For some reason, both my online, and desktop compiling environments think that everything is invalid syntax all of the sudden! here is my code:
def sumn(n): #summation for sigma
return ((n + 1) * n) / 2
ipt = raw_input('How In Depth Would You Like To Go? ')
ipt = int(ipt)
pi = sumn(ipt) * ((4 / (8 * ipt + 1)) - (2 / (8 * ipt + 4)) - (1 / (8 * ipt + 5)) - (1 / (8 * ipt + 6)) * (1 / (16 ^ ipt))
print pi
Your indent for the function sumn(n)is invalid.
Your syntax for line : ipt = int(raw_input('How In Depth Would You Like To Go? ')) is incorrect.
For power use ** instead of ^.
I've got your code working with proper indentation and syntax for pi:
>>> def sumn(n): #summation for sigma
return ((n + 1) * n) / 2
>>> ipt = float(raw_input('How In Depth Would You Like To Go? '))
How In Depth Would You Like To Go? 4
>>> pi = sumn(ipt) * ((4 / (8 * ipt + 1)) - (2 / (8 * ipt + 4)) - (1 / (8 * ipt + 5)) - (1 / (8 * ipt + 6)) * (1 / (16 ** ipt)))
>>> print pi
0.386291370825
Yayy, I finally solved it!
from decimal import *
def sumn(n): #summation for sigma
return ((n + 1) * n) / 2
ipt = int(raw_input('How In Depth Would You Like To Go? '))
getcontext().prec = ipt
def factorial(n):
if n<1:
return 1
else:
return n * factorial(n-1)
def BBP(n):
pi = Decimal(0)
k = 0
while k < n:
pi += (Decimal(1) / (16 ** k)) * ((Decimal(4) / (8 * k + 1)) - (Decimal(2) / (8 * k + 4)) - (Decimal(1) / (8 * k + 5)) - (Decimal(1) / (8 * k + 6)))
k += 1
return pi
print BBP(ipt)
ex.
How In Depth Would You Like To Go? 1000
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420193
The last two values are usually wrong though, and I am still working on trying to truncate them. Then all I have to do is add 2 to ipt, which would take me all of 5 seconds ☺
Related
int f(int x)
{
if (x < 1)
return 0;
else
return f(x - 3) + x;
}
Can some one explain how it calculates the final result for f(10)=22?
step by step please
Working through this step by step is not difficult.
f(10)
f(10 - 3) + 10
(f(7 - 3) + 7) + 10
((f(4 - 3) + 4) + 7) + 10
(((f(1 - 3) + 1) + 4) + 7) + 10
(((0 + 1) + 4) + 7) + 10
((1 + 4) + 7) + 10
(5 + 7) + 10
12 + 10
22
f(10) = f(7) + 10 = f(4) + 10 + 7 = f(1) + 10 + 7 + 4 = f(-2) + 10 + 7 + 4 + 1 = 0 + 10 + 7 + 4 + 1
I need to evaluate the finite sum of numbers which are increasing in absolute value, but are alternate. Problem is: the abolute values grow too fast and it starts accumulating numerical errors. These the functions definitions, one (Gj_full) straight to it and the other (Gj) recursively. fact_quo is a simple factorial function.
def fact_quo(n, m=1):
if (type(n) != int) or (type(m) != int):
raise TypeError("Arguments must be integers.")
if (n < 0) or (m < 0):
raise ValueError("n=" + str(n) + "\t m=" + str(m))
if (n == 0) or (n == 1) or (n == m):
return 1
if (n < m):
raise ValueError("n=" + str(n) + "\t m=" + str(m))
f = n
while n > (m+1):
n -= 1
f *= n
return f
def Gj_full(X, Y, Xl, Yl, j=0, coef=1):
if (X - Y + Xl + Yl) % 2 or X < Y or Y < j:
raise ValueError
u = (X - Y + Xl + Yl) // 2
v = coef * (2 ** (X - Y) * fact_quo(X, Y-j) * fact_quo(u+j, j) *
4 ** j * (-1) ** j)
w = 3 ** (u+j) * fact_quo(X-Y+j)
den2 = fact_quo(X) * fact_quo(Xl) * fact_quo(Yl)
z = (np.sqrt(fact_quo(X)) * np.sqrt(fact_quo(Y))
* np.sqrt(fact_quo(Xl)) * np.sqrt(fact_quo(Yl)))
return (v / (den2 * w)) * z
def Gj(X, Y, Xl, Yl, j=0, coef=1):
if (X - Y + Xl + Yl) % 2 or X < Y or Y < j:
raise ValueError
kX, kY, kXl, kYl, kj = X % 2, Y % 2, Xl % 2, Yl % 2, 0
g = coef * Gj_full(kX, kY, kXl, kYl, kj)
while kX < X:
u = (kX - kY + kXl + kYl) // 2
v = 4 * (u + kj + 1)
w = 3 * (kX + 2 - kY + kj) * (kX + 1 - kY + kj)
g *= (v / w) * np.sqrt(kX + 2) * np.sqrt(kX + 1)
kX += 2
while kXl < Xl:
u = (kX - kY + kXl + kYl) // 2
v = u + kj + 1
w = 3 * (kXl + 2) * (kXl + 1)
g *= (v / w) * np.sqrt(kXl + 2) * np.sqrt(kXl + 1)
kXl += 2
while kYl < Yl:
u = (kX - kY + kXl + kYl) // 2
v = u + kj + 1
w = 3 * (kYl + 2) * (kYl + 1)
g *= (v / w) * np.sqrt(kYl + 2) * np.sqrt(kYl + 1)
kYl += 2
while kY < Y:
u = (kX - kY + kXl + kYl) // 2
v = 3 * (kX - kY + kj) * (kX - kY - 1 + kj)
w = 4 * (kY + 2 - kj) * (kY + 1 - kj) * (u + kj)
g *= (v / w) * np.sqrt(kY + 2) * np.sqrt(kY + 1)
kY += 2
while kj < j:
u = (kX - kY + kXl + kYl) // 2
v = -4 * (kY - kj) * (u + kj + 1)
w = 3 * (kX - kY + kj + 1) * (kj + 1)
g *= (v / w)
kj += 1
return g
The (4/3) ** j and the factorials quicly increase the absolute value of the summing terms. The sum, however, are supposed to be smaller than 1. In fact, for X = Y and Xl = Yl = 0, the sum equals to (-1/3) ** X.
The precision for infinitely large numbers for floats are not available yet without using a lib. Therefore you should look into the decimal lib, you can even set the precision. Eg.
import decimal
decimal.getcontext().prec = 100
def pi():
pi = decimal.Decimal(0)
for k in range(350):
pi += (decimal.Decimal(4)/(decimal.Decimal(8)*decimal.Decimal(k+1))...)
If you manage to force all the numbers to be integers, you don't need to worry about it
Are and correctly written?
if(x>0.1){z = 2*n+x}
if else(x <= 0.1){x(pow,n) - 1 / sqrt(n(pow,2)) + x(pow,2) * n}
x = n / 2 * n(pow,2) + 3 * n - 2
x is correctly translated to C++.
As for the rest of the code, it would be:
if(x > 0.1) {
z = 2 * n + x;
}
else {
z = pow(x, n) - 1 / sqrt(pow(n, 2) + pow(x, 2) * n);
}
I have a code on C++ it's b-spline curve that has 4 points if I want to change it to 6 point what shall I change in the code?
You can check the code:
#include "graphics.h"
#include <math.h>
int main(void) {
int gd, gm, page = 0;
gd = VGA;
gm = VGAMED;
initgraph(&gd, &gm, "");
point2d pontok[4] = { 100, 100, 150, 200, 170, 130, 240, 270 }; //pontok means points
int ap;
for (;;) {
setactivepage(page);
cleardevice();
for (int i = 0; i < 4; i++)
circle(integer(pontok[i].x), integer(pontok[i].y), 3);
double t = 0;
moveto((1.0 / 6) * (pontok[0].x * pow(1 - t, 3) +
pontok[1].x * (3 * t * t * t - 6 * t * t + 4) +
pontok[2].x * (-3 * t * t * t + 3 * t * t + 3 * t + 1) +
pontok[3].x * t * t * t),
(1.0 / 6) * (pontok[0].y * pow(1 - t, 3) +
pontok[1].y * (3 * t * t * t - 6 * t * t + 4) +
pontok[2].y * (-3 * t * t * t + 3 * t * t + 3 * t + 1) +
pontok[3].y * t * t * t));
for (t = 0; t <= 1; t += 0.01)
lineto(
(1.0 / 6) * (pontok[0].x * pow(1 - t, 3) +
pontok[1].x * (3 * t * t * t - 6 * t * t + 4) +
pontok[2].x * (-3 * t * t * t + 3 * t * t + 3 * t + 1) +
pontok[3].x * t * t * t),
(1.0 / 6) * (pontok[0].y * pow(1 - t, 3) +
pontok[1].y * (3 * t * t * t - 6 * t * t + 4) +
pontok[2].y * (-3 * t * t * t + 3 * t * t + 3 * t + 1) +
pontok[3].y * t * t * t));
/* Egerkezeles */ //Egerkezeles means mouse event handling
if (!balgomb)
ap = getactivepoint((point2d *)pontok, 4, 5);
if (ap >= 0 && balgomb) { //balgomb means left mouse button
pontok[ap].x = egerx; //eger means mouse
pontok[ap].y = egery;
}
/* Egerkezeles vege */
setvisualpage(page);
page = 1 - page;
if (kbhit())
break;
}
getch();
closegraph();
return 0;
}
From your formula, it looks like you are trying to draw a cubic Bezier curve. But the formula does not seem entirely correct. You can google "cubic Bezier curve" to find the correct formula. The Wikipedia page contains the formula for any degree of Bezier curve. You can find the "6-points" formula from there by using degree = 5.
I wonder what's the algorithm of make_heap in in C++ such that the complexity is 3*N? Only way I can think of to make a heap by inserting elements have complexity of O(N Log N). Thanks a lot!
You represent the heap as an array. The two elements below the i'th element are at positions 2i+1 and 2i+2. If the array has n elements then, starting from the end, take each element, and let it "fall" to the right place in the heap. This is O(n) to run.
Why? Well for n/2 of the elements there are no children. For n/4 there is a subtree of height 1. For n/8 there is a subtree of height 2. For n/16 a subtree of height 3. And so on. So we get the series n/22 + 2n/23 + 3n/24 + ... = (n/2)(1 * (1/2 + 1/4 + 1/8 + . ...) + (1/2) * (1/2 + 1/4 + 1/8 + . ...) + (1/4) * (1/2 + 1/4 + 1/8 + . ...) + ...) = (n/2) * (1 * 1 + (1/2) * 1 + (1/4) * 1 + ...) = (n/2) * 2 = n. Or, formatted maybe more readably to see the geometric series that are being summed:
n/2^2 + 2n/2^3 + 3n/2^4 + ...
= (n/2^2 + n/2^3 + n/2^4 + ...)
+ (n/2^3 + n/2^4 + ...)
+ (n/2^4 + ...)
+ ...
= n/2^2 (1 + 1/2 + 1/2^4 + ...)
+ n/2^3 (1 + 1/2 + 1/2^3 + ...)
+ n/2^4 (1 + 1/2 + 1/2^3 + ...)
+ ...
= n/2^2 * 2
+ n/2^3 * 2
+ n/2^4 * 2
+ ...
= n/2 + n/2^2 + n/2^3 + ...
= n(1/2 + 1/4 + 1/8 + ...)
= n
And the trick we used repeatedly is that we can sum the geometric series with
1 + 1/2 + 1/4 + 1/8 + ...
= (1 + 1/2 + 1/4 + 1/8 + ...) (1 - 1/2)/(1 - 1/2)
= (1 * (1 - 1/2)
+ 1/2 * (1 - 1/2)
+ 1/4 * (1 - 1/2)
+ 1/8 * (1 - 1/2)
+ ...) / (1 - 1/2)
= (1 - 1/2
+ 1/2 - 1/4
+ 1/4 - 1/8
+ 1/8 - 1/16
+ ...) / (1 - 1/2)
= 1 / (1 - 1/2)
= 1 / (1/2)
= 2
So the total number of "see if I need to fall one more, and if so which way do I fall? comparisons comes to n. But you get round-off from discretization, so you always come out to less than n sets of swaps to figure out. Each of which requires at most 3 comparisons. (Compare root to each child to see if it needs to fall, then the children to each other if the root was larger than both children.)