Code Golf: The wave

code-golflanguage-agnosticrosetta-stone

The challenge

The shortest code by character count to generate a wave from the input string.

A wave is generated by elevating (line-1) a higher character, and degrading (line+1) a lower character. Equal characters are kept on the same line (no elevating or degrading done).

Input is made of lower case characters and numbers only, letters are considered higher than numbers.

Test cases:

Input:
    1234567890qwertyuiopasdfghjklzxcvbnm

Output:
                                 z
                                l x v n
                               k   c b m
                              j
                             h
                            g
                   y   p s f
                  t u o a d
               w r   i
            9 q e
           8 0
          7
         6
        5
       4
      3
     2
    1

Input:
    31415926535897932384626433832795028841971693993751058209749445923078164062862

Output:
                9 9   8 6 6
         9 6   8 7 3 3 4 2 4  8   9   88
    3 4 5 2 5 5     2       33 3 7 5 2  4 9   9 99 7
     1 1     3                  2   0    1 7 6 3  3 5   8              8 6
                                            1        1 5 2 9      9 3 7 1 4 6 8
                                                      0   0 7 9  5 2 0     0 2 6
                                                             4 44               2

Code count includes input/output (i.e full program).

Best Answer

x86 machine code (37 bytes)

Hexdump:

6800B807BF8007BE8200B40EAC3C0D741338D8740A720481EF400181C7A000AB86C3EBE8C3

Run in MS-DOS with 50 line console, the input is taken from the command line.

E.g.

wave.com 1234567890qwertyuiopasdfghjklzxcvbnm

Download binary here

Update: Shaved off three bytes thanks to jrandomhacker

Related Solutions

What and where are the stack and heap

The stack is the memory set aside as scratch space for a thread of execution. When a function is called, a block is reserved on the top of the stack for local variables and some bookkeeping data. When that function returns, the block becomes unused and can be used the next time a function is called. The stack is always reserved in a LIFO (last in first out) order; the most recently reserved block is always the next block to be freed. This makes it really simple to keep track of the stack; freeing a block from the stack is nothing more than adjusting one pointer.

The heap is memory set aside for dynamic allocation. Unlike the stack, there's no enforced pattern to the allocation and deallocation of blocks from the heap; you can allocate a block at any time and free it at any time. This makes it much more complex to keep track of which parts of the heap are allocated or freed at any given time; there are many custom heap allocators available to tune heap performance for different usage patterns.

Each thread gets a stack, while there's typically only one heap for the application (although it isn't uncommon to have multiple heaps for different types of allocation).

To answer your questions directly:

To what extent are they controlled by the OS or language runtime?

The OS allocates the stack for each system-level thread when the thread is created. Typically the OS is called by the language runtime to allocate the heap for the application.

What is their scope?

The stack is attached to a thread, so when the thread exits the stack is reclaimed. The heap is typically allocated at application startup by the runtime, and is reclaimed when the application (technically process) exits.

What determines the size of each of them?

The size of the stack is set when a thread is created. The size of the heap is set on application startup, but can grow as space is needed (the allocator requests more memory from the operating system).

What makes one faster?

The stack is faster because the access pattern makes it trivial to allocate and deallocate memory from it (a pointer/integer is simply incremented or decremented), while the heap has much more complex bookkeeping involved in an allocation or deallocation. Also, each byte in the stack tends to be reused very frequently which means it tends to be mapped to the processor's cache, making it very fast. Another performance hit for the heap is that the heap, being mostly a global resource, typically has to be multi-threading safe, i.e. each allocation and deallocation needs to be - typically - synchronized with "all" other heap accesses in the program.

A clear demonstration:
_{Image source: vikashazrati.wordpress.com}

Code Golf: Seven Segments

Perl, 134 characters

All linebreaks may be removed.

@s=unpack"C*",~"P\xdbLI\xc3a`[\@AB\xe0t\xc8df";
$_=<>;for$s(6,3,0){print map$s[hex$&]&1<<$_+$s?$_%2?"_":"|":" ",
0..2while/./g;print$/}

Explanation

@s stores the bits for each segment. The entries are in order from 0 to F (thanks to hex()) and the bits map to segments in this order:

6 7 x
3 4 5
0 1 2

with 0 being the LSB. (Bit 6 is unused). The values are stored packed in the string bit-inverted so there are a lot more printable characters; the ~ operator flips the bits and unpack gives me numbers (perl's bitwise operators are much clumsier when it comes to strings).

With the data in hand I read the input and proceed to loop over it three times; the only difference between the three loops is the bitmask required. For each character of input three characters of output are printed. The character to be printed is

$s[ hex $& ] & (1 << ($_ + $s) )
? ($_ % 2 ? "_" : "|" )
: " "

where @s is the lookup table, $s is the shift in effect depending on the row, and $_ is whether we're printing the 1st, 2nd, or 3rd character in the row. If the right bit in the lookup table entry is false it prints a space; otherwise it prints a "|" on the sides or a "_" in the middle.