During lunch I came across a reference to the fizzbuzz question (ie for the integers between 1-100 … print fizz if the number is divisble by 3, buzz if divisible by 5 and fizzbuzz if divisible by both) in reading someone’s blog. I have always said to myself, I need to see how long it will take me to solve that. Javascript:
var fizzbuzz = function(){ document.writeln(“Matthews Fizzbuzz”); document.writeln("
"); //an array to hold my numbers and numses var nums = new Array(); for (x = 1; x < 101; ++x){ nums.push(x); } for (z=0; z < nums.length; ++z){ if(nums[z] % 3 == 0 && nums[z] % 5 == 0){ document.writeln(“fizzbuzz” + "
"); }else if(nums[z] % 3 == 0){ document.writeln(“fizz” + "
"); }else if(nums[z] % 5 == 0){ document.writeln(“buzz” + "
"); } else{ document.writeln(nums[z] + "
"); } } } var start = new Date(); start = start.getTime(); // returns millisecs since epoch which is what I want since this operation takes less than a second start += parseInt( start ); fizzbuzz(); var end = new Date(); end = end.getTime(); end += parseInt( end ); document.writeln(“This took ” + (end - start) + ” milliseconds to run”);
Notes: it took me about five minutes give or take and I have seen it report between 2-4 milliseconds to run. The current program is 33 lines of code. Emacs Lisp:
;;print out a list of numbers, one per line, 1 - 100, if divisible by 3 print fizz, if divisble by 5 print buzz if divisible by both print fizzbuzz ;;time how long it takes to run fizzbuzz (defvar *emacs-fb-start* (current-time)) (setq nums 1) (while (< nums 101) (if (and (= 0 (% nums 3)) (= 0 (% nums 5))) (print “fizzbuzz”) (if (= 0 (% nums 3)) (print “fizz”) (if (= 0 (% nums 5)) (print “buzz”) (print nums)))) (setq nums (+ nums 1))) ;;this must be last line to calculate fizzbuzz runtime (message “Fizzbuzz ran in %dms” (destructuring-bind (hi lo ms) (current-time) (- (+ hi lo) (+ (first *emacs-load-start*) (second *emacs-fb-start*)))))
Notes: this took me a bit longer since I am used to stealing my emacs lisp code from google. I had to go in and read the emacs-lisp-intro info doc included with emacs to figure out how to run the while if and multiple conditionals for the divisible by five and 3 check. I still need to put a timer on it but I can steal that code (stolen from someone elses .emacs file) from my .emacs file which times emacs when I start it up. I borrowed the timer code from .emacs and i think its output is wrong: Fizzbuzz ran in 186ms. The current program is 21 lines of code with blank lines and comments, cool! bash
#!/bin/bash before=”$(date -j -f “%a %b %d %T %Z %Y” “`date`” ”+%s”)” echo “=========================Starting Fizzbuzz: $before=========================” num=1 while [ $num -lt 101 ]; do #if % 3&&5 print fizzbuzz if [ $(( $num % 5 )) -eq 0 ] && [ $(( $num % 3 )) -eq 0 ] ; then echo “fizzbuzz” elif [ $(( $num % 5 )) -eq 0 ] ; then echo “buzz” elif [ $(( $num %3 )) -eq 0 ] ; then echo fizz else echo $num fi let num=num+1 done after=”$(date -j -f “%a %b %d %T %Z %Y” “`date`” ”+%s”)” elapsed=”$(expr $after - $before)” echo “Time elapsed: $elapsed” echo “=========================End Fizzbuzz: $after=========================” exit 0
Notes: This was pretty straight forward as I could use the javascript logic flow and just substitute the bash syntax. The time does’t work very well since it only resolves to seconds and I think this goes faster than that. I would need to find a way to get microseconds and I didn’t see anything like that in the data man page. Lines of code: 26 total, no comments.
