Collide against walls

Question

Statement

Given an input like this 35 15 0 0 10 that matches the following constrains:

• Width in characters of the grid (5-50)
• Height in characters of the grid (5-25)
• Initial X position of the ball (0-(Width-1)
• Initial Y position of the ball (0-(Height-1)
• Number of movements to simulate

Generate a grid that shows the trajectory of the ball, knowing it's direction is initially DR (down-right) and that colliding with a wall will reverse the increment being applied to that axis. In other words, if we are going DR and hit the bottom we will now be going UR until we collide again.

The trajectory is not affected by passing twice over the same point since it's not the game snake, it's the trajectory of an object in 2d.

The input will be a string such as the one above and the output should be similar to this, not caring about which characters represent the wall, trajectory or ball, as long as they're consistent.

Points

Your score is the length of your submission, in characters.

Winners by language

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<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

Answer 1

MATL, 35 34 characters

Z"42Gi:qi!+lGqE!/tYo-KM!*|QYoZ}X](

Try it online!

Explanation

This takes care of the reflections using the formulas

nx = 2*(Nx-1)
px = abs(t/nx-round(t/nx))*nx

where t is a time variable that increases in steps of 1, Nx is the maximum size along the x direction and px is the computed x position. Analogous formulas are used for the y coordinate.

A char matrix of spaces is initially created, and then it is filled with character * at the computed x and y positions.

Z"     % Implicitly take [width height]. Push char matrix of spaces with that size
42     % Push 42. ASCII for character `*`
G      % Push first input again, that is, [width height]
i      % Take number of movements, M
:q     % Range [0 1 ... M-1]
i!     % Take initial position [x y]. Transpose into column vector
+      % Add element-wise with broadcast. Gives 2-row vector
lG     % Push first input, [width height]
qE!    % Transform into column vector [2*width-2; 2*height-2]
/      % Divide element-wise with broadcast
tYo    % Duplicate. Round to closest integer
-      % Subtract
KM!    % Push [2*width-2; 2*height-2] again
*      % Multiply element-wise with broadcast
|      % Absolute value
Q      % Add 1
Yo     % Round to closest integer
Z}     % Separate 2-row array into two row arrays. These contain row and column
       % indices, respectively, of the ball positions 
X]     % Convert row and colummn indices to linear index
(      % Assign 42 to those positions of the matrix of spaces

Answer 2

Python 180 172 171 163 145 Bytes

(-2 Bytes with Python3)

def r(m,n,x,y,t):
 s=([' ']*m+['\n'])*n;a=b=1
 for c in'o'*(t-1)+'x':s[x+y*m+y],x,y=c,x+a,y+b;a*=2*(0<x<m-1)-1;b*=2*(0<y<n-1)-1
 print''.join(s)

To save 2 more bytes with Python3 (thanks to shooqie) replace second line with

 s=[*' '*m,'\n']*n

Had to use a list of characters because python does not support string item assignement. The x+y*m+y part is from x+y*(m+1) because of the '\n' add each line end

Edit1: right after posting, saw how to get rid of the overwriting 'x'

Edit2: space after print

Edit3: using < and > instead ==, also tabs for 2nd indentation

Edit4: Replaced the if statements with assignements, daisy-chaining less-than and placing the loop in one line (thanks to RootTwo)

Answer 3

JavaScript, 293 279

Nothing too clever going on here.

EDIT: Saved 14 characters by initialising the 2D array differently.

p=prompt().split(' ')
a=[];for(i=0;i++<p[1];)a.push(new Array(~~p[0]).fill(' '))
x=~~p[2];y=~~p[3]
d=e=1
for(i=0;i++<p[4];){a[y][x]=i==p[4]?'x':'o'
if(x<1&&d<0||x>p[0]-2&&d>0)d*=-1
if(y<1&&e<0||y>p[1]-2&&e>0)e*=-1
x+=d;y+=e}o=''
for(i=0;i++<p[1];)o+=a[i-1].join('')+'\n'
alert(o)

Answer 4

R, 156 bytes/chars

As an unnamed function. It's quite long but I quite like the method. There's probably a better way to do the 1 to w to 1 vector.

function(w,h,x,y,i)cat(c(' ','o')[1:(w*h)%in%(rep(c(c(1:w),c((w-1):2)),i)[x:((i=i-1)+x)+1]+(rep(c(c(1:h),c((h-1):2)),i)*w-w)[y:(i+y)+1])*1+1],fill=w,sep='')

Brief explanation of function internals

cat(                                                 # output to STDOUT
  c(' ','o')                                         # vector of characters
    [1:(w*h)%in%(                                    # number of characters in the board
      rep(c(c(1:w),c((w-1):2)),i)[x:((i=i-1)+x)+1]+  # oscillating vector of x to w to 1 to x
      (rep(c(c(1:h),c((h-1):2)),i)*w-w)[y:(i+y)+1]   # similar to above but in steps of w
    *1+1]                                            # cause a vector of 1 or 2 to pick char
  ,fill=w,sep='')                                    # print width of w, no separator

Like the other current answers this doesn't output the border, but I will add one if required.

Answer 5

Javascript ES6 202 181 177 chars

(w,h,x,y,s)=>(a=(r=Array)(h).fill().map(_=>r(w).fill` `),eval("for(vx=vy=1;s--;x+=vx,y+=vy,a){a[y][x]=0;if(!x||x==w-1)vx*=-1;if(!y||y==h-1)vy*=-1;}"),a.map(r=>r.join``).join`
`)

Not very golfed, I take a another look at it once I get home.

Answer 6

C++11, 355 Bytes

Takes input string as command line arguments

#include <iostream>
#define A(n,i) int n=atoi(a[i]);
int main(int,char**a) {
A(w,1)A(h,2)A(x,3)A(y,4)A(n,5)
int d=1,r=1;
auto b=new int64_t[h];
while(--n>1){
b[y]|=1<<x;
if(r)x++;else x--;
if(d)y++;else y--;
if(x==0||x==w-1)r=!r;
if(y==0||y==h-1)d=!d;
}
for(int i=0;i<h;++i){
for(int j=0;j<w;++j){
std::cout<<(x==j&&y==i?'x':b[i]&(1<<j)?'o':' ');
}
std::cout<<'\n';
}  
}

Don't worry about the newlines, they are already removed from the count.

Storing the rows as 64-Bit integer is okay since maximum width is 50.

Answer 7

Q, 94 Bytes (kx.com)

{[w;h;i;j;n]./[(1+h,w)#" ";+{z#y_+\0,-1 1@{x=2*_x%2}'&(-_-(z+y)%x)#x}.'(h,j,n;w,i,n);:;"*"]}.

Defines a lambda with parameters w (width), h (height), i (initial x), j (initial y), and n (number of movements).

Lambda is delimited with {}

Final . (after }) allows to use a sequence as argument. Standard function invocation is f[arg;arg;..;arg], but we can also invoque as f . sequence. An integer sequence is v v .. v, so we can invoque as f . w h i j n. If we define f:{..}. we can invoque as f w h i j n (as required by the question)

Solution uses 'K mode' of Q. K is the functional inner core of Q (Q is more readable, verbose and with a lot of additional libraries, but K is compact). To execute in K mode we must use scripts with k extension (vs q extension for q mode). Interactive interpreter starts in Q mode (-> q) prompt). We can load code from k script, and execute test directly on interactive interpreter (even in q mode, as shown in test). There is also a k mode at interactive interpreter level, but it's not worth in this case.

TEST

NOTE.- to avoid display cluttering, we assume f name associated to lambda (f:{..}.)

q)f 35 15 0 0 49
"*                             *     "
" *                           * *    "
"  *                         *   *   "
"   *                       *     *  "
"    *                     *       * "
"     *                   *         *"
"      *                 *         * "
"       *               *         *  "
"        *             *         *   "
"         *           *         *    "
"          *         *         *     "
"           *       *         *      "
"            *     *   *     *       "
"             *   *     *   *        "
"              * *       * *         "
"               *         *          "
q)f 35 15 4 2 49
"                                *   "
"                               * *  "
"    *                         *   * "
"     *                       *     *"
"      *                     *     * "
"       *                   *     *  "
"        *                 *     *   "
"         *               *     *    "
"          *             *     *     "
"           *           *     *      "
"            *     *   *     *       "
"             *     * *     *        "
"              *     *     *         "
"               *   * *   *          "
"                * *   * *           "
"                 *     *            "

EXPLANATION

We define a lambda for 1D movement that accepts args [max;init;n] (legal positions 0..max-1, initial pos init, n movements) and returns a sequence of positions (after computing rebounds). Later apply that lambda to y and x axes (with args [h;j;n] and [w;i;n] respectively).

• Initially assume starting at 0 pos. We calculate number of rebounds to accomplish n movements, and generate a sequence with repeated increasing numbers (0 0 ...0; 1 1 .. 1; 2 2 .. 2; ..).

• Each subsequence has w items, so there are rebounds just between subsequences. For each number determine if is even, and sustitutes even results by 1 and even results by -1. (1 1 .. 1; -1 -1 .. -1; 1 1 .. 1; ..)

• We join subsequences and calculate accumulated sums starting with 0 -> 0 1 2 3 .. 35 34 33 32 .. 0 1 2 ..

• finally we drop starting i movements and take exactly n movements

Last, we define a (h x w) char matrix filled with blanks, and for each position (resulting of applying 1D lambda on y and x) we replace character with "*"