Any idea why the else if statment will be never executed ? The value of difference is constantly changing when the program runs.
double difference = abs(reale_x[0] - reale_x[1]);
if (0 <= difference < 45) {
timer_counter += 1;
if (timer_counter == 30) {
cout << "CLICK" << '\n';
}
}
else if (difference > 50) {
timer_counter = 0;
}
That is not how comparation works in c++.
What this code
if (0 <= difference < 45) {
does is it first compares if 0 is smaller or equal to difference. It is then "replaced" by a bool value either true or false. And then a bool value (so either 1 or 0) is compared to 45. And it will always be smaller than 45. What you have there is an always true statement.
So the way you would write this if statement is
if (difference >= 0 && difference < 45){
Note that because of your else if statement it will not execute if the difference is >44 and <51
if (0 <= difference < 45) will be executed as if ((0 <= difference) < 45), which will be either 0<45 or 1<45 and will always be true. That's why the else part is not getting executed.
in mathematics, we see and write 0 <= x < 45 or something like that to define the range of the variable x. But in order to tell the computer the same thing, you have to tell more clearly. Saying, to have to tell the compiler, that the value of x is greater than or equal to zero and at the same time, that value will be less than 45, and you can tell the compiler by this statement: difference >= && difference < 45 . the && is an 'AND' operator in most of the languages.
I am using deff to define piecewise functions in Scilab. I use if-else statements to define the function in different intervals. All of them seem to work except for this one.
This is my approach:
deff('a=f(p)', ["if(q >= -2 & q <= -1) then"; "a=1"; "elseif(q >= 1 & q <= 2) then"; "a=-1"; "else"; "a=0"; "end";])
The same approach worked on similar functions such as this.
eff('a=f(p)', ["if((p > 0 & p <= 2)) then"; "a=p/2"; "else"; "a=0"; "end";])
I have defined q as so:
q = -3:Ts:3;
where Ts is some floating point number. What am I doing wrong?
You cannot use your function as-is when the input is a vector. A possible workaround is to use feval:
a = feval(q,f)
Another (better) solution is to code f in a vectorized manner:
function a = f(q)
a = 1*(q >= -2 & q <= -1) - 1*(q >= 1 & q <= 2)
end
the 1* are not necessary unless you need to change the values.
I am making a program for nth Fibonacci number. I made the following program using recursion and memoization.
The main problem is that the value of n can go up to 10000 which means that the Fibonacci number of 10000 would be more than 2000 digit long.
With a little bit of googling, I found that i could use arrays and store every digit of the solution in an element of the array but I am still not able to figure out how to implement this approach with my program.
#include<iostream>
using namespace std;
long long int memo[101000];
long long int n;
long long int fib(long long int n)
{
if(n==1 || n==2)
return 1;
if(memo[n]!=0)
return memo[n];
return memo[n] = fib(n-1) + fib(n-2);
}
int main()
{
cin>>n;
long long int ans = fib(n);
cout<<ans;
}
How do I implement that approach or if there is another method that can be used to achieve such large values?
One thing that I think should be pointed out is there's other ways to implement fib that are much easier for something like C++ to compute
consider the following pseudo code
function fib (n) {
let a = 0, b = 1, _;
while (n > 0) {
_ = a;
a = b;
b = b + _;
n = n - 1;
}
return a;
}
This doesn't require memoisation and you don't have to be concerned about blowing up your stack with too many recursive calls. Recursion is a really powerful looping construct but it's one of those fubu things that's best left to langs like Lisp, Scheme, Kotlin, Lua (and a few others) that support it so elegantly.
That's not to say tail call elimination is impossible in C++, but unless you're doing something to optimise/compile for it explicitly, I'm doubtful that whatever compiler you're using would support it by default.
As for computing the exceptionally large numbers, you'll have to either get creative doing adding The Hard Way or rely upon an arbitrary precision arithmetic library like GMP. I'm sure there's other libs for this too.
Adding The Hard Way™
Remember how you used to add big numbers when you were a little tater tot, fresh off the aluminum foil?
5-year-old math
1259601512351095520986368
+ 50695640938240596831104
---------------------------
?
Well you gotta add each column, right to left. And when a column overflows into the double digits, remember to carry that 1 over to the next column.
... <-001
1259601512351095520986368
+ 50695640938240596831104
---------------------------
... <-472
The 10,000th fibonacci number is thousands of digits long, so there's no way that's going to fit in any integer C++ provides out of the box. So without relying upon a library, you could use a string or an array of single-digit numbers. To output the final number, you'll have to convert it to a string tho.
(woflram alpha: fibonacci 10000)
Doing it this way, you'll perform a couple million single-digit additions; it might take a while, but it should be a breeze for any modern computer to handle. Time to get to work !
Here's an example in of a Bignum module in JavaScript
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s, Number) .reverse ()
, toString: (b) =>
b .reverse () .join ("")
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
We can verify that the Wolfram Alpha answer above is correct
const { fromInt, toString, add } =
Bignum
const bigfib = (n = 0) =>
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n > 0) {
_ = a
a = b
b = add (b, _)
n = n - 1
}
return toString (a)
}
bigfib (10000)
// "336447 ... 366875"
Expand the program below to run it in your browser
const Bignum =
{ fromInt: (n = 0) =>
n < 10
? [ n ]
: [ n % 10, ...Bignum.fromInt (n / 10 >> 0) ]
, fromString: (s = "0") =>
Array.from (s) .reverse ()
, toString: (b) =>
b .reverse () .join ("")
, add: (b1, b2) =>
{
const len = Math.max (b1.length, b2.length)
let answer = []
let carry = 0
for (let i = 0; i < len; i = i + 1) {
const x = b1[i] || 0
const y = b2[i] || 0
const sum = x + y + carry
answer.push (sum % 10)
carry = sum / 10 >> 0
}
if (carry > 0) answer.push (carry)
return answer
}
}
const { fromInt, toString, add } =
Bignum
const bigfib = (n = 0) =>
{
let a = fromInt (0)
let b = fromInt (1)
let _
while (n > 0) {
_ = a
a = b
b = add (b, _)
n = n - 1
}
return toString (a)
}
console.log (bigfib (10000))
Try not to use recursion for a simple problem like fibonacci. And if you'll only use it once, don't use an array to store all results. An array of 2 elements containing the 2 previous fibonacci numbers will be enough. In each step, you then only have to sum up those 2 numbers. How can you save 2 consecutive fibonacci numbers? Well, you know that when you have 2 consecutive integers one is even and one is odd. So you can use that property to know where to get/place a fibonacci number: for fib(i), if i is even (i%2 is 0) place it in the first element of the array (index 0), else (i%2 is then 1) place it in the second element(index 1). Why can you just place it there? Well when you're calculating fib(i), the value that is on the place fib(i) should go is fib(i-2) (because (i-2)%2 is the same as i%2). But you won't need fib(i-2) any more: fib(i+1) only needs fib(i-1)(that's still in the array) and fib(i)(that just got inserted in the array).
So you could replace the recursion calls with a for loop like this:
int fibonacci(int n){
if( n <= 0){
return 0;
}
int previous[] = {0, 1}; // start with fib(0) and fib(1)
for(int i = 2; i <= n; ++i){
// modulo can be implemented with bit operations(much faster): i % 2 = i & 1
previous[i&1] += previous[(i-1)&1]; //shorter way to say: previous[i&1] = previous[i&1] + previous[(i-1)&1]
}
//Result is in previous[n&1]
return previous[n&1];
}
Recursion is actually discommanded while programming because of the time(function calls) and ressources(stack) it consumes. So each time you use recursion, try to replace it with a loop and a stack with simple pop/push operations if needed to save the "current position" (in c++ one can use a vector). In the case of the fibonacci, the stack isn't even needed but if you are iterating over a tree datastructure for example you'll need a stack (depends on the implementation though). As I was looking for my solution, I saw #naomik provided a solution with the while loop. That one is fine too, but I prefer the array with the modulo operation (a bit shorter).
Now concerning the problem of the size long long int has, it can be solved by using external libraries that implement operations for big numbers (like the GMP library or Boost.multiprecision). But you could also create your own version of a BigInteger-like class from Java and implement the basic operations like the one I have. I've only implemented the addition in my example (try to implement the others they are quite similar).
The main idea is simple, a BigInt represents a big decimal number by cutting its little endian representation into pieces (I'll explain why little endian at the end). The length of those pieces depends on the base you choose. If you want to work with decimal representations, it will only work if your base is a power of 10: if you choose 10 as base each piece will represent one digit, if you choose 100 (= 10^2) as base each piece will represent two consecutive digits starting from the end(see little endian), if you choose 1000 as base (10^3) each piece will represent three consecutive digits, ... and so on. Let's say that you have base 100, 12765 will then be [65, 27, 1], 1789 will be [89, 17], 505 will be [5, 5] (= [05,5]), ... with base 1000: 12765 would be [765, 12], 1789 would be [789, 1], 505 would be [505]. It's not the most efficient, but it is the most intuitive (I think ...)
The addition is then a bit like the addition on paper we learned at school:
begin with the lowest piece of the BigInt
add it with the corresponding piece of the other one
the lowest piece of that sum(= the sum modulus the base) becomes the corresponding piece of the final result
the "bigger" pieces of that sum will be added ("carried") to the sum of the following pieces
go to step 2 with next piece
if no piece left, add the carry and the remaining bigger pieces of the other BigInt (if it has pieces left)
For example:
9542 + 1097855 = [42, 95] + [55, 78, 09, 1]
lowest piece = 42 and 55 --> 42 + 55 = 97 = [97]
---> lowest piece of result = 97 (no carry, carry = 0)
2nd piece = 95 and 78 --> (95+78) + 0 = 173 = [73, 1]
---> 2nd piece of final result = 73
---> remaining: [1] = 1 = carry (will be added to sum of following pieces)
no piece left in first `BigInt`!
--> add carry ( [1] ) and remaining pieces from second `BigInt`( [9, 1] ) to final result
--> first additional piece: 9 + 1 = 10 = [10] (no carry)
--> second additional piece: 1 + 0 = 1 = [1] (no carry)
==> 9542 + 1 097 855 = [42, 95] + [55, 78, 09, 1] = [97, 73, 10, 1] = 1 107 397
Here is a demo where I used the class above to calculate the fibonacci of 10000 (result is too big to copy here)
Good luck!
PS: Why little endian? For the ease of the implementation: it allows to use push_back when adding digits and iteration while implementing the operations will start from the first piece instead of the last piece in the array.
The project I am working on needs to find some way of verifying that a variable after the modulus operation is either number != 0, number > 0, or number < (0 < x < 1). I have the first two understood, however employing the mod operator to accomplish the third is difficult.
Essentially what I am looking to do is to be able to catch a value similar to something like this:
a) 2 % 6
b) flag it and store the fact that .333 is less than 1 in a variable (bool)
c) perform a follow up action on the basis that the variable returned a value less than 1.
I have a feeling that the mod operator cannot perform this by itself. I'm looking for a way to utilize its ability to find remainders in order to produce a result.
edit: Here is some context. Obveously the below code will not give me what I want.
if (((inGameTotalCoins-1) % (maxPerTurn+1)) < 0){
computerTakenCoins = (inGameTotalCoins - 1);
inGameTotalCoins = 1;
The quotient is 0(2/6) with the fractional part discarded.The fractional part is .3333 ... So you are basically talking about the fractional part of the quotient , not the modulus value. Modulus can be calculated as follows :
(a / b) * b + (a % b) = a
(2 / 6) * 6 + (2 % 6) = 2
0 * 6 + (2 % 7) = 2
(2 % 6) = 2
*6 goes into 2 zero times with 2 left over.
How about this:-
int number1 = 2;
int number2 = 6;
float number3 = static_cast<float>(number1) / static_cast<float>(number2);
bool isfraction = number3 > 0 && number3 < 1;
if(isfraction){
std :: cout << "true\n" << number3;
}
else{
std :: cout << "false" << number3;
}
number != 0 includes number > 0 and number < (0 x < 1). And number > 0 includes number < (0 x < 1). Generally we do not classify so. For example, people classify number > 0, number == 0 and number < 0.
If you do the modulous operation, you get remainder. Remainder's definition is not one thing. You can see it at https://en.m.wikipedia.org/wiki/Remainder
this is a broad question, with a specific example. My question is, how do you figure out how to evaluate an algorithm or expression in a loop?
In the example beblow, I figured out how to use a loop to calculate a square root using a counter with the babylonian algorithm and it appears to be working ok.
But I can't figure out how evaluate this to a certain degree, say within .01 without using a counter. I see other people on the web doing it without a counter but the abililty to do it eludes me.
Here is my counter version:
int count(20);
while (count > 0 )
{
if (guess <= (guess * p) + guess)
prevGuess = guess;
else
r = input / guess;
guess = (guess + r) / 2.0;
count -= 1;
cout << "\nGuess = " << guess;
}
I've tried doing things like:
while ((guess - lastGuess) / lastGuess)
and
while(guess - lastGuess * .01)
But I either get just one interation or an infinite loop.
What is a good method to figure how to evaluate any algorithm inside a loop?
And I don't mean just for this little program, I mean for anything, even something like x + y = 2.
Would you do something like:
while(x + y !=2) {}
Thanks!
Normally this kind of computation can stop when the next iteration doesn't change the value by much; for example
for(;;) {
double next_v = (v + (n / v)) / 2;
if (fabs(v - next_v) < eps) break; // We got there
v = next_v;
}
where eps is the accuracy you want in the result.
You need to compare your error to your determined error limit, instead of zero as your attempts do.
You would use
while (abs((guess - lastGuess) / lastGuess) > .01)
for a relative error of .01, and
while(abs(guess - lastGuess) > .01)
for an absolute one.
In while (and every loop) there is condition which tells program when to stop it. This conditions must be some expression which return true or false value. Simpler, it's just an equation, or something which gives as clear answer: true or false. In your example expression like : ((guess - lastGuess) / lastGuess) isn't returning either true nor false. It returns some numbers etc.
Expression like ((guess - lastGuess) / lastGuess) == 4 will return true if ((guess - lastGuess) / lastGuess) will be equal to 4 and false if not. This example will work: while(x + y !=2). Program will call loop until x+y is equal 2.