
Here you go:
#include <stdio.h>
int countBits(unsigned char byte);
int main(){
FILE* out = fopen( "bitcount.c" ,"w");
int i;
fprintf(out, "#include <stdio.h>\n#include <stdlib.h>\n#include <time.h>\n\n");
fprintf(out, "int bitcount[256] = {");
for(i=0;i<256;i++){
fprintf(out, "%i", countBits((unsigned char)i));
if( i < 255 ) fprintf(out, ", ");
}
fprintf(out, "};\n\n");
fprintf(out, "int main(){\n");
fprintf(out, "srand ( time(NULL) );\n");
fprintf(out, "\tint num = rand() %% 256;\n");
fprintf(out, "\tprintf(\"The byte %%i has %%i bits set to ON.\\n\", num, bitcount[num]);\n");
fprintf(out, "\treturn 0;\n");
fprintf(out, "}\n");
fclose(out);
return 0;
}
int countBits(unsigned char byte){
unsigned char mask = 1;
int count = 0;
while(mask){
if( mask&byte ) count++;
mask <<= 1;
}
return count;
}
The classic Bit Hacks page has several (really very clever) ways to do this, but it's all in C. Any language derived from C syntax (notably Java) will likely have similar methods. I'm sure we'll get some Haskell versions in this thread ;)
byte ReverseByte(byte b)
{
return b ^ 0xff;
}
That works if ^ is XOR in your language, but not if it's AND, which it often is.
And here's a version directly cut and pasted from OpenJDK, which is interesting because it involves no loop. On the other hand, unlike the Scheme version I posted, this version only works for 32-bit and 64-bit numbers. :-)
32-bit version:
public static int reverse(int i) {
// HD, Figure 7-1
i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
i = (i << 24) | ((i & 0xff00) << 8) |
((i >>> 8) & 0xff00) | (i >>> 24);
return i;
}
64-bit version:
public static long reverse(long i) {
// HD, Figure 7-1
i = (i & 0x5555555555555555L) << 1 | (i >>> 1) & 0x5555555555555555L;
i = (i & 0x3333333333333333L) << 2 | (i >>> 2) & 0x3333333333333333L;
i = (i & 0x0f0f0f0f0f0f0f0fL) << 4 | (i >>> 4) & 0x0f0f0f0f0f0f0f0fL;
i = (i & 0x00ff00ff00ff00ffL) << 8 | (i >>> 8) & 0x00ff00ff00ff00ffL;
i = (i << 48) | ((i & 0xffff0000L) << 16) |
((i >>> 16) & 0xffff0000L) | (i >>> 48);
return i;
}
pseudo code..
while (Read())
Write(0);
I'm probably misremembering, but I thought that Joel's question was about counting the "on" bits rather than reversing them.
Here's the obligatory Haskell soln for complementing the bits, it uses the library function, complement:
import Data.Bits
import Data.Int
i = 123::Int
i32 = 123::Int32
i64 = 123::Int64
var2 = 123::Integer
test1 = sho i
test2 = sho i32
test3 = sho i64
test4 = sho var2 -- Exception
sho i = putStrLn $ showBits i ++ "\n" ++ (showBits $complement i)
showBits v = concatMap f (showBits2 v) where
f False = "0"
f True = "1"
showBits2 v = map (testBit v) [0..(bitSize v - 1)]
If the question means to flip all the bits, and you aren't allowed to use C-like operators such as XOR and NOT, then this will work:
bFlipped = -1 - bInput;
I'd modify palmsey's second example, eliminating a bug and eliminating the eval:
n = 0b11001100
n.to_s(2).rjust(8, '0').reverse.to_i(2)
The rjust is important if the number to be bitwise-reversed is a fixed-length bit field -- without it, the reverse of 0b00101010 would be 0b10101 rather than the correct 0b01010100. (Obviously, the 8 should be replaced with the length in question.) I just got tripped up by this one.
Asking here so you can show me how to do in a Non C way (if possible)
Say you have the number 10101010. To change 1s to 0s (and vice versa) you just use XOR:
10101010
^11111111
--------
01010101
Doing it by hand is about as "Non C" as you'll get.
However from the wording of the question it really sounds like it's only turning off "ON" bits... In which case the answer is zero (as has already been mentioned) (unless of course the question is actually asking to swap the order of the bits).
Since the question asked for a non-C way, here's a Scheme implementation, cheerfully plagiarised from SLIB:
(define (bit-reverse k n)
(do ((m (if (negative? n) (lognot n) n) (arithmetic-shift m -1))
(k (+ -1 k) (+ -1 k))
(rvs 0 (logior (arithmetic-shift rvs 1) (logand 1 m))))
((negative? k) (if (negative? n) (lognot rvs) rvs))))
(define (reverse-bit-field n start end)
(define width (- end start))
(let ((mask (lognot (ash -1 width))))
(define zn (logand mask (arithmetic-shift n (- start))))
(logior (arithmetic-shift (bit-reverse width zn) start)
(logand (lognot (ash mask start)) n))))
Rewritten as C (for people unfamiliar with Scheme), it'd look something like this (with the understanding that in Scheme, numbers can be arbitrarily big):
int
bit_reverse(int k, int n)
{
int m = n < 0 ? ~n : n;
int rvs = 0;
while (--k >= 0) {
rvs = (rvs << 1) | (m & 1);
m >>= 1;
}
return n < 0 ? ~rvs : rvs;
}
int
reverse_bit_field(int n, int start, int end)
{
int width = end - start;
int mask = ~(-1 << width);
int zn = mask & (n >> start);
return (bit_reverse(width, zn) << start) | (~(mask << start) & n);
}
Reversing the bits.
For example we have a number represented by 01101011 . Now if we reverse the bits then this number will become 11010110. Now to achieve this you should first know how to do swap two bits in a number.
Swapping two bits in a number:-
XOR both the bits with one and see if results are different. If they are not then both the bits are same otherwise XOR both the bits with XOR and save it in its original number;
Now for reversing the number
FOR I less than Numberofbits/2
swap(Number,I,NumberOfBits-1-I);