Matlab, 112

The core idea is generating a list of possible next positions and then uniformly drawing one of them. If we are e.g. at position $l=1$ the possible steps would be [-1,0,1,2,3] Of course if we would chose -1 that would be invalid and we would have to stay at the same position. Thats why we replace the invalid positions with the current position, [1,0,1,2,3], and then randomly chose an element from this updated list.

The OP asked us tho draw the program, so here we go:

The transcription:

function c(n,s); l=19; %initialize position for k=1:n; disp([ones(1,l)*32,'.']); %print the line z=(-2:2)+l; %get vector of possible next steps (for l=1 we get [-1,0,1,2,3]) z(z<0)=l; %prune invalids: here we just replace the the invalid positions with the current position z(z>39)=l; % this ensures the same behaivour as staying in the same spot when going outside of the range l=z(randi(5)); %draw random sample of those pause(s); end 

Perl, 136 128 116 106 101 90 86

$p=19;map{say$"x$p.".";sleep $ARGV[0];$x=rand(5)+$p-2;$p=$x>0&&$x<40?$x:$p}1..$ARGV[1] 

Requires the seconds to be an integer.

Run with perl <filename> <second delay> <number of steps>.

There may be more golfing potential here, although honestly, I'm surprised it got this far. (Come on, only 6 more bytes to beat the bash answer...)

Changes

• Saved 8 bytes by removing unneeded parentheses and spelling out ARGV (it's actually shorter that way)
• Saved 12 more bytes by removing $s and $n and just using the plain $ARGV[0] and $ARGV[1]
• Saved another 10 bytes when I realized I could use $" and didn't need to specifically define $u as $undef.
• Saved 5 more bytes by rearranging the ternary and how $x is used and by using map instead of for.
• Saved 11 bytes by no longer accepting the seconds as decimals (challenge spec says it's OK.)
• Saved another 5 bytes by using say instead of print.

Python 2, 124 119 bytes

@janrn and @Steve Eckert: I don't have enough reputation to comment your answer but here's essentially your version shortened down. The task is to draw a program or a function, so by using f(s,x) you can save quite some bits, as well you can use max(0,min(x,39)) to avoid an extra if clause. Got it down to:

import time,random as r def f(s,x): n=19 while x:print' '*n+'.';time.sleep(s);n=max(0,min(n+r.randint(-2,2),39));x-=1 

Bash, 81

for((s=20;i++<$1;t=s,s+=RANDOM%5-2,s=s<0|s>39?t:s)){ printf %${s}s.\\n sleep $2 } 

Edit: If the step takes the walker outsite the range 1 to 40 it should be just ignored and the walker position stays the same handled correctly.

Input from command-line options. E.g.:

$ ./randwalk.sh 5 0.5 . . . . . $ 

Ruby, 84

def w(s,n)q=19;n.times{puts ' '*q+'.';sleep s;q+=rand(5)-2;q=[[q,0].max,39].min};end 

Python 2.7, 198 162 143 133

import time,random as r;p=20;s=0;d=input();w=input() while s<d: print' '*p+'.';s+=1;p=max(1,min(p+r.randint(-2,2),40));time.sleep(w) 

When calling the script with python script.py, the first input is the amount of steps, second input the time between steps (accepts float or int). Any suggestions for improving?

Edits

• saved 36 bytes due to now using print ' '*p+'.', thanks to @corsiKlause Ho Ho Ho
• down another 19 bytes by removing tab indents, replaced them with one space or ; where possible
• 10 bytes less thanks to @Bruce_Forte idea with p=max(1,min(p+r.randint(-2,2),40)) (I can't comment on your answer as well, but thanks; don't want to copy it completely)

Processing, 150 147

void w(int n,int s){int x=20,i,y,c=0;for(;c<n;c++){x+=y=int(random(5)-2);if(x>40||x<0)x-=y;for(i=1;i<x;i++)print(" ");println(".");delay(s*1000);}} 

Usage:

void setup() { w(10,1); } 

Note: 1000 can't be changed to 1e3 for type reasons.

Lua, 140 Bytes

Note: This program requires the LuaSocket package.

require"socket"p=19 for i=1,arg[2]+0 do print((" "):rep(p)..".")p=p+math.random(-2,2)p=p<0 and 0 or p>39 and 39 or p socket.sleep(arg[1])end 

Perl 6, 92 bytes

my (\n,\s)=@*ARGS;$/=19;for (-2..2).roll(n) {put ' 'x($/+=(40>$/+$_>=0??$_!!0)),'.';sleep s} # 92 

my (\n,\s)=@*ARGS; $/=19; for (-2..2).roll(n) { put ' 'x($/+=(40>$/+$_>=0??$_!!0)),'.'; sleep s } 

Usage:

$ perl6 -e 'my (\n,\s)=@*ARGS;$/=19;for (-2..2).roll(n) {put " "x($/+=(40>$/+$_>=0??$_!!0)),".",;sleep s}' 10 0.001 . . . . . . . . . . 

JavaScript (ES6), 125 bytes

(s,n)=>(t=setTimeout)(c=`console.log(" ".repeat(p)+".");p+=m=Math.random()*5|0;p-=p>41|p<2?m:2;--i&&t(c,d)`,d=s*1e3,p=19,i=n) 

Explanation

(s,n)=> (t=setTimeout)( // set inital timeout c=` // c = code for timeout to execute console.log(" ".repeat(p)+"."); // print the current line p+=m=Math.random()*5|0; // move walker 0 - 4 positions to the right p-=p>41|p<2? // if walker was moved out of bounds (2 - 41 instead // of 0 - 39 because the random move of 0 - 4 to // the right has not had the 2 subtracted yet) m: // undo the move 2; // else subtract 2 to make the move -2 to 2 --i&&t(c,d) // while we have steps remaining schedule next one `, d=s*1e3, // d = milliseconds to wait between steps p=19, // p = position of walker (0 indexed) i=n // i = steps remaining (needed to make n global) ) 

Test

console.log = s => result.textContent += s + "\n"; var solution = (s,n)=>(t=setTimeout)(c=`console.log(" ".repeat(p)+".");p+=m=Math.random()*5|0;p-=p>41|p<2?m:2;--i&&t(c,d)`,d=s*1e3,p=19,i=n)

<input type="number" id="seconds" value="0.1" /><br /> <input type="number" id="steps" value="30" /><br /> <button onclick="solution(+seconds.value,+steps.value)">Go</button> <pre id="result"></pre>

k4, 61 characters

f:{y{-1((y:1|40&y+-2+*1?5)#" "),".";."\\sleep ",$x;y}[x]\20;} 

sample run:

 f[.1]30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

Mathematica, 122 117 bytes

$RecursionLimit=∞;If[#2>0,Print[Indent[a=Min[#3+RandomInteger@{-2,2},39]~Max~0],"."];Pause@#;#0[#,#2-1,a]]&[##,19]& 

Recursive anonymous function, takes input in the order specified. Could probably be golfed further.

Python 3, 154 bytes

import time from random import* w=int(input()) z=int(input()) g=19 c=6 s=" " while c: c-=1 s+=s while z: z-=1 if -1<g<40: print(s[:g]+'.') time.sleep(w) g+=randint(-2,2) 

Generate a string of spaces greater than the maximum required length, then print that string ONLY up to the char at index 'g', then print a '.'. Finish by incrementing g by a random value in the range [-2:2], and repeat.

If anyone could help me golf down that horrific input block, I'd appreciate it.

C funtion, 114

s=20,i,t;w(int n,float f){for(;i++<n;t=s,s+=rand()%5-2,s=s<0||s>39?t:s,usleep((int)(f*1e6)))printf("%*c.\n",s,0);} 

Pretty much a direct translation of my bash answer.

Full test program:

s=20,i,t;w(int n,float f){for(;i++<n;t=s,s+=rand()%5-2,s=s<0||s>39?t:s,usleep((int)(f*1e6)))printf("%*c.\n",s,0);} int main (int argc, char **argv) { w(10, 0.2); return 0; } 

APL (Dyalog Unicode), 35 bytes

⍬⊣⎕{⎕←⌽⍵↑'.'⋄1⌈40⌊⍵+3-?5⊣⎕DL⍺}⍣⎕⊢20 

Try it online!

A full program which takes n as first input and s as the second.

⍬⊣⎕{⎕←⌽⍵↑'.'⋄1⌈40⌊⍵+3-?5⊣⎕DL⍺}⍣⎕⊢20 { }⍣⎕ do the following n times ⎕ with s as constant left argument ⍺ ⊢20 and 20 as right argument ⍵ ⍵↑'.' take ⍵ characters from '.' ⌽ reverse to pad with spaces ⎕← and display it ⎕DL⍺ delay s seconds ⊣ suppress it's return value with ?5 a random number in 1-5 3- subtract from 3 to get negative possibilities ⍵+ add current ⍵ to it 40⌊ minimum of that and 40 1⌈ maximum of that and 1 ⍬⊣ suppress the final return value.