Labyrinth pattern

while ( true ) ; do if [ $(expr $RANDOM % 2 ) -eq 0 ] ; then echo -ne "\xE2\x95\xB1" ; else echo -ne "\xE2\x95\xB2" ; fi ; done
Generates labyrinth-like pattern on UTF-8 terminal in bash. For fun ;)
Sample Output

By: tobi
2015-01-17 12:46:37

11 Alternatives + Submit Alt

What Others Think

The output's pretty, but the command could be improved. . First, let's swap (true) for : The colon operator is a null-operation (but it returns true). So this is common: while :; do COMMAND; done . Next, tidy the test condition and get rid of expr (an external command). Bash supports doing maths in $((...)) if [ $(($RANDOM % 2)) -eq 0 ]; then ... Personally, I'd change the "-eq 0" to "= 0" Yes, that's a string comparison, but it's nicer to read. . Next, replace the if statement with some boolean operators. '[' is a command in its own right: man test(1) If you've only got two one-command options, you can write: [ TEST-CONDITION ] && TRUE-COMMAND || FALSE-COMMAND . Combine all the above and you get: while :; do [ $(($RANDOM % 2)) = 0 ] && echo -ne "\xE2\x95\xB1" || echo -ne "\xE2\x95\xB2"; done . Because it only uses bash builtin commands it's much quicker, too.
flatcap · 359 weeks and 4 days ago
Hmm... I wasn't content with my previous answer. It's not 'fu' enough :-) The output is one of: echo -ne "\xE2\x95\xB1" echo -ne "\xE2\x95\xB2" . So I need to substitute a 1 or a 2 into an echo. . Adding 1 to the previous random number: echo $((RANDOM%2+1)) . And substitute for the test condition: while :; do echo -ne "\xE2\x95\xB$((RANDOM%2+1))"; done . Much nicer.
flatcap · 359 weeks and 4 days ago
Well, really fast!
tobi · 359 weeks and 4 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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