https://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Print_Version
rebelmachina rebelmachina
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| # Helper function to plot a decision boundary. | |
| # If you don't fully understand this function don't worry, it just generates the contour plot below. | |
| def plot_decision_boundary(pred_func): | |
| # Set min and max values and give it some padding | |
| x_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5 | |
| y_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5 | |
| h = 0.01 | |
| # Generate a grid of points with distance h between them | |
| xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h)) | |
| # Predict the function value for the whole gid |
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| M[16],X=16,W,k;main(){T(system("stty cbreak") | |
| );puts(W&1?"WIN":"LOSE");}K[]={2,3,1};s(f,d,i | |
| ,j,l,P){for(i=4;i--;)for(j=k=l=0;k<4;)j<4?P=M | |
| [w(d,i,j++)],W|=P>>11,l*P&&(f?M[w(d,i,k)]=l<< | |
| (l==P):0,k++),l=l?P?l-P?P:0:l:P:(f?M[w(d,i,k) | |
| ]=l:0,++k,W|=2*!l,l=0);}w(d,i,j){return d?w(d | |
| -1,j,3-i):4*i+j;}T(i){for(i=X+rand()%X;M[i%X] | |
| *i;i--);i?M[i%X]=2<<rand()%2:0;for(W=i=0;i<4; | |
| )s(0,i++);for(i=X,puts("\e[2J\e[H");i--;i%4|| | |
| puts(""))printf(M[i]?"%4d|":" |",M[i]);W-2 |
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| %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% | |
| % Structured General Purpose Assignment | |
| % LaTeX Template | |
| % | |
| % This template has been downloaded from: | |
| % http://www.latextemplates.com | |
| % | |
| % Original author: | |
| % Ted Pavlic (http://www.tedpavlic.com) | |
| % |
grantslatton
/ fizzbuzz.c
Last active
August 19, 2022 11:20
FizzBuzz solved using only bit twiddling. It essentially uses two deterministic finite automata for divisibility testing.
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| #include <stdio.h> | |
| int f0(unsigned int x) { return x? (x&(1<<31)? f1(x<<1) : f0(x<<1)) : 1; } | |
| int f1(unsigned int x) { return x? (x&(1<<31)? f3(x<<1) : f2(x<<1)) : 0; } | |
| int f2(unsigned int x) { return x? (x&(1<<31)? f0(x<<1) : f4(x<<1)) : 0; } | |
| int f3(unsigned int x) { return x? (x&(1<<31)? f2(x<<1) : f1(x<<1)) : 0; } | |
| int f4(unsigned int x) { return x? (x&(1<<31)? f4(x<<1) : f3(x<<1)) : 0; } | |
| int t0(unsigned int x) { return x? (x&(1<<31)? t1(x<<1) : t0(x<<1)) : 1; } | |
| int t1(unsigned int x) { return x? (x&(1<<31)? t0(x<<1) : t2(x<<1)) : 0; } | |
| int t2(unsigned int x) { return x? (x&(1<<31)? t2(x<<1) : t1(x<<1)) : 0; } |
norioxkimura
/ arg.py
Created
April 30, 2012 14:54
argparse: exclusive options and variable number of arguments
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| import argparse | |
| import sys | |
| parser = argparse.ArgumentParser() | |
| g = parser.add_mutually_exclusive_group(required= True) | |
| g.add_argument('--a', action= 'store_const', const='a', dest= 'subcommand') | |
| g.add_argument('--b', action= 'store_const', const='b', dest= 'subcommand') | |
| parser.add_argument('bar', nargs= '*') | |
| print parser.parse_args(sys.argv[1:]) |
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| #!/bin/env python | |
| """ | |
| 9.15 Value Iteration 3 | |
| ============================== | |
| We got the following world:: | |
| +--+-----+-----+-----+-----+ | |
| | | 1 | 2 | 3 | 4 | | |
| +--+-----+-----+-----+-----+ |