This is valid:
M = [1,2,3]
for M[0] in range(0,10):
print M[0]
2013-05-14
The star operator opens tuples and other sequences:
| # Solution to https://twitter.com/iwontoffendyou/status/1704935240907518367 | |
| # Tobin Fricke 2023-09-22 | |
| import numpy as np | |
| import scipy.optimize | |
| def dist_from_circle_center_to_curve(R): | |
| # Center of circle of radius R that is tangent to the y axis and the unit circle. | |
| x0 = R | |
| # Solution to the following puzzle: | |
| # Invert three boolean signals Using only two NOT gates, | |
| # and as many AND or OR gates as needed, | |
| # | |
| # Tobin Fricke 2020-09-03 | |
| for x in [True, False]: | |
| for y in [True, False]: | |
| for z in [True, False]: | |
| # If we assume that there is only one solution, then, when searching for the input |
| #include <stdio.h> | |
| #include <stdlib.h> | |
| int main(int argc, char **argv) { | |
| const int baz = rand(); // ok! | |
| baz = 99; // compile-time error. | |
| static int goo = rand(); // compile-time error. | |
| goo = 99; // ok. |
| def exact_sqrt(x): | |
| """Calculate the square root of an arbitrarily large integer. | |
| The result of exact_sqrt(x) is a tuple (a, r) such that a**2 + r = x, where | |
| a is the largest integer such that a**2 <= x, and r is the "remainder". If | |
| x is a perfect square, then r will be zero. | |
| The algorithm used is the "long-hand square root" algorithm, as described at | |
| http://mathforum.org/library/drmath/view/52656.html |
| % Plot the frequency response of an ideal active twin-T notch, versus | |
| % bootstrap feedback gain \alpha. | |
| % | |
| % Tobin Fricke <[email protected]> | |
| % 2013-11-21 | |
| % Pick the color map (to make the plot pretty) | |
| cm = jet(); | |
| % Choose the frequency axis |
| A PDH error signal can be calibrated (into Hz per volt) simply by | |
| observing the peak-to-peak amplitude of the PDH signal as the | |
| cavity is swept through resonance. Here's the math: | |
| An impedance-matched, lossless cavity has amplitude reflectivity of | |
| r_c(f) = i f / (i f + pcav) | |
| where pcav is the cavity pole, and f is the detuning frequency (i.e. | |
| the difference between the frequency of the light and the resonant |
This is valid:
M = [1,2,3]
for M[0] in range(0,10):
print M[0]
2013-05-14
The star operator opens tuples and other sequences:
| % Program to solve a sort of magic square, where only the rows and columns | |
| % must sum to the magic constant, and not the diagonals. Some elements of | |
| % the square are given in advance while others (indicated with 0) are | |
| % unknown. | |
| % | |
| % TF Time-Wasting Project 04-2013 | |
| function result = fu_puzzle | |
| puzzle = ... | |
| [ 1 0 0 0 22 |
| f_mod = 35.5e6 | |
| pcav = 300e3 | |
| i = {0, 1} | |
| conj(z) = real(z) - i*imag(z) | |
| r_c(f) = i*f / (i*f + pcav) | |
| pdh(f) = imag(r_c(f)*conj(r_c(f+f_mod))-conj(r_c(f))*r_c(f-f_mod)) | |
| set samples 501 |