Zeckendorf arithmetic: Difference between revisions
Thundergnat (talk | contribs) (→{{header|Perl 6}}: Converted to a completely string based implementation) |
(Promoted) |
||
| Line 1: | Line 1: | ||
{{ |
{{task}} |
||
This task is a total immersion zeckendorf task, using decimal numbers will attract serious disapprobation. |
This task is a total immersion zeckendorf task, using decimal numbers will attract serious disapprobation. |
||
| Line 382: | Line 382: | ||
[[Category:Arithmetic operations]] |
[[Category:Arithmetic operations]] |
||
[http://www.example.com link title] |
|||
Revision as of 11:16, 10 April 2013
You are encouraged to solve this task according to the task description, using any language you may know.
This task is a total immersion zeckendorf task, using decimal numbers will attract serious disapprobation.
The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions.
- Addition
Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case.
Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100;
- Subtraction
10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg:
abcde
10100 -
1000
_____
100 borrow 1 from a leaves 100
+ 100 add the carry
_____
1001
A larger example:
abcdef
100100 -
1000
______
1*0100 borrow 1 from b
+ 100 add the carry
______
1*1001
Sadly we borrowed 1 from b which didn't have it to lend. So now b borrows from a:
1001
+ 1000 add the carry
____
10100
- Multiplication
Here you teach your computer its zeckendorf tables. eg. 101 * 1001:
a = 1 * 101 = 101 b = 10 * 101 = a + a = 10000 c = 100 * 101 = b + a = 10101 d = 1000 * 101 = c + b = 101010 1001 = d + a therefore 101 * 1001 = 101010 + 101 ______ 1000100
- Division
Lets try 1000101 divided by 101, so we can use the same table used for addition.
1000101 -
101010 subtract d (1000 * 101)
_______
1000 -
101 b and c are too large to subtract, so subtract a
____
1 so 1000101 divided by 101 is d + a (1001) remainder 1
C++
<lang cpp>// For a class N which implements Zeckendorf numbers: // I define an increment operation ++() // I define a comparison operation <=(other N) // I define an addition operation +=(other N) // I define a subtraction operation -=(other N) // Nigel Galloway October 28th., 2012
- include <iostream>
enum class zd {N00,N01,N10,N11}; class N { private:
int dVal = 0, dLen;
void _a(int i) {
for (;; i++) {
if (dLen < i) dLen = i;
switch ((zd)((dVal >> (i*2)) & 3)) {
case zd::N00: case zd::N01: return;
case zd::N10: if (((dVal >> ((i+1)*2)) & 1) != 1) return;
dVal += (1 << (i*2+1)); return;
case zd::N11: dVal &= ~(3 << (i*2)); _b((i+1)*2);
}}}
void _b(int pos) {
if (pos == 0) {++*this; return;}
if (((dVal >> pos) & 1) == 0) {
dVal += 1 << pos;
_a(pos/2);
if (pos > 1) _a((pos/2)-1);
} else {
dVal &= ~(1 << pos);
_b(pos + 1);
_b(pos - ((pos > 1)? 2:1));
}}
void _c(int pos) {
if (((dVal >> pos) & 1) == 1) {dVal &= ~(1 << pos); return;}
_c(pos + 1);
if (pos > 0) _b(pos - 1); else ++*this;
return;
}
public:
N(char const* x = "0") {
int i = 0, q = 1;
for (; x[i] > 0; i++);
for (dLen = --i/2; i >= 0; i--) {dVal+=(x[i]-48)*q; q*=2;
}}
const N& operator++() {dVal += 1; _a(0); return *this;}
const N& operator+=(const N& other) {
for (int GN = 0; GN < (other.dLen + 1) * 2; GN++) if ((other.dVal >> GN) & 1 == 1) _b(GN);
return *this;
}
const N& operator-=(const N& other) {
for (int GN = 0; GN < (other.dLen + 1) * 2; GN++) if ((other.dVal >> GN) & 1 == 1) _c(GN);
for (;((dVal >> dLen*2) & 3) == 0 or dLen == 0; dLen--);
return *this;
}
const N& operator*=(const N& other) {
N Na = other, Nb = other, Nt, Nr;
for (int i = 0; i <= (dLen + 1) * 2; i++) {
if (((dVal >> i) & 1) > 0) Nr += Nb;
Nt = Nb; Nb += Na; Na = Nt;
}
return *this = Nr;
}
const bool operator<=(const N& other) const {return dVal <= other.dVal;}
friend std::ostream& operator<<(std::ostream&, const N&);
}; N operator "" N(char const* x) {return N(x);} std::ostream &operator<<(std::ostream &os, const N &G) {
const static std::string dig[] {"00","01","10"}, dig1[] {"","1","10"};
if (G.dVal == 0) return os << "0";
os << dig1[(G.dVal >> (G.dLen*2)) & 3];
for (int i = G.dLen-1; i >= 0; i--) os << dig[(G.dVal >> (i*2)) & 3];
return os;
} </lang>
Testing
The following tests addtition: <lang cpp>int main(void) {
N G; G = 10N; G += 10N; std::cout << G << std::endl; G += 10N; std::cout << G << std::endl; G += 1001N; std::cout << G << std::endl; G += 1000N; std::cout << G << std::endl; G += 10101N; std::cout << G << std::endl; return 0;
}</lang>
- Output:
101 1001 10101 100101 1010000
The following tests subtraction: <lang cpp>int main(void) {
N G; G = 1000N; G -= 101N; std::cout << G << std::endl; G = 10101010N; G -= 1010101N; std::cout << G << std::endl; return 0;
}</lang>
- Output:
1 1000000
The following tests multiplication: <lang cpp> int main(void) {
N G = 1001N; G *= 101N; std::cout << G << std::endl;
G = 101010N; G += 101N; std::cout << G << std::endl; return 0;
}</lang>
- Output:
1000100 1000100
Perl 6
This is a somewhat limited implementation of Zeckendorf arithmetic operators. They only handle positive integer values. There are no actual calculations, everything is done with string manipulations, so it doesn't matter what glyphs you use for 1 and 0.
Implemented arithmetic operators:
addition: +z subtraction: -z multiplication: *z division: /z (more of a divmod really) post increment: ++z post decrement: --z
Comparison operators:
equal eqz not equal nez greater than gtz less than ltz
<lang perl6>my $z1 = '1'; # glyph to use for a '1' my $z0 = '0'; # glyph to use for a '0'
- helper sub to translate constants into the particular glyphs you used
sub z($a) { $a.trans([<1 0>] => [$z1, $z0]) };
- Zeckendorf comparison operators #########
- less than
sub infix:<ltz>($a, $b) { ($z0 lt $z1) ?? ($a lt $b) !!
($a.trans([$z1, $z0] => [<1 0>]) lt $b.trans([$z1, $z0] => [<1 0>]))
};
- greater than
sub infix:<gtz>($a, $b) { ($z0 lt $z1) ?? ($a gt $b) !!
($a.trans([$z1, $z0] => [<1 0>]) gt $b.trans([$z1, $z0] => [<1 0>]))
};
- equal
sub infix:<eqz>($a, $b) { $a eq $b };
- not equal
sub infix:<nez>($a, $b) { $a ne $b };
- Operators for Zeckendorf arithmetic ########
- post increment
sub postfix:<++z>($a is rw) {
$a = ("$z0$z0"~$a).subst(/("$z0$z0")($z1+ %% $z0)?$/, -> $/ { "$z0$z1" ~ $z0 x $1.chars });
$a ~~ s/^$z0+//;
$a
}
- post decrement
sub postfix:<--z>($a is rw) {
$a.=subst(/$z1($z0*)$/, -> $/ {$z0 ~ "$z1$z0" x $0.chars div 2 ~ $z1 x $0.chars mod 2});
$a ~~ s/^$z0+(.+)$/$0/;
$a
}
- addition
sub infix:<+z>($a is copy, $b is copy) { $a++z while $b--z nez $z0; $a };
- subtraction
sub infix:<-z>($a is copy, $b is copy) { $a--z while $b--z nez $z0; $a };
- multiplication
sub infix:<*z>($a, $b) {
return $z0 if $a eq $z0 or $b eq $z0;
return $a if $b eq $z1;
return $b if $a eq $z1;
my $c = $a;
my $d = $z1;
repeat {
my $e = $z0;
repeat { $c++z; $e++z } until $e eqz $a;
$d++z;
} until $d eqz $b;
$c
};
- division (really more of a div mod)
sub infix:</z>($a is copy, $b is copy) {
fail "Divide by zero" if $b eqz $z0;
return $a if $a eqz $z0 or $b eqz $z1;
my $c = $z0;
repeat {
my $d = $b +z ($z1 ~ $z0);
$c++z;
$a--z while $d--z nez $z0
} until $a ltz $b;
$c ~= " remainder $a" if $a nez $z0;
$c
};
- Testing ######################
say "Using the glyph '$z1' for 1 and '$z0' for 0\n";
my $fmt = "%-22s = %15s %s\n";
my $zeck = $z1;
printf( $fmt, "$zeck++z", $zeck++z, '# increment' ) for 1 .. 10;
printf $fmt, "$zeck +z {z('1010')}", $zeck +z= z('1010'), '# addition';
printf $fmt, "$zeck -z {z('100')}", $zeck -z= z('100'), '# subtraction';
printf $fmt, "$zeck *z {z('100101')}", $zeck *z= z('100101'), '# multiplication';
printf $fmt, "$zeck /z {z('100')}", $zeck /z= z('100'), '# division';
printf( $fmt, "$zeck--z", $zeck--z, '# decrement' ) for 1 .. 5;
printf $fmt, "$zeck *z {z('101001')}", $zeck *z= z('101001'), '# multiplication';
printf $fmt, "$zeck /z {z('100')}", $zeck /z= z('100'), '# division'; </lang>
Testing Output
Using the glyph '1' for 1 and '0' for 0 1++z = 10 # increment 10++z = 100 # increment 100++z = 101 # increment 101++z = 1000 # increment 1000++z = 1001 # increment 1001++z = 1010 # increment 1010++z = 10000 # increment 10000++z = 10001 # increment 10001++z = 10010 # increment 10010++z = 10100 # increment 10100 +z 1010 = 100101 # addition 100101 -z 100 = 100010 # subtraction 100010 *z 100101 = 100001000001 # multiplication 100001000001 /z 100 = 101010001 # division 101010001--z = 101010000 # decrement 101010000--z = 101001010 # decrement 101001010--z = 101001001 # decrement 101001001--z = 101001000 # decrement 101001000--z = 101000101 # decrement 101000101 *z 101001 = 101010000010101 # multiplication 101010000010101 /z 100 = 1001010001001 remainder 10 # division
Output using 'X' for 1 and 'O' for 0:
Using the glyph 'X' for 1 and 'O' for 0 X++z = XO # increment XO++z = XOO # increment XOO++z = XOX # increment XOX++z = XOOO # increment XOOO++z = XOOX # increment XOOX++z = XOXO # increment XOXO++z = XOOOO # increment XOOOO++z = XOOOX # increment XOOOX++z = XOOXO # increment XOOXO++z = XOXOO # increment XOXOO +z XOXO = XOOXOX # addition XOOXOX -z XOO = XOOOXO # subtraction XOOOXO *z XOOXOX = XOOOOXOOOOOX # multiplication XOOOOXOOOOOX /z XOO = XOXOXOOOX # division XOXOXOOOX--z = XOXOXOOOO # decrement XOXOXOOOO--z = XOXOOXOXO # decrement XOXOOXOXO--z = XOXOOXOOX # decrement XOXOOXOOX--z = XOXOOXOOO # decrement XOXOOXOOO--z = XOXOOOXOX # decrement XOXOOOXOX *z XOXOOX = XOXOXOOOOOXOXOX # multiplication XOXOXOOOOOXOXOX /z XOO = XOOXOXOOOXOOX remainder XO # division