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By: makka

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Showing 1-10 of 22 entries

In forum "Kickass Programmers General Discussion"

How bad is your keyboard?

the quick brown fox jumped over the lazy dog
both shifts activate capslock for me ;-)

Code Golf -- Don't Google "Google"

Originally posted by Spunky Sam:
I tried in python...
a=raw_input() print len(a)

First off, welcome to the community!
And secondly, your answer is not quite correct. You're supposed to crash if the user inputted "Google" otherwise print the length. Read the original link I posted for more information.

On that note, here's one for DLang

D, 26 chars

n=>n.length/(n!="Google");

Code Golf -- Implement Bogosort

C++14, 158 bytes

Probably would be less bytes if the requirement were not for it to be uniform (could just use std::random_shuffle and not have to include random nor specify an engine or source of entropy).

include <algorithm> #include <random> [](int*a,int s){std::random_device r;for(std::knuth_b g(r());!std::is_sorted(a,a+s);std::shuffle(a,a+s,g));return a;};

Haven't done a codegolf in awhile... and this one is just too good to pass.[codegolf.stackexchange.com]
Write the shortest function to implement bogosort. In specific, your function should:

Take an array (or your language's equivalent) as input
Check if its elements are in sorted order; if so, return the array
If not, shuffle the elements, and start again

GO GO GO GO

How bad is your keyboard?

THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG
Got my mechanical keyboard

Code Golf -- Print N Squared

Python is cheating ;)

Code Golf -- Read a crossword

C++14, 209 bytes

using S=vector<string>;[](S m){S t=m;int C=size(m[0]),j=0;for(;j<C*C;++j)t[j%C][j/C]=m[j/C][j%C];for(j=0;j<C+C;++j){stringstream Z(j<C?m[j]:t[j-C]);string s;while(getline(Z,s,' '))cout<<(size(s)>1?s+' ':"");}}

Transpose matrix, split string. Easy. Too bad no native function for transposing in C++14.

Ungolfed:

using S=vector<string>; [](S m){ S t=m; int C=size(m[0]),j=0; for(;j<C*C;++j)t[j%C][j/C]=m[j/C][j%C]; // Transpose for(j=0;j<C+C;++j){ // since rectangle matrix, we can iterate like so stringstream Z(j<C?m[j]:t[j-C]); // Get string from m or t string s; while(getline(Z,s,' ')) cout<<(size(s)>1?s+' ':""); } }

How to use it (note you must enforce padding as the question states so that the matrix is a rectangle):

using S = vector<string>;[](S m) { ... }({"pies", " not", " no", "wasp"});

Take a 2D string (of any format, you may assume it's padded with spaces to form a rectangle) and output each word (>1 character) found both horizontally and vertically. The code golf link is here[codegolf.stackexchange.com]

Example:
word
e e
step
t d
--------
word step west reed

Feel free to discuss and post your answer both here and on code golf.

Code Golf -- Print N Squared

C++14, 156 chars

#define f for(i=0;i++<n;c<<t); [](string t){auto&c=cout;int n=stoi(t),i;f c<<'\n';for(i=0;++i<n-1;c<<t,c.width(~-n*size(t)+1),c.fill(0),c<<t+'\n');if(n-1)f}

This one is pretty cool though it took a long time of fidgeting to get it to work. Unpacked:

#define f for ( i = 0; i++ < n; c << t ); [](string t) { auto& c = cout; int n = stoi(t), i; f // print first row c << '\n'; // kind of annoying but no way to get rid of (yes I tried // c << '\n'+t instead of c << t+'\n') for ( i = 0; ++i < n - 1; ) { c << t; // output the number // then we we get the width of necessary spaces c.width(~-n*size(t)+1); // Equivalent to (n-1)*size(t) + 1, but we save // two characters because ~- takes precedence over // the multiplication c.fill(0); // fill with spaces, ' ' == 0 c << t+'\n'; } if ( n-1 ) f // This if statement is dissapointing }

And like always... to run it you need to use [](string t) { ... }("10");

C++14, 162 chars

The above was my first attempt at the solution and took about 20 minutes. This next one took close to an hour, but unfortunately was not as good as my first attempt. It's a lot more sane, but 6 more chars, if it were not for edge cases 1 & 0 this could match my first solution. I figured I'd post it anyways since I spent so much time on it.

#define f for(i=-2;i++<n;c<<t);c<<'\n'; [](string t){auto&c=cout;int n=stoi(t)-2,i;auto g=t;g.append(abs(n)*size(t),' ');g+=t+'\n';f;for(i=0;i++<n;c<<g);if(n+1)f;}

and unpacked:

[](string t) { auto&c = cout; int n = stoi(t)-2,i; auto g = t; // creating "num ... num" // this abs(n) is very unfortunate, blame edge cases 1 & 0 g.append(abs(n)*size(t),' '); g += t+'\n'; f; for ( i = 0; i++ < n; c << g); if ( n+1 ) f; }

Write a function that takes in a non-negative integer N and prints out or returns a hollow ASCII-art square whose sides are made of N copies of the number N. Link to full challenge, details and unit tests here[codegolf.stackexchange.com]

An example for if N were 10, the output would be:

10101010101010101010 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10101010101010101010

Discuss and leave solutions in the thread.

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