5
0
My teacher gave me a simple problem which I should solve with a code tiny as possible. I should print all odd numbers from 0 to 100000 into a HTML document using javascript.
for(var y = 0; y < 100000; y++){
if(!(y % 2 == 0)){
document.write(y + " ");
}
};
Is it possible to do that shorter?
Max Krause
Posted 2014-12-25T12:08:00.353
Reputation: 161
13
Yet another way to do this, shortest of all.
This supposedly prints each number on a new line using \n. Now as the data is being written to an HTML document, \n is not of any use and the numbers appear to be separated by space only.
for(i=1;i<1e5;i+=2)document.writeln(i)
Optimizer
Posted 2014-12-25T12:08:00.353
Reputation: 25 836
13Just edit your existing answer(s) – proud haskeller – 2014-12-25T17:53:24.933
3@proudhaskeller more rep this way ;) – hobbs – 2014-12-26T04:10:54.260
Or the alternative for(i=0;++i<1e5;)document.writeln(i++). I've found plenty, but none beats 38. Good job. – Domino – 2015-11-02T19:05:44.133
7
If order doesn't matter:
for(i=1e5;i--;)document.writeln(i--)
Writes the numbers in reverse order.
Noyo
Posted 2014-12-25T12:08:00.353
Reputation: 943
6
another way to do it in 38 bytes:
for(i=0;i<5e4;)document.writeln(i+++i)
explanation:i+++i means the same as (i++)+i, with the second i resolving to its new value (as effected by the ++) so it goes like this for the results:
135 etc...
Jasen
Posted 2014-12-25T12:08:00.353
Reputation: 413
Can you explain the i+++i condition? Thanks – NiCk Newman – 2015-11-01T12:21:03.150
1@NiCkNewman see edit. – Jasen – 2015-11-02T04:12:30.947
Learn something new everyday, thanks Jasen. – NiCk Newman – 2015-11-02T13:32:38.017
5
for(i=1;i<1e5;i+=2)document.write(i+" ")The code is fairly straight forward.
i=1 to i=1e5. 1e5 is nothing but 1 followed by 5 0, so 100000.i and increment i by 2. Thus printing only every other number starting from 1 and till 100000.This prints all the odd numbers from 1 to 100000.
Optimizer
Posted 2014-12-25T12:08:00.353
Reputation: 25 836
4
Another way to do it!
for(i=0;i<5e4;)document.writeln(2*++i-1)The code is fairly straight forward.
i = 0 to i = 5e4. 5e4 is nothing but 5 followed by 4 0, so 50000.2*i - 1 after incrementing i by 1. Thus printing only odd numbers from 1 to 99999. Optimizer
Posted 2014-12-25T12:08:00.353
Reputation: 25 836
0
Using while 42
i=-1;while((i+=2)<10e5)document.writeln(i)
Yahor Zhylinski
Posted 2014-12-25T12:08:00.353
Reputation: 301
By shorter do you mean Javascript only or any language? – Sp3000 – 2014-12-25T12:09:35.310
The problem should be solved in JS only. – Max Krause – 2014-12-25T12:10:33.267
17-1 for needlessly restriction on the language – Johannes Kuhn – 2014-12-25T17:52:05.510
7
@JohannesKuhn language-specific questions asking for golfing advice are perfectly on-topic
– Martin Ender – 2014-12-25T18:16:26.3133-1 for asking us to do your homework better. – R-D – 2014-12-27T01:36:58.210
The non-golfed Perl6 answer would be:
say "{ 1, 3, 5, 7 ...^ *>100000 }"The golfed version would be:say "{1,3...9 x 5}"orsay "{1,3...99999}"– Brad Gilbert b2gills – 2014-12-29T15:02:47.243Read this: http://codegolf.stackexchange.com/questions/2682/tips-for-golfing-in-javascript
– anatolyg – 2014-12-30T21:13:17.8631Did you teacher specify that only the odd numbers were to be printed? – Flygenring – 2015-01-09T02:16:53.653