I am currently working on project using and arduino, a gyro, an accelerometer, and a Bluetooth chip to try to model some data. I am currently trying to gather data, package it up and send it to a phone via Bluetooth. The issue is the Bluetooth chip I am using is a low energy one and so it can only send messages of 20 bytes at a time. I am trying to get past this issue by storing the data collected for a certain amount of time then send it all in 20 byte bursts. I am currently testing this method without sending the data and just printing the data to the serial monitor. This is where my issue is arising, when printing the data in real time everything works but when I try to store it in an array I get this:
593,575,567,0,0,0
592,575,567,0,0,0
592,575,567,0,0,0
592,575,567,0,0,0
592,575,567,0,0,0
593,575,567,0,0,0
586,576,568,0,0,0
0,0,0
0,0
0,0
,0,0,0
0,0,0
As you can see it seems to just break. If anyone could help me out it would be great!
Here is the relevant code chunk
for(int i = 0; i < loopVal; i++)
{
yawGyroValDouble = 0;
pitchGyroValDouble = 0;
rollGyroValDouble = 0;
totalClicksY = 0;
angleY = 0;
totalClicksP = 0;
angleP = 0;
totalClicksR = 0;
angleR = 0;
xRe = 0;
yRe = 0;
zRe = 0;
s = "";
int starttime = millis(); // get start time
int endtime = starttime; // init end time
while ((endtime - starttime) < time)
{
getGyroValues(); // This will update rollGyroVal, pitchGyroVal, and yawGyroVal with new values
yawGyroValDouble =yawGyroVal;
if(abs(yawGyroValDouble) > abs(gyroNoiseThresh)){ // ignore noise
totalClicksY+=yawGyroValDouble; // update runsum
}
pitchGyroValDouble =pitchGyroVal;
if(abs(yawGyroValDouble) > abs(gyroNoiseThresh)){ // ignore noise
totalClicksP+=pitchGyroValDouble; // update runsum
}
rollGyroValDouble =rollGyroVal;
if(abs(yawGyroValDouble) > abs(gyroNoiseThresh)){ // ignore noise
totalClicksR+=rollGyroValDouble; // update runsum
}
xRe = analogRead(pinX);
yRe = analogRead(pinY);
zRe = analogRead(pinZ);
delay (gyroDelayTime);
endtime = millis();
}
angleY = totalClicksY / clicksPerDegCCW;
angleP = totalClicksP / clicksPerDegCCW;
angleR = totalClicksR / clicksPerDegCCW;
String yawSend = String(angleY);
String pitchSend = String(angleP);
String rollSend = String(angleR);
String xSend = String(xRe);
String ySend = String(yRe);
String zSend = String(zRe);
//s = "Accel - X: " + xSend + " Y: " + ySend + " Z: " + zSend + "\n" + "Gyro - Yaw: " + yawSend + " Pitch: " + pitchSend + " Roll: " + rollSend;
s = "" + xSend + "," + ySend + "," + zSend + "," + yawSend + "," + pitchSend + "," + rollSend;
Serial.println(s);
res[i] = s;
}
You didn't show where totalClicksY, totalClicksP, totalClicksR, and clicksPerDegCCW are declared, but I'm betting they are declared as integer types (int or long). If so, the result of your maths:
angleY = totalClicksY / clicksPerDegCCW;
angleP = totalClicksP / clicksPerDegCCW;
angleR = totalClicksR / clicksPerDegCCW;
will be integers. And if the results of those divisions are less than 1, they will be truncated to 0.
Try declaring totalClicksY, totalClicksP, totalClicksR and clicksPerDegCCW as double. That, or cast them when you do the math, like this:
angleY = (double)totalClicksY / (double)clicksPerDegCCW;
angleP = (double)totalClicksP / (double)clicksPerDegCCW;
angleR = (double)totalClicksR / (double)clicksPerDegCCW;
(I'm also assuming that angleY, angleP, and angleR are also declared as doubles - if not they definitely should be).
Related
There is one game on the Unreal Engine 2 (pirate MMORPG with anti-cheat). I'm using reshade library to hook some functions including dinput/window input handling.
With the SendInput function i can send input but only in the active input field. F1-F12 or 1-9 keys are bind to the panel of skills. I tried to send these keys but nothing happens.
int SendKey(const wchar_t *text)
{
INPUT *keystroke;
UINT i, character_count, keystrokes_to_send, keystrokes_sent;
assert(text != NULL);
//Fill in the array of keystrokes to send.
character_count = wcslen(text);
keystrokes_to_send = character_count * 2;
keystroke = new INPUT[keystrokes_to_send];
for (i = 0; i < character_count; ++i)
{
keystroke[i * 2].type = INPUT_KEYBOARD;
keystroke[i * 2].ki.wVk = 0;
keystroke[i * 2].ki.wScan = text[i];
keystroke[i * 2].ki.dwFlags = KEYEVENTF_UNICODE;
keystroke[i * 2].ki.time = 0;
keystroke[i * 2].ki.dwExtraInfo = GetMessageExtraInfo();
keystroke[i * 2 + 1].type = INPUT_KEYBOARD;
keystroke[i * 2 + 1].ki.wVk = 0;
keystroke[i * 2 + 1].ki.wScan = text[i];
keystroke[i * 2 + 1].ki.dwFlags = KEYEVENTF_UNICODE | KEYEVENTF_KEYUP;
keystroke[i * 2 + 1].ki.time = 0;
keystroke[i * 2 + 1].ki.dwExtraInfo = GetMessageExtraInfo();
}
//Send the keystrokes.
keystrokes_sent = SendInput((UINT)keystrokes_to_send, keystroke, sizeof(*keystroke));
delete[] keystroke;
return 0;
}
Or I can somehow use dinput to send keys?
I know that anti-cheat uses kbdHook (From Kernel-Mode) but how can he keep track of where the keys were sent in the game? (if the active input field works for me)?
PS: If I'm missing something please correct me, thanks!
After some image processing using ImgaeJ macro, I have a ‘Results’ tab which contains 2 columns A and B. Lets’s say I have 50 rows of data.
Now I want to subtract the value of last row from all other 49 rows above in column B.
After this, I want to write all the values in “.csv” file (column A, B and C with 49 values each).
Below is the part of the code. I think the only problem is fetching the values from arrays that the script can write to the csv file.
Array.getStatistics command only exports the mean, std values for a given column. I'm interested in fetching all 49 values.
directory = getDirectory("Choose a Directory");
resultFilename = directory + Dialog.getString() + ".csv";
A = newArray(nResults() - 1);
B = newArray(nResults() - 1);
D = getResult("B", nResults() - 1);
for (i = 0; i < nResults() - 2; i++) {
A[i] = getResult("A", i);
B[i] = getResult("B", i);
C[i] = A[i] - D;
}
Any idea about what is the command to get the values of A[i], B[i] and C[i]?
Looking forward for some help here.
Thank you.
One solution is to write to the file as you do the calculations. I have modified your example (untested) to show how this works.
directory = getDirectory("Choose a Directory");
resultFilename = directory + Dialog.getString() + ".csv";
f = File.open(resultFilename);
A = newArray(nResults() - 1);
B = newArray(nResults() - 1);
// no C array is made so:
C = newArray(nResults() - 1);
D = getResult("B", nResults() - 1);
for (i = 0; i < nResults() - 2; i++) {
A[i] = getResult("A", i);
B[i] = getResult("B", i);
C[i] = A[i] - D;
// should the line above should be C[i] = B[i] - D;
print(f, d2s(A[i],6) + " \t" + d2s(B[i],6) + " \t" + d2s(C[i],6));
}
File.close(f);
Note that you don't need to make the arrays at all and can just write to the file (again this is untested):
directory = getDirectory("Choose a Directory");
resultFilename = directory + Dialog.getString() + ".csv";
f = File.open(resultFilename);
D = getResult("B", nResults() - 1);
for (i = 0; i < nResults() - 2; i++) {
ai = getResult("A", i);
bi = getResult("B", i);
ci = ai - D;
// should the line above should be ci = bi - D;
print(f, d2s(ai,6) + " \t" + d2s(bi,6) + " \t" + d2s(ci,6));
}
File.close(f);
I have used " \t" (tab character) as a separator not comma.
I'm trying to add together two large numbers, stored as strings.
Here's what I have so far:
function addBigNums(a,b){
c = ""; // output
o = 0; // carryover
startLen = a.length-1;
for(i = startLen; i >= 0; i--) {
sum = parseInt(a[i], 10) + parseInt(b[i], 10) + o;
c = (sum % 10) + c;
o = sum >= 10;
}
if(o === true) c = "1" + c;
return c;
}
I'm running into two issues:
1 ) my carry is not always functioning properly, primarily when:
2 ) the numbers length differ.
Right now I think I would have to prepend 0's onto the shorter number in order to get this to function as expected.
Any better alternatives to this?
Simple, straightforward integer addition like you would do it manually:
a = "123456"; // input a
b = "123456"; // input b
c = ""; // target-string
o = 0; // overflow-bit
// traverse string from right to left
for(i = a.length - 1; i >= 0; i--) {
// do the calculation (with overflow bit)
sum = parseInt(a[i]) + parseInt(b[i]) + o;
// prepend resulting digit to target
c = (sum % 10) + c;
// set overflow bit for next round
o = sum >= 10;
}
// prepend another "1" if last overflow-bit is true
if(o == true) c = "1" + c;
If strings a and b are not equal length (but you stated that they are), you should prepend the shorter string with zeros before calculation.
Consider both numbers to be an array of digits. Add them up right-to-left handling overflow flag. Demo. Assuming your numbers are of the same length
function getNumber(len) {
return Array.apply(null, new Array(len)).map(function(){
return Math.floor(Math.random()*9);
}).join('');
}
var len = 600,
a = getNumber(len), //use your numbers here
b = getNumber(len),
flag = 0;
var c = [].reduceRight.call(a, function(acc, val, idx) {
val = +val + (+b.charAt(idx)) + flag;
flag = val / 10 | 0;
val %= 10;
return val + acc;
}, '');
c = (flag ? 1: '') + c;
console.log(a, b, c);
I want to change color from 0x008000 (green) to 0x0000FF (blue).
If I multiply 0x008000 * 256 = 0x800000 (Google search acts as a calculator).
I need to find the correct multiplier so the result would be 0x0000FF.
To answer people below - I am doing this in order to make a color transition on a rectangle in pixi.js.
From what I've gathered, RGB color code is divided into 3 parts - red, green and blue in a scale of 0-FF expressed in hex, or 0-255 in decimal. But how to multiply correctly to get desired result?
If you want linear change from one color to another, i recommend something like this:
int startColor = 0x008000;
int endColor = 0x0000FF;
int startRed = (startColor >> 16) & 0xFF;
int startGreen = (startColor >> 8) & 0xFF;
int startBlue = startColor & 0xFF;
int endRed, endGreen, endBlue; //same code
int steps = 24;
int[] result = new int[steps];
for(int i=0; i<steps; i++) {
int newRed = ( (steps - 1 - i)*startRed + i*endRed ) / (steps - 1);
int newGreen, newBlue; //same code
result[i] = newRed << 16 | newGreen << 8 | newBlue;
}
This is for JavaScript:
var startColor = 0x008000;
var endColor = 0x0000FF;
var startRed = (startColor >> 16) & 0xFF;
var startGreen = (startColor >> 8) & 0xFF;
var startBlue = startColor & 0xFF;
var endRed = (endColor >> 16) & 0xFF;
var endGreen = (endColor >> 8) & 0xFF;
var endBlue = endColor & 0xFF;
var steps = 24;
var result = [];
for (var i = 0; i < steps; i++) {
var newRed = ((steps - 1 - i) * startRed + i * endRed) / (steps - 1);
var newGreen = ((steps - 1 - i) * startGreen + i * endGreen) / (steps - 1);
var newBlue = ((steps - 1 - i) * startBlue + i * endBlue) / (steps - 1);
var comb = newRed << 16 | newGreen << 8 | newBlue;
console.log(i + " -> " + comb.toString(16));
result.push(comb);
}
console.log(result);
I'm looking for an algorithm that finds short tandem repeats in a genome sequence.
Basically, given a really long string which can only consist of the 4 characters 'ATCG', I need to find short repeats between 2-5 characters long that are next to each other.
ex:
TACATGAGATCATGATGATGATGATGGAGCTGTGAGATC
would give ATGATGATG or ATG repeated 3 times
The algorithm needs to scale up to a string of 1 million characters so I'm trying to get as close to linear runtime as possible.
My current algorithm:
Since the repeats can be 2-5 characters long, I check the string character by character and see if the Nth character is the same as the N+Xth character, with X being 2 through 5. With a counter for each X that counts sequential matches and resets at a mismatch, we know if there is a repeat when X = the counter. The subsequent repeats can then be checked manually.
You are looking at each character which gives you O(n), since you compare on each character the next (maximum) five characters this gives you a constant c:
var data = get_input();
var compare = { `A`, `T`, `G`, `A`, `T` } // or whatever
var MAX_LOOKAHEAD = compare.length
var n
var c
for(n = data_array.length; n < size; i++) { // Has runtime O(n)
for(c = 0; c < MAX_LOOKAHEAD; c++) { // Maximum O(c)
if( compare[c] != data[i+c] ) {
break;
} else {
report( "found match at position " + i )
}
}
}
It is easy to see that this runs O(n*c) times. Since c is very small it can be ignored - and I think one can not get rid of that constant - which results in a total runtime of O(n).
The good news:
You can speed this up with parallelization. E.g. you could split this up in k intervals and let multiple threads do the job for you by giving them appropriate start and end indices. This could give you a linear speedup.
If you do that make sure that you treat the intersections as special cases since you could miss a match if your intervals split a match in two.
E.g. n = 50000:
Partition for 4 threads: (n/10000) - 1 = 4. The 5th thread won't have a lot to do since it just handles the intersections which is why we don't need to consider its (in our case tiny) overhead.
1 10000 20000 40000 50000
|-------------------|-------------------|-------------------|-------------------|
| <- thread 1 -> | <- thread 2 -> | <- thread 3 -> | <- thread 4 -> |
|---| |---| |---|
|___________________|___________________|
|
thread 5
And this is how it could look like:
var data;
var compare = { `A`, `T`, `G`, `A`, `T` };
var MAX_LOOKAHEAD = compare.length;
thread_function(args[]) {
var from = args[0];
var to = args[1];
for(n = from ; n < to ; i++) {
for(c = 0; c < MAX_LOOKAHEAD; c++) {
if( compare[c] != data[i+c] ) {
break;
} else {
report( "found match at position " + i )
}
}
}
}
main() {
var data_size = 50000;
var thread_count = 4;
var interval_size = data_size / ( thread_count + 1) ;
var tid[]
// This loop starts the threads for us:
for( var i = 0; i < thread_count; i++ ) {
var args = { interval_size * i, (interval_size * i) + interval_size };
tid.add( create_thread( thread_function, args ) );
}
// And this handles the intersections:
for( var i = 1; i < thread_count - 1; i++ ) {
var args = { interval_size * i, (interval_size * i) + interval_size };
from = (interval_size * i) - compare.length + 1;
to = (interval_size * i) + compare.length - 1;
for(j = from; j < to ; j++) {
for(k = 0; k < MAX_LOOKAHEAD; k++) {
if( compare[k] != data[j+k] ) {
break;
} else {
report( "found match at position " + j )
}
}
}
}
wait_for_multiple_threads( tid );
}