I've been learning elixir this month, and was in a situation where I wanted to convert a binary object into a list of bits, for pattern matching.
My research led me here, to an article showing a method for doing so. However, I don't fully understand one of the arguments passed to the extract function.
I could just copy and paste the code, but I'd like to understand what's going on under the hood here.
The argument is this: <<b :: size(1), bits :: bitstring>>.
What I understand
I understand that << x >> denotes a binary object x. Logically to me, it looks as though this is similar to performing: [head | tail] = list on a List, to get the first element, and then the remaining ones as a new list called tail.
What I don't understand
However, I'm not familiar with the syntax, and I have never seen :: in elixir, nor have I ever seen a binary object separated by a comma: ,. I also, haven't seen size(x) used in Elixir, and have never encountered a bitstring.
The Bottom Line
If someone, could explain exactly how the syntax for this argument breaks down, or point me towards a resource I would highly appreciate it.
For your convenience, the code from that article:
defmodule Bits do
# this is the public api which allows you to pass any binary representation
def extract(str) when is_binary(str) do
extract(str, [])
end
# this function does the heavy lifting by matching the input binary to
# a single bit and sends the rest of the bits recursively back to itself
defp extract(<<b :: size(1), bits :: bitstring>>, acc) when is_bitstring(bits) do
extract(bits, [b | acc])
end
# this is the terminal condition when we don't have anything more to extract
defp extract(<<>>, acc), do: acc |> Enum.reverse
end
IO.inspect Bits.extract("!!") # => [0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1]
IO.inspect Bits.extract(<< 99 >>) #=> [0, 1, 1, 0, 0, 0, 1, 1]
Elixir pattern matching seems mind blowingly easy to use for
structured binary data.
Yep. You can thank the erlang inventors.
According to the documentation, <<x :: size(y)>> denotes a bitstring,
whos decimal value is x and is represented by a string of bits that is
y in length.
Let's dumb it down a bit: <<x :: size(y)>> is the integer x inserted into y bits. Examples:
<<1 :: size(1)>> => 1
<<1 :: size(2)>> => 01
<<1 :: size(3)>> => 001
<<2 :: size(3)>> => 010
<<2 :: size(4)>> => 0010
The number of bits in the binary type is divisible by 8, so a binary type has a whole number of bytes (1 byte = 8 bits). The number of bits in a bitstring is not divisible by 8. That's the difference between the binary type and the bitstring type.
I understand that << x >> denotes a binary object x. Logically to me,
it looks as though this is similar to performing: [head | tail] = list
on a List, to get the first element, and then the remaining ones as a
new list called tail.
Yes:
defmodule A do
def show_list([]), do: :ok
def show_list([head|tail]) do
IO.puts head
show_list(tail)
end
def show_binary(<<>>), do: :ok
def show_binary(<<char::binary-size(1), rest::binary>>) do
IO.puts char
show_binary(rest)
end
end
In iex:
iex(6)> A.show_list(["a", "b", "c"])
a
b
c
:ok
iex(7)> "abc" = <<"abc">> = <<"a", "b", "c">> = <<97, 98, 99>>
"abc"
iex(9)> A.show_binary(<<97, 98, 99>>)
a
b
c
:ok
Or you can interpret the integers in the binary as plain old integers:
def show(<<>>), do: :ok
def show(<<ascii_code::integer-size(8), rest::binary>>) do
IO.puts ascii_code
show(rest)
end
In iex:
iex(6)> A.show(<<97, 98, 99>>)
97
98
99
:ok
The utf8 type is super useful because it will grab as many bytes as required to get a whole utf8 character:
def show(<<>>), do: :ok
def show(<<char::utf8, rest::binary>>) do
IO.puts char
show(rest)
end
In iex:
iex(8)> A.show("ۑ")
8364
235
:ok
As you can see, the uft8 type returns the unicode codepoint of the character. To get the character as a string/binary:
def show(<<>>), do: :ok
def show(<<codepoint::utf8, rest::binary>>) do
IO.puts <<codepoint::utf8>>
show(rest)
end
You take the codepoint(an integer) and use it to create the binary/string <<codepoint::utf8>>.
In iex:
iex(1)> A.show("ۑ")
€
ë
:ok
You can't specify a size for the utf8 type, though, so if you want to read multiple utf8 characters, you have to specify multiple segments.
And of course, the segment rest::binary, i.e. a binary type with no size specified, is super useful. It can only appear at the end of a pattern, and rest::binary is like the greedy regex: (.*). The same goes for rest::bitstring.
Although the elixir docs don't mention it anywhere, the total number of bits in a segment, where a segment is one of those things:
| | |
v v v
<< 1::size(8), 1::size(16), 1::size(1) >>
is actually unit * size, where each type has a default unit. The default type for a segment is integer, so the type for each segment above defaults to integer. An integer has a default unit of 1 bit, so the total number of bits in the first segment is: 8 * 1 bit = 8 bits. The default unit for the binary type is 8 bits, so a segment like:
<< char::binary-size(6)>>
has a total size of 6 * 8 bits = 48 bits. Equivalently, size(6) is just the number of bytes. You can specify the unit just like you can the size, e.g. <<1::integer-size(2)-unit(3)>>. The total bit size of that segment is: 2 * 3 bits = 6 bits.
However, I'm not familiar with the syntax
Check this out:
def bitstr2bits(bitstr) do
for <<bit::integer-size(1) <- bitstr>>, do: bit
end
In iex:
iex(17)> A.bitstr2bits <<1::integer-size(2), 2::integer-size(2)>>
[0, 1, 1, 0]
Equivalently:
iex(3)> A.bitstr2bits(<<0b01::integer-size(2), 0b10::integer-size(2)>>)
[0, 1, 1, 0]
Elixir tends to abstract away recursion with library functions, so usually you don't have to come up with your own recursive definitions like at your link. However, that link shows one of the standard, basic recursion tricks: adding an accumulator to the function call to gather results that you want the function to return. That function could also be written like this:
def bitstr2bits(<<>>), do: []
def bitstr2bits(<<bit::integer-size(1), rest::bitstring>>) do
[bit | bitstr2bits(rest)]
end
The accumulator function at the link is tail recursive, which means it takes up a constant (small) amount of memory--no matter how many recursive function calls are needed to step through the bitstring. A bitstring with 10 million bits? Requiring 10 million recursive function calls? That would only require a small amount of memory. In the old days, the alternate definition I posted could potentially crash your program because it would take up more and more memory for each recursive function call, and if the bitstring were long enough the amount of memory needed would be too large, and you would get stackoverflow and your program would crash. However, erlang has optimized away the disadvantages of recursive functions that are not tail recursive.
You can read about all these here, short answer:
:: is similar as guard, like a when is_integer(a), in you case size(1) expect a 1 bit binary
, is a separator between matching binaries, like | in [x | []] or like comma in [a, b]
bitstring is a superset over binaries, you can read about it here and here, any binary can be respresented as bitstring
iex> ?h
104
iex> ?e
101
iex> ?l
108
iex> ?o
111
iex> <<104, 101, 108, 108, 111>>
"hello"
iex> [104, 101, 108, 108, 111]
'hello'
but not vice versa
iex> <<1, 2, 3>>
<<1, 2, 3>>
After some research, I realized I overlooked some important information located at: elixir-lang.
According to the documentation, <<x :: size(y)>> denotes a bitstring, whos decimal value is x and is represented by a string of bits that is y in length.
Furthermore, <<binary>> will always attempt to conglomerate values in a left-first direction, into bytes or 8-bits, however, if the number of bits is not divisible by 8, there will by x bytes, followed by a bitstring.
For example:
iex> <<3::size(5), 5::size(6)>> # <<00011, 000101>>
<<24, 5::size(3)>> # automatically shifted to: <<00011000(24) , 101>>
Now, elixir also lets us pattern match binaries, and bitstrings like so:
iex> <<3 :: size(2), b :: bitstring>> = <<61 :: size(6)>> # [11] [1101]
iex> b
<<13 :: size(4)>> # [1101]
So, i completly misunderstood binaries and biststrings, and pattern matching between the two.
Not really the answer to the question stated, but I’d put it here for the sake of formatting. In elixir we usually use Kernel.SpecialForms.for/1 comprehension for bitstring generation.
for << b :: size(1) <- "!!" >>, do: b
#⇒ [0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1]
for << b :: size(1) <- <<99>> >>, do: b
#⇒ [0, 1, 1, 0, 0, 0, 1, 1]
I wanted to use the bits, in an 8 bit binary to toggle conditions. So
[b1, b2, ...] = extract(<<binary>>)
I then wanted to say:
if b1, do: x....
if b2, do: y...
Is there a better way to do what I'm trying to do, instead of pattern
matching?
First of all, the only terms that evaluate to false in elixir are false and nil (just like in ruby):
iex(18)> x = 1
1
iex(19)> y = 0
0
iex(20)> if x, do: IO.puts "I'm true."
I'm true.
:ok
iex(21)> if y, do: IO.puts "I'm true."
I'm true.
:ok
Although, the fix is easy:
if b1 == 1, do: ...
Extracting the bits into a list is unnecessary because you can just iterate the bitstring:
def check_bits(<<>>), do: :ok
def check_bits(<<bit::integer-size(1), rest::bitstring>>) do
if bit == 1, do: IO.puts "bit is on"
check_bits(rest)
end
In other words, you can treat a bitstring similarly to a list.
Or, instead of performing the logic in the body of the function to determine whether the bit is 1, you can use pattern matching in the head of the function:
def check_bits(<<>>), do: :ok
def check_bits(<< 1::integer-size(1), rest::bitstring >>) do
IO.puts "The bit is 1."
check_bits(rest)
end
def check_bits(<< 0::integer-size(1), rest::bitstring >>) do
IO.puts "The bit is 0."
check_bits(rest)
end
Instead of using a variable, bit, for the match like here:
bit::integer-size(1)
...you use a literal value, 1:
1::integer-size(1)
The only thing that can match a literal value is the literal value itself. As a result, the pattern:
<< 1::integer-size(1), rest::bitstring >>
will only match a bitstring where the first bit, integer-size(1), is 1. The literal matching employed there is similar to doing the following with a list:
def list_contains_4([4|_tail]) do
IO.puts "found a 4"
true #end the recursion and return true
end
def list_contains_4([head|tail]) do
IO.puts "#{head} is not a 4"
list_contains_4(tail)
end
def list_contains_4([]), do: false
The first function clause tries to match the literal 4 at the head of the list. If the head of the list is not 4, there's no match; so elixir moves on to the next function clause, and in the next function clause the variable head will match anything in the list.
Using pattern matching in the head of a function rather than performing logic in the body of a function is considered more stylish and efficient in erlang.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
The puzzle
A little puzzle I heard while I was in high school went something like this...
The questioner would ask me to give him a number;
On hearing the number, the questioner would do some sort of transformation on it repeatedly (for example, he might say ten is three) until eventually arriving at the number 4 (at which point he would finish with four is magic).
Any number seems to be transformable into four eventually, no matter what.
The goal was to try to figure out the transformation function and then be able to reliably proctor this puzzle yourself.
The solution
The transformation function at any step was to
Take the number in question,
Count the number of letters in its English word representation, ignoring a hyphen or spaces or "and" (e.g., "ten" has 3 letters in it, "thirty-four" has 10 letters in it, "one hundred forty-three" has 20 letters in it).
Return that number of letters.
For all of the numbers I have ever cared to test, this converges to 4. Since "four" also has four letters in it, there would be an infinite loop here; instead it is merely referred to as magic by convention to end the sequence.
The challenge
Your challenge is to create a piece of code that will read a number from the user and then print lines showing the transformation function being repeatedly applied until "four is magic" is reached.
Specifically:
Solutions must be complete programs in and of themselves. They cannot merely be functions which take in a number-- factor in the input.
Input must be read from standard input. (Piping from "echo" or using input redirection is fine since that also goes from stdin)
The input should be in numeric form.
For every application of the transformation function, a line should be printed: a is b., where a and b are numeric forms of the numbers in the transformation.
Full stops (periods) ARE required!
The last line should naturally say, 4 is magic..
The code should produce correct output for all numbers from 0 to 99.
Examples:
> 4
4 is magic.
> 12
12 is 6.
6 is 3.
3 is 5.
5 is 4.
4 is magic.
> 42
42 is 8.
8 is 5.
5 is 4.
4 is magic.
> 0
0 is 4.
4 is magic.
> 99
99 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
The winner is the shortest submission by source code character count which is also correct.
BONUS
You may also try to write a version of the code which prints out the ENGLISH NAMES for the numbers with each application of the transformation function. The original input is still numeric, but the output lines should have the word form of the number.
(Double bonus for drawing shapes with your code)
(EDIT) Some clarifications:
I do want the word to appear on both sides in all applicable cases, e.g. Nine is four. Four is magic.
I don't care about capitalization, though. And I don't care how you separate the word tokens, though they should be separated: ninety-nine is okay, ninety nine is okay, ninetynine is not okay.
I'm considering these a separate category for bonus competition with regard to the challenge, so if you go for this, don't worry about your code being longer than the numeric version.
Feel free to submit one solution for each version.
Perl, about 147 char
Loosely based on Platinum Azure's solution:
chop
($_.=
<>);#
u="433
5443554
366 887
798 866
555 766
"=~ /\d
/gx ;#4
sub r{4
-$_ ?$_
<20 ?$u
[$_ ]:(
$'? $u[
$'] :0)
+$u[18+$&]:magic}print"
$_ is ",$_=r(),'.'while
/\d
/x;
444
GolfScript - 101 96 93 92 91 90 94 86 bytes
90 → 94: Fixed output for multiples of 10.
94 → 86: Restructured code. Using base 100 to remove non-printable characters.
86 → 85: Shorter cast to string.
{n+~."+#,#6$DWOXB79Bd")base`1/10/~{~2${~1$+}%(;+~}%++=" is "\".
"1$4$4-}do;;;"magic."
Common Lisp 157 Chars
New more conforming version, now reading form standard input and ignoring spaces and hyphens:
(labels((g (x)(if(= x 4)(princ"4 is magic.")(let((n(length(remove-if(lambda(x)(find x" -"))(format nil"~r"x)))))(format t"~a is ~a.~%"x n)(g n)))))(g(read)))
In human-readable form:
(labels ((g (x)
(if (= x 4)
(princ "4 is magic.")
(let ((n (length (remove-if (lambda(x) (find x " -"))
(format nil "~r" x)))))
(format t"~a is ~a.~%" x n)
(g n)))))
(g (read)))
And some test runs:
>24
24 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
>23152436
23152436 is 64.
64 is 9.
9 is 4.
4 is magic.
And the bonus version, at 165 chars:
(labels((g(x)(if(= x 4)(princ"four is magic.")(let*((f(format nil"~r"x))(n(length(remove-if(lambda(x)(find x" -"))f))))(format t"~a is ~r.~%"f n)(g n)))))(g(read)))
Giving
>24
twenty-four is ten.
ten is three.
three is five.
five is four.
four is magic.
>234235
two hundred thirty-four thousand two hundred thirty-five is forty-eight.
forty-eight is ten.
ten is three.
three is five.
five is four.
four is magic.
Python 2.x, 144 150 154 166 chars
This separates the number into tens and ones and sum them up. The undesirable property of the pseudo-ternary operator a and b or c that c is returned if b is 0 is being abused here.
n=input()
x=0x4d2d0f47815890bd2
while n-4:p=n<20and x/10**n%10or 44378/4**(n/10-2)%4+x/10**(n%10)%10+4;print n,"is %d."%p;n=p
print"4 is magic."
The previous naive version (150 chars). Just encode all lengths as an integer.
n=input()
while n-4:p=3+int('1yrof7i9b1lsi207bozyzg2m7sclycst0zsczde5oks6zt8pedmnup5omwfx56b29',36)/10**n%10;print n,"is %d."%p;n=p
print"4 is magic."
C - with number words
445 431 427 421 399 386 371 359* 356 354† 348 347 characters
That's it. I don't think I can make this any shorter.
All newlines are for readability and can be removed:
i;P(x){char*p=",one,two,three,four,five,six,sM,eight,nine,tL,elM,twelve,NP,4P,
fifP,6P,7P,8O,9P,twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,en,evL,thir,eL,tO,ty, is ,.\n,
4RmagicS,zero,";while(x--)if(*++p-44&&!x++)*p>95|*p<48?putchar(*p),++i:P(*p-48);
}main(c){for(scanf("%d",&c);c+(i=-4);P(34),P(c=i),P(35))P(c?c>19?P(c/10+18),
(c%=10)&&putchar(45):0,c:37);P(36);}
Below, it is somewhat unminified, but still pretty hard to read. See below for a more readable version.
i;
P(x){
char*p=",one,two,three,four,five,six,sM,eight,nine,tL,elM,twelve,NP,4P,fifP,6P,7P,8O,9P,twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,en,evL,thir,eL,tO,ty, is ,.\n,4RmagicS,zero,";
while(x--)
if(*++p-44&&!x++)
*p>95|*p<48?putchar(*p),++i:P(*p-48);
}
main(c){
for(scanf("%d",&c);c+(i=-4);P(34),P(c=i),P(35))
P(c?
c>19?
P(c/10+18),
(c%=10)&&
putchar(45)
:0,
c
:37);
P(36);
}
Expanded and commented:
int count; /* type int is assumed in the minified version */
void print(int index){ /* the minified version assumes a return type of int, but it's ignored */
/* see explanation of this string after code */
char *word =
/* 1 - 9 */
",one,two,three,four,five,six,sM,eight,nine,"
/* 10 - 19 */
"tL,elM,twelve,NP,4P,fifP,6P,7P,8O,9P,"
/* 20 - 90, by tens */
"twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,"
/* lookup table */
"en,evL,thir,eL,tO,ty, is ,.\n,4RmagicS,zero,";
while(index >= 0){
if(*word == ',')
index--;
else if(index == 0) /* we found the right word */
if(*word >= '0' && *word < 'a') /* a compression marker */
print(*word - '0'/*convert to a number*/);
else{
putchar(*word); /* write the letter to the output */
++count;
}
++word;
}
}
int main(int argc, char **argv){ /* see note about this after code */
scanf("%d", &argc); /* parse user input to an integer */
while(argc != 4){
count = 0;
if(argc == 0)
print(37/*index of "zero"*/);
else{
if(argc > 19){
print(argc / 10/*high digit*/ + 20/*offset of "twenty"*/ - 2/*20 / 10*/);
argc %= 10; /* get low digit */
if(argc != 0) /* we need a hyphen before the low digit */
putchar('-');
}
print(argc/* if 0, then nothing is printed or counted */);
}
argc = count;
print(34/*" is "*/);
print(argc); /* print count as word */
print(35/*".\n"*/);
}
print(36/*"four is magic.\n"*/);
}
About the encoded string near the beginning
The names of the numbers are compressed using a very simple scheme. Frequently used substrings are replaced with one-character indices into the name array. A "lookup table" of extra name entries is added to the end for substrings not used in their entirety in the first set. Lookups are recursive: entries can refer to other entries.
For instance, the compressed name for 11 is elM. The print() function outputs the characters e and l (lower-case 'L', not number '1') verbatim, but then it finds the M, so it calls itself with the index of the 29th entry (ASCII 'M' - ASCII '0') into the lookup table. This string is evL, so it outputs e and v, then calls itself again with the index of the 28th entry in the lookup table, which is en, and is output verbatim. This is useful because en is also used in eL for een (used after eight in eighteen), which is used in tO for teen (used for every other -teen name).
This scheme results in a fairly significant compression of the number names, while requiring only a small amount of code to decompress.
The commas at the beginning and end of the string account for the simplistic way that substrings are found within this string. Adding two characters here saves more characters later.
About the abuse of main()
argv is ignored (and therefore not declared in the compressed version), argc's value is ignored, but the storage is reused to hold the current number. This just saves me from having to declare an extra variable.
About the lack of #include
Some will complain that omitting #include <stdio.h> is cheating. It is not at all. The given is a completely legal C program that will compile correctly on any C compiler I know of (albeit with warnings). Lacking protoypes for the stdio functions, the compiler will assume that they are cdecl functions returning int, and will trust that you know what arguments to pass. The return values are ignored in this program, anyway, and they are all cdecl ("C" calling convention) functions, and we do indeed know what arguments to pass.
Output
Output is as expected:
0
zero is four.
four is magic.
1
one is three.
three is five.
five is four.
four is magic.
4
four is magic.
20
twenty is six.
six is three.
three is five.
five is four.
four is magic.
21
twenty-one is nine.
nine is four.
four is magic.
* The previous version missed the mark on two parts of the spec: it didn't handle zero, and it took input on the command line instead of stdin. Handling zeros added characters, but using stdin instead of command line args, as well as a couple of other optimzations saved the same number of characters, resulting in a wash.
† The requirements have been changed to make clear that the number word should be printed on both sides of " is ". This new version meets that requirement, and implements a couple more optimizations to (more than) account for the extra size necessary.
J, 107 112 characters
'4 is magic.',~}:('.',~":#{.,' is ',":#{:)"1]2&{.\.
(]{&(#.100 4$,#:3 u:ucp'䌵䐵吶梇禈榛ꪛ멩鮪鮺墊馊誙誩墊馊ꥺ겻곋榛ꪛ멩鮪鮺'))^:a:
(Newline for readability only)
Usage and output:
'4 is magic.',~}:('.',~":#{.,' is ',":#{:)"1]2&{.\.(]{&(#.100 4$,#:3 u:ucp'䌵䐵吶梇禈榛ꪛ멩鮪鮺墊馊誙誩墊馊ꥺ겻곋榛ꪛ멩鮪鮺'))^:a:12
12 is 6.
6 is 3.
3 is 5.
5 is 4.
4 is magic.
T-SQL, 413 451 499 chars
CREATE FUNCTION d(#N int) RETURNS int AS BEGIN
Declare #l char(50), #s char(50)
Select #l='0066555766',#s='03354435543668877987'
if #N<20 return 0+substring(#s,#N+1,1) return 0+substring(#l,(#N/10)+1,1) + 0+(substring(#s,#N%10+1,1))END
GO
CREATE proc M(#x int) as BEGIN
WITH r(p,n)AS(SELECT p=#x,n=dbo.d(#x) UNION ALL SELECT p=n,n=dbo.d(n) FROM r where n<>4)Select p,'is',n,'.' from r print '4 is magic.'END
(Not that I'm seriously suggesting you'd do this... really I just wanted to write a CTE)
To use:
M 95
Returns
p n
----------- ---- -----------
95 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
Java (with boilerplate), 308 290 286 282 280 characters
class A{public static void main(String[]a){int i=4,j=0;for(;;)System.out.printf("%d is %s.%n",i=i==4?new java.util.Scanner(System.in).nextInt():j,i!=4?j="43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:".charAt(i)-48:"magic");}}
I'm sure Groovy would get rid of much of that.
Explanation and formatting (all comments, newlines and leading/trailing whitespace removed in count):
Reasonably straight forward, but
//boilerplate
class A{
public static void main(String[]a){
//i is current/left number, j right/next number. i=4 signals to start
//by reading input
int i=4,j=0;
for(;;)
//print in the form "<left> is <right>."
System.out.printf(
"%d is %s.%n",
i=i==4?
//<left>: if i is 4 <left> will be a new starting number
new java.util.Scanner(System.in).nextInt():
//otherwise it's the next val
j,
i!=4?
//use string to map number to its length (:;< come after 9 in ASCII)
//48 is value of '0'. store in j for next iteration
j="43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:".charAt(i)-48:
//i==4 is special case for right; print "magic"
"magic");
}
}
Edit: No longer use hex, this is less keystrokes
Windows PowerShell: 152 153 184 bytes
based on the previous solution, with more influence from other solutions
$o="03354435543668877988"
for($input|sv b;($a=$b)-4){if(!($b=$o[$a])){$b=$o[$a%10]-48+"66555766"[($a-$a%10)/10-2]}$b-=48-4*!$a
"$a is $b."}'4 is magic.'
C, 158 characters
main(n,c){char*d="03354435543668877988";for(scanf("%d",&n);n-4;n=c)printf("%d is %d.\n",n,c=n?n<19?d[n]-48:d[n%10]-"_,**+++)**"[n/10]:4);puts("4 is magic.");}
(originally based on Vlad's Python code, borrowed a trick from Tom Sirgedas' C++ solution to squeeze out a few more characters)
expanded version:
main(n, c) {
char *d = "03354435543668877988";
for (scanf("%d",&n); n-4; n = c)
printf("%d is %d.\n", n, c = n ? n<19 ? d[n]-48 : d[n%10] - "_,**+++)**"[n/10] : 4);
puts("4 is magic.");
}
Python, 129 133 137 148 chars
As a warm-up, here is my first version (improves couple of chars over previous best Python).
PS. After a few redactions now it is about twenty char's shorter:
n=input()
while n-4:p=(922148248>>n/10*3&7)+(632179416>>n%10*3&7)+(737280>>n&1)+4*(n<1);print n,'is %d.'%p;n=p
print'4 is magic.'
C#: 210 Characters.
Squished:
using C=System.Console;class B{static void Main(){int
x=0,y=int.Parse(C.ReadLine());while(x!=4)C.Write((x=y)+" is {0}.\n",x==4?"magic":""+(y=x==0?4:"03354435543668877988"[x<20?x:x%10]+"0066555766"[x/10]-96));}}
Expanded:
using C=System.Console;
class B
{
static void Main()
{
int x=0,y=int.Parse(C.ReadLine());
while(x!=4)
C.Write((x=y)+" is {0}.\n",
x==4?
"magic":
""+(y= x==0?
4:
"03354435543668877988"[x<20?x:x%10]+
"0066555766"[x/10]-96)
);
}
}
Tricks this approach uses:
Create a lookup table for number name lengths based on digits that appear in the number.
Use character array lookup on a string, and char arithmetic instead of a numeric array.
Use class name aliasing to short Console. to C.
Use the conditional (ternary) operator (?:) instead of if/else.
Use the \n with Write escape code instead of WriteLine
Use the fact that C# has a defined order of evaluation to allow assignments inside the Write function call
Use the assignment expressions to eliminate extra statements, and thus extra braces
Perl: 148 characters
(Perl: 233 181 212 206 200 199 198 185 179 149 148 characters)
Moved exceptions hash into unit array. This resulted in my being able to cut a lot of characters :-)
mobrule pointed out a nasty bug. Quick fix adds 31 characters, ouch!
Refactored for zero special case, mild golfing done as well.
Direct list access for single use rather than storing to array? Hell yes!
SO MUCH REFACTORING for just ONE bloody character. This, truly, is the life of a golfer. :-(
Oops, easy whitespace fix. 198 now.
Refactored some redundant code.
Last return keyword in r is unnecessary, shaved some more off.
Massive refactoring per comments; unfortunately I could only get it to 149 because I had to fix a bug that was present in both my earlier code and the commenters' versions.
Trying bareword "magic".
Let's get this ball rolling with a modest attempt in Perl.
#u=split'','4335443554366887798866555766';$_=<>;chop;print"$_ is ".($_=$_==4?0:$_<20?$u[$_]:($u[$_/10+18]+($_%10&&$u[$_%10]))or magic).".
"while$_
Tricks:
Too many!
JavaScript 1.8 (SpiderMonkey) - 153 Chars
l='4335443554366887798866555766'.split('')
for(b=readline();(a=+b)-4;print(a,'is '+b+'.'))b=a<20?l[a]:+l[18+a/10|0]+(a%10&&+l[a%10])
print('4 is magic.')
Usage: echo 42 | js golf.js
Output:
42 is 8.
8 is 5.
5 is 4.
4 is magic.
With bonus - 364 chars
l='zero one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty thirty fourty fifty sixty seventy eighty ninety'.split(' ')
z=function(a)a<20?l[a]:l[18+a/10|0]+(a%10?' '+l[a%10]:'')
for(b=+readline();(a=b)-4;print(z(a),'is '+z(b)+'.'))b=z(a).replace(' ','').length
print('four is magic.')
Output:
ninety nine is ten.
ten is three.
three is five.
five is four.
four is magic.
Haskell, 224 270 characters
o="43354435543668877988"
x!i=read[x!!i]
n x|x<20=o!x|0<1="0066555766"!div x 10+o!mod x 10
f x=zipWith(\a b->a++" is "++b++".")l(tail l)where l=map show(takeWhile(/=4)$iterate n x)++["4","magic"]
main=readLn>>=mapM putStrLn.f
And little more readable -
ones = [4,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8]
tens = [0,0,6,6,5,5,5,7,6,6]
n x = if x < 20 then ones !! x else (tens !! div x 10) + (ones !! mod x 10)
f x = zipWith (\a b -> a ++ " is " ++ b ++ ".") l (tail l)
where l = map show (takeWhile (/=4) (iterate n x)) ++ ["4", "magic"]
main = readLn >>= mapM putStrLn . f
C++ Stdio version, minified: 196 characters
#include <cstdio>
#define P;printf(
char*o="43354435543668877988";main(int p){scanf("%d",&p)P"%d",p);while(p!=4){p=p<20?o[p]-48:"0366555966"[p/10]-96+o[p%10]P" is %d.\n%d",p,p);}P" is magic.\n");}
C++ Iostreams version, minified: 195 characters
#include <iostream>
#define O;std::cout<<
char*o="43354435543668877988";main(int p){std::cin>>p;O p;while(p!=4){p=p<20?o[p]-48:"0366555966"[p/10]-96+o[p%10]O" is "<<p<<".\n"<<p;}O" is magic.\n";}
Original, un-minified: 344 characters
#include <cstdio>
int ones[] = { 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8 };
int tens[] = { 0, 3, 6, 6, 5, 5, 5, 9, 6, 6 };
int n(int n) {
return n<20 ? ones[n] : tens[n/10] + ones[n%10];
}
int main(int p) {
scanf("%d", &p);
while(p!=4) {
int q = n(p);
printf("%i is %i\n", p, q);
p = q;
}
printf("%i is magic\n", p);
}
Delphi: 329 characters
Single Line Version:
program P;{$APPTYPE CONSOLE}uses SysUtils;const S=65;A='EDDFEEDFFEDGGIIHHJII';B='DGGFFFJGG';function Z(X:Byte):Byte;begin if X<20 then Z:=Ord(A[X+1])-S else Z:=(Ord(B[X DIV 10])-S)+Z(X MOD 10)end;var X,Y:Byte;begin Write('> ');ReadLn(X);repeat Y:=Z(X);WriteLn(Format('%d is %d.',[X,Y]));X:=Y;until X=4;WriteLn('4 is magic.');end.
Formated:
program P;
{$APPTYPE CONSOLE}
uses
SysUtils;
const
S = 65;
A = 'EDDFEEDFFEDGGIIHHJII';
B = 'DGGFFFJGG';
function Z(X:Byte):Byte;
begin
if X<20
then Z := Ord(A[X+1])-S
else Z := (Ord(B[X DIV 10])-S) + Z(X MOD 10);
end;
var
X,Y: Byte;
begin
Write('> ');
ReadLn(X);
repeat
Y:=Z(X);
WriteLn(Format('%d is %d.' , [X,Y]));
X:=Y;
until X=4;
WriteLn('4 is magic.');
end.
Probably room for some more squeezing... :-P
C# 314 286 283 274 289 273 252 chars.
Squished:
252
Normal:
using C = System.Console;
class P
{
static void Main()
{
var x = "4335443554366877798866555766";
int m, o, v = int.Parse(C.ReadLine());
do {
C.Write("{0} is {1}.\n", o = v, v == 4 ? (object)"magic" : v = v < 20 ? x[v] - 48 : x[17 + v / 10] - 96 + ((m = v % 10) > 0 ? x[m] : 48));
} while (o != 4);
C.ReadLine();
}
}
Edit Dykam: Did quite some carefull insertions and changes:
Changed the l.ToString() into a cast to object of the string "magic".
Created a temporary variable o, so I could move the break outside the for loop, that is, resulting in a do-while.
Inlined the o assignment, aswell the v assignment, continueing in inserting the calculation of l in the function arguments altogether, removing the need for l. Also inlined the assignment of m.
Removed a space in int[] x, int[]x is legit too.
Tried to transform the array into a string transformation, but the using System.Linq was too much to make this an improvement.
Edit 2 Dykam
Changed the int array to a char array/string, added proper arithmics to correct this.
Lua, 176 Characters
o={[0]=4,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8}t={3,6,6,5,5,5,7,6,6}n=0+io.read()while n~=4 do a=o[n]or o[n%10]+t[(n-n%10)/10]print(n.." is "..a..".")n=a end print"4 is magic."
or
o={[0]=4,3,3,5,4,4
,3,5,5,4,3,6,6,8,8
,7,7,9,8,8}t={3,6,
6,5,5,5,7,6,6}n=
0+io.read()while
n ~= 4 do a= o[n
]or o[n%10]+t[(n
-n%10)/10]print(
n.." is "..a.."." )n=a
end print"4 is magic."
C - without number words
180 175* 172 167 characters
All newlines are for readability and can be removed:
i;V(x){return"\3#,#6$:WOXB79B"[x/2]/(x%2?1:10)%10;}main(c){for(scanf("%d",&c);
c-4;)i=c,printf("%d is %d.\n",i,c=c?c>19?V(c/10+19)+V(c%10):V(c):4);puts(
"4 is magic.");}
Slightly unminified:
i;
V(x){return"\3#,#6$:WOXB79B"[x/2]/(x%2?1:10)%10;}
main(c){
for(scanf("%d",&c);c-4;)
i=c,
printf("%d is %d.\n",i,c=c?c>19?V(c/10+19)+V(c%10):V(c):4);
puts("4 is magic.");
}
* The previous version missed the mark on two parts of the spec: it didn't handle zero, and it took input on the command line instead of stdin. Handling zero added characters, but using stdin instead of command line args saved even more, resulting in a net savings.
perl, 123 122 characters
Just realized that there is no requirement to output to STDOUT, so output to STDERR instead and knock off another character.
#u='0335443554366887798866555766'=~/./g;$_+=<>;warn"$_ is ",$_=$_-4?$_<20?$u[$_]||4:$u[chop]+$u[$_+18]:magic,".\n"until/g/
And, a version that returns spelled out numbers:
279 278 276 280 characters
#p=(Thir,Four,Fif,Six,Seven,Eigh,Nine);#n=("",One,Two,Three,Four,Five,#p[3..6],Ten,Eleven,Twelve,map$_.teen,#p);s/u//for#m=map$_.ty,Twen,#p;$n[8].=t;sub n{$n=shift;$n?$n<20?$n[$n]:"$m[$n/10-2] $n[$n%10]":Zero}$p+=<>;warnt$m=n($p)," is ",$_=$p-4?n$p=()=$m=~/\w/g:magic,".\n"until/c/
While that meets the spec, it is not 100% well formatted. It returns an extra space after numbers ending in zero. The spec does say:
"I don't care how you separate the word tokens, though they should be separated"
That's kind of weaselly though.
A more correct version at
282 281 279 283 characters
#p=(Thir,Four,Fif,Six,Seven,Eigh,Nine);#n=("\x8",One,Two,Three,Four,Five,#p[3..6],Ten,Eleven,Twelve,map$_.teen,#p);s/u//for#m=map$_.ty,Twen,#p;$n[8].=t;sub n{$n=shift;$n?$n<20?$n[$n]:"$m[$n/10-2]-$n[$n%10]":Zero}$p+=<>;warn$m=n($p)," is ",$_=$p-4?n$p=()=$m=~/\w/g:magic,".\n"until/c/
Python:
#!/usr/bin/env python
# Number of letters in each part, we don't count spaces
Decades = ( 0, 3, 6, 6, 6, 5, 5, 7, 6, 6, 0 )
Smalls = ( 0, 3, 3, 5, 4, 4, 3, 5, 5, 4 )
Teens = ( 6, 6, 8, 8, 7, 7, 9, 8, 8 )
def Count(n):
if n > 10 and n < 20: return Teens[n-11]
return Smalls[n % 10 ] + Decades [ n / 10 ]
N = input()
while N-4:
Cnt = Count(N)
print "%d is %d" % ( N, Cnt)
N = Cnt
print "4 is magic"
C++, 171 characters (#include omitted)
void main(){char x,y,*a="03354435543668877988";scanf("%d",&x);for(;x-4;x=y)y=x?x<19?a[x]-48:"_466555766"[x/10]+a[x%10]-96:4,printf("%d is %d.\n",x,y);puts("4 is magic.");}
Ruby, 164 characters
n=gets.to_i;s="03354435543668877987";if n==0;puts"0 is 4.";else;puts"#{n} is #{n=(n<20)?s[n]-48:"0066555766"[n/10]-48+s[n%10]-48}." until n==4;end;puts"4 is magic."
decoded:
n = gets.to_i
s = "03354435543668877987"
if n == 0
puts "0 is 4."
else
puts "#{n} is #{n = (n < 20) ? s[n] - 48 : "0066555766"[n / 10] - 48 + s[n % 10] - 48}." until n == 4
end
puts "4 is magic."
Lua 185 190 199
added periods, added io.read, removed ()'s on last print
n=io.read();while(n~=4)do m=('43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:'):sub(n+1):byte()-48;print(n,' is ',m,'.')n=m;end print'4 is magic.'
with line breaks
n=io.read()
while (n~=4) do
m=('43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:'):sub(n+1):byte()-48;
print(n,' is ',m,'.')
n=m;
end
print'4 is magic.'
PhP Code
function get_num_name($num){
switch($num){
case 1:return 'one';
case 2:return 'two';
case 3:return 'three';
case 4:return 'four';
case 5:return 'five';
case 6:return 'six';
case 7:return 'seven';
case 8:return 'eight';
case 9:return 'nine';
}
}
function num_to_words($number, $real_name, $decimal_digit, $decimal_name){
$res = '';
$real = 0;
$decimal = 0;
if($number == 0)
return 'Zero'.(($real_name == '')?'':' '.$real_name);
if($number >= 0){
$real = floor($number);
$decimal = number_format($number - $real, $decimal_digit, '.', ',');
}else{
$real = ceil($number) * (-1);
$number = abs($number);
$decimal = number_format($number - $real, $decimal_digit, '.', ',');
}
$decimal = substr($decimal, strpos($decimal, '.') +1);
$unit_name[1] = 'thousand';
$unit_name[2] = 'million';
$unit_name[3] = 'billion';
$unit_name[4] = 'trillion';
$packet = array();
$number = strrev($real);
$packet = str_split($number,3);
for($i=0;$i<count($packet);$i++){
$tmp = strrev($packet[$i]);
$unit = $unit_name[$i];
if((int)$tmp == 0)
continue;
$tmp_res = '';
if(strlen($tmp) >= 2){
$tmp_proc = substr($tmp,-2);
switch($tmp_proc){
case '10':
$tmp_res = 'ten';
break;
case '11':
$tmp_res = 'eleven';
break;
case '12':
$tmp_res = 'twelve';
break;
case '13':
$tmp_res = 'thirteen';
break;
case '15':
$tmp_res = 'fifteen';
break;
case '20':
$tmp_res = 'twenty';
break;
case '30':
$tmp_res = 'thirty';
break;
case '40':
$tmp_res = 'forty';
break;
case '50':
$tmp_res = 'fifty';
break;
case '70':
$tmp_res = 'seventy';
break;
case '80':
$tmp_res = 'eighty';
break;
default:
$tmp_begin = substr($tmp_proc,0,1);
$tmp_end = substr($tmp_proc,1,1);
if($tmp_begin == '1')
$tmp_res = get_num_name($tmp_end).'teen';
elseif($tmp_begin == '0')
$tmp_res = get_num_name($tmp_end);
elseif($tmp_end == '0')
$tmp_res = get_num_name($tmp_begin).'ty';
else{
if($tmp_begin == '2')
$tmp_res = 'twenty';
elseif($tmp_begin == '3')
$tmp_res = 'thirty';
elseif($tmp_begin == '4')
$tmp_res = 'forty';
elseif($tmp_begin == '5')
$tmp_res = 'fifty';
elseif($tmp_begin == '6')
$tmp_res = 'sixty';
elseif($tmp_begin == '7')
$tmp_res = 'seventy';
elseif($tmp_begin == '8')
$tmp_res = 'eighty';
elseif($tmp_begin == '9')
$tmp_res = 'ninety';
$tmp_res = $tmp_res.' '.get_num_name($tmp_end);
}
break;
}
if(strlen($tmp) == 3){
$tmp_begin = substr($tmp,0,1);
$space = '';
if(substr($tmp_res,0,1) != ' ' && $tmp_res != '')
$space = ' ';
if($tmp_begin != 0){
if($tmp_begin != '0'){
if($tmp_res != '')
$tmp_res = 'and'.$space.$tmp_res;
}
$tmp_res = get_num_name($tmp_begin).' hundred'.$space.$tmp_res;
}
}
}else
$tmp_res = get_num_name($tmp);
$space = '';
if(substr($res,0,1) != ' ' && $res != '')
$space = ' ';
$res = $tmp_res.' '.$unit.$space.$res;
}
$space = '';
if(substr($res,-1) != ' ' && $res != '')
$space = ' ';
if($res)
$res .= $space.$real_name.(($real > 1 && $real_name != '')?'s':'');
if($decimal > 0)
$res .= ' '.num_to_words($decimal, '', 0, '').' '.$decimal_name.(($decimal > 1 && $decimal_name != '')?'s':'');
return ucfirst($res);
}
//////////// testing ////////////////
$str2num = 12;
while($str2num!=4){
$str = num_to_words($str2num, '', 0, '');
$str2num = strlen($str)-1;
echo $str . '=' . $str2num .'<br/>';
if ($str2num == 4)
echo 'four is magic';
}
////// Results /////////
Twelve =6
Six =3
Three =5
Five =4
four is magic
Perl - 130 chars
5.12.1 (130 chars) 121 123 132 136 140
# 1 2 3 4 5 6 7 8 9 100 11 12 13 14
#23456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123
#u='4335443554366887798866555766'=~/./g;$_=pop;say"$_ is ",$_=$_-4?$_<20?$u[$_]:$u[$_/10+18]+(($_%=10)&&$u[$_]):magic,"."until/\D/
5.10.1 (134 chars) 125 127 136 140 144
# 1 2 3 4 5 6 7 8 9 100 11 12 13 14
#23456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 1234
#u='4335443554366887798866555766'=~/./g;$_=pop;print"$_ is ",$_=$_-4?$_<20?$u[$_]:$u[$_/10+18]+(($_%=10)&&$u[$_]):magic,".\n"until/\D/
Change History:
20100714:2223 - reverted change at the attention of mobrule, but ($_%10&&$u[$_%10]) → (($_%=10)&&$u[$_]), which is the same # of chars, but I did it in case someone might see a way to improve it
20100714:0041 - split//,'...' → '...'=~/./g
20100714:0025 - ($_%10&&$u[$_%10]) → $u[$_%10]
20100713:2340 - while$_ → until/\D/ + removed unnecessary parentheses
20100713:xxxx - $=<>;chop; → $_=pop; - courtesy to mobrule
Note: I was tired of improving others' answers in comments, so now I'm being greedy and can just add my changes here :) This is a split off from Platinum Azure's answer - credit in part to Hobbs, mobrule, and Platinum Azure.
Shameless Perl with Number Words (329 characters)
Adapted fairly directly from P Daddy's C code, with some tweaks to p() to make it do the same thing using Perl primitives instead of C ones, and a mostly-rewritten mainloop. See his for an explanation. Newlines are all optional.
#t=(qw(zero one two three four five six sM eight nine
tL elM twelve NP 4P fifP 6P 7P 8O 9P twLQ NQ forQ fifQ
6Q 7Q 8y 9Q en evL thir eL tO ty 4SmagicT)," is ",".\n");
sub p{local$_=$t[pop];1while s/[0-Z]/$t[-48+ord$&]/e;
print;length}$_=<>;chop;while($_-4){
$_=($_>19?(p($_/10+18),$_&&print("-"),$_%=10)[0]:0)+p$_;
p 35;p$_;p 36}p 34
Side note: it's too bad that perl print just returns true/false; if it returned a count it would save me 7 strokes.
Ruby, 141 chars:
n=gets.to_i;m="4335443554366887798866555766";loop{s=n;n=n>20?m[18+n/10]+m[n%10]-96: m[n]-48;puts"#{s} is #{n==s ? 'magic': n}.";n==s &&break}
while(true)
{
string a;
ReadLine(a)
WriteLine(4);
}