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fix: UPPER->lower instead of lower->UPPER (thanks to Firefly for catch & correction).

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edited Oct 6, 2013 at 16:43

J - 30

'`z''@Z'(]-32*1=I]+32*1=I.)&.(a.&i.)1!:1]1

J is read right-to-left, so to break this down:

Prompt user for input: 1!:1]1
Perform algorithm in code-point-space: &.(a.&i.)
Identify character range for each letter; the characters between codepoints "`""@" and "z""Z" are considered lowercaseuppercase: 1=I..
For each lower-caseuppercase codepoint, subtractadd 32: `']-32* ...']+32* ...
Note that step (2) creates an implicit step (5): we started out by projecting from character to integer domain, so now that we're finished, we map those integers back onto characters.

Obviously this particular implementation only considers ASCII; but the approach could be extended to at least the basic multilingual plane in Unicode.

J - 30

'`z'(]-32*1=I.)&.(a.&i.)1!:1]1

J is read right-to-left, so to break this down:

Prompt user for input: 1!:1]1
Perform algorithm in code-point-space: &.(a.&i.)
Identify character range for each letter; the characters between codepoints "`" and "z" are considered lowercase: 1=I..
For each lower-case codepoint, subtract 32: `']-32* ...'
Note that step (2) creates an implicit step (5): we started out by projecting from character to integer domain, so now that we're finished, we map those integers back onto characters.

Obviously this particular implementation only considers ASCII; but the approach could be extended to at least the basic multilingual plane in Unicode.

J - 30

'@Z'(]+32*1=I.)&.(a.&i.)1!:1]1

J is read right-to-left, so to break this down:

Prompt user for input: 1!:1]1
Perform algorithm in code-point-space: &.(a.&i.)
Identify character range for each letter; the characters between codepoints "@" and "Z" are considered uppercase: 1=I..
For each uppercase codepoint, add 32: ]+32* ...
Note that step (2) creates an implicit step (5): we started out by projecting from character to integer domain, so now that we're finished, we map those integers back onto characters.

Obviously this particular implementation only considers ASCII; but the approach could be extended to at least the basic multilingual plane in Unicode.

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created Oct 6, 2013 at 16:14

Dan Bron

J - 30

'`z'(]-32*1=I.)&.(a.&i.)1!:1]1

J is read right-to-left, so to break this down:

Prompt user for input: 1!:1]1
Perform algorithm in code-point-space: &.(a.&i.)
Identify character range for each letter; the characters between codepoints "`" and "z" are considered lowercase: 1=I..
For each lower-case codepoint, subtract 32: `']-32* ...'
Note that step (2) creates an implicit step (5): we started out by projecting from character to integer domain, so now that we're finished, we map those integers back onto characters.

Obviously this particular implementation only considers ASCII; but the approach could be extended to at least the basic multilingual plane in Unicode.