Long multiplication
Explicitly implement long multiplication.
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
This is one possible approach to arbitrary-precision integer algebra.
For output, display the result of 264 * 264.
Optionally, verify your result against builtin arbitrary precision support.
The decimal representation of 264 is:
18,446,744,073,709,551,616
The output of 264 * 264 is 2128, and is:
340,282,366,920,938,463,463,374,607,431,768,211,456
11l
<lang 11l>F add_with_carry(&result, =addend, =addendpos)
L
L result.len < addendpos + 1
result.append(‘0’)
V addend_result = String(Int(addend) + Int(result[addendpos]))
V addend_digits = Array(addend_result)
result[addendpos] = addend_digits.pop()
I addend_digits.empty
L.break
addend = addend_digits.pop()
addendpos++
F longhand_multiplication(multiplicand, multiplier)
[Char] result
L(multiplicand_digit) reversed(multiplicand)
V multiplicand_offset = L.index
L(multiplier_digit) reversed(multiplier)
V multiplier_offset = L.index + multiplicand_offset
V multiplication_result = String(Int(multiplicand_digit) * Int(multiplier_digit))
L(result_digit_addend) reversed(multiplication_result)
V addend_offset = L.index + multiplier_offset
add_with_carry(&result, result_digit_addend, addend_offset)
result.reverse() R result.join(‘’)
V sixtyfour = ‘18446744073709551616’ print(longhand_multiplication(sixtyfour, sixtyfour))</lang>
- Output:
340282366920938463463374607431768211456
360 Assembly
For maximum compatibility, we use only the basic 370 instruction set (use of MVCL). Pseudo-macro instruction XPRNT can be replaced by a WTO. <lang 360asm>LONGINT CSECT
USING LONGINT,R13
SAVEAREA B PROLOG-SAVEAREA(R15)
DC 17F'0'
DC CL8'LONGINT'
PROLOG STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
MVC XX(1),=C'1'
MVC LENXX,=H'1' xx=1
LA R2,64
LOOPII ST R2,RLOOPII do for 64
MVC X-2(LL+2),XX-2 x=xx
MVC Y(1),=C'2'
MVC LENY,=H'1' y=2
BAL R14,LONGMULT
MVC XX-2(LL+2),Z-2 xx=longmult(xx,2) xx=xx*2
L R2,RLOOPII
ELOOPII BCT R2,LOOPII loop
MVC X-2(LL+2),XX-2
MVC Y-2(LL+2),XX-2
BAL R14,LONGMULT
MVC YY-2(LL+2),Z-2 yy=longmult(xx,xx) yy=xx*xx
XPRNT XX,LL output xx
XPRNT YY,LL output yy
RETURN L R13,4(0,R13) epilog
LM R14,R12,12(R13)
XR R15,R15 set return code
BR R14 return to caller
RLOOPII DS F
LONGMULT EQU * function longmult z=(x,y)
MVC LENSHIFT,=H'0' shift=
MVC LENZ,=H'0' z=
LH R6,LENX
LA R6,1(R6) from lenx
XR R8,R8
BCTR R8,0 by -1
LA R9,0 to 1
LOOPI BXLE R6,R8,ELOOPI do i=lenx to 1 by -1
LA R2,X
AR R2,R6 +i
BCTR R2,0
MVC CI,0(R2) ci=substr(x,i,1)
IC R0,CI ni=integer(ci)
N R0,=X'0000000F'
STH R0,NI
MVC LENT,=H'0' t=
SR R0,R0
STH R0,CARRY carry=0
LH R7,LENY
LA R7,1(R7) from lenx
XR R10,R10
BCTR R10,0 by -1
LA R11,0 to 1
LOOPJ1 BXLE R7,R10,ELOOPJ1 do j=leny to 1 by -1
LA R2,Y
AR R2,R7 +j
BCTR R2,0
MVC CJ,0(R2) cj=substr(y,j,1)
IC R0,CJ
N R0,=X'0000000F'
STH R0,NJ nj=integer(cj)
LH R2,NI
MH R2,NJ
AH R2,CARRY
STH R2,NKR nkr=ni*nj+carry
LH R2,NKR
LA R1,10
SRDA R2,32
DR R2,R1
STH R2,NK nk=nkr//10
STH R3,CARRY carry=nkr/10
LH R2,NK
O R2,=X'000000F0'
STC R2,CK ck=string(nk)
MVC TEMP,T
MVC T(1),CK
MVC T+1(LL-1),TEMP
LH R2,LENT
LA R2,1(R2)
STH R2,LENT t=ck!!t
B LOOPJ1 next j
ELOOPJ1 EQU *
LH R2,CARRY
O R2,=X'000000F0'
STC R2,CK ck=string(carry)
MVC TEMP,T
MVC T(1),CK
MVC T+1(LL-1),TEMP
LH R2,LENT
LA R2,1(R2)
STH R2,LENT t=ck!!t
LA R2,T
AH R2,LENT
LH R3,LENSHIFT
LA R4,SHIFT
LH R5,LENSHIFT
MVCL R2,R4
LH R2,LENT
AH R2,LENSHIFT
STH R2,LENT t=t!!shift
IF1 LH R4,LENZ
CH R4,LENT if lenz>lent
BNH ELSE1
LH R2,LENZ then
LA R2,1(R2)
STH R2,L l=lenz+1
B EIF1
ELSE1 LH R2,LENT else
LA R2,1(R2)
STH R2,L l=lent+1
EIF1 EQU *
MVI TEMP,C'0' to
MVC TEMP+1(LL-1),TEMP
LA R2,TEMP
AH R2,L
SH R2,LENZ
LH R3,LENZ
LA R4,Z
LH R5,LENZ
MVCL R2,R4
MVC LENZ,L
MVC Z,TEMP z=right(z,l,'0')
MVI TEMP,C'0' to
MVC TEMP+1(LL-1),TEMP
LA R2,TEMP
AH R2,L
SH R2,LENT
LH R3,LENT
LA R4,T
LH R5,LENT
MVCL R2,R4
MVC LENT,L
MVC T,TEMP t=right(t,l,'0')
MVC LENW,=H'0' w=
SR R0,R0
STH R0,CARRY carry=0
LH R7,L
LA R7,1(R7) from l
XR R10,R10
BCTR R10,0 by -1
LA R11,0 to 1
LOOPJ2 BXLE R7,R10,ELOOPJ2 do j=l to 1 by -1
LA R2,Z
AR R2,R7 +j
BCTR R2,0
MVC CZ,0(R2) cz=substr(z,j,1)
IC R0,CZ
N R0,=X'0000000F'
STH R0,NZ nz=integer(cz)
LA R2,T
AR R2,R7 -j
BCTR R2,0
MVC CT,0(R2) ct=substr(t,j,1)
IC R0,CT
N R0,=X'0000000F'
STH R0,NT nt=integer(ct)
LH R2,NZ
AH R2,NT
AH R2,CARRY
STH R2,NKR nkr=nz+nt+carry
LH R2,NKR
LA R1,10
SRDA R2,32
DR R2,R1
STH R2,NK
STH R3,CARRY nk=nkr//10; carry=nkr/10
LH R2,NK
O R2,=X'000000F0'
STC R2,CK ck=string(nk)
MVC TEMP,W
MVC W(1),CK
MVC W+1(LL-1),TEMP
LH R2,LENW
LA R2,1(R2)
STH R2,LENW w=ck!!w
B LOOPJ2 next j
ELOOPJ2 EQU *
LH R2,CARRY
O R2,=X'000000F0'
STC R2,CK ck=string(carry)
MVC Z(1),CK
MVC Z+1(LL-1),W
LH R2,LENW
LA R2,1(R2)
STH R2,LENZ z=ck!!w
LA R7,0 from 1
LA R10,1 by 1
LH R11,LENZ to lenz
LOOPJ3 BXH R7,R10,ELOOPJ3 do j=1 to lenz
LA R2,Z
AR R2,R7 j
BCTR R2,0
MVC ZJ(1),0(R2) zj=substr(z,j,1)
CLI ZJ,C'0' if zj^='0'
BNE ELOOPJ3 then leave j
B LOOPJ3 next j
ELOOPJ3 EQU * IF2 CH R7,LENZ if j>lenz
BNH EIF2
LH R7,LENZ then j=lenz
EIF2 EQU *
LA R2,TEMP to
LH R3,LENZ
SR R3,R7 -j
LA R3,1(R3)
STH R3,LENTEMP
LA R4,Z from
AR R4,R7 +j
BCTR R4,0
LR R5,R3
MVCL R2,R4
MVC Z-2(LL+2),TEMP-2 z=substr(z,j)
LA R2,SHIFT
AH R2,LENSHIFT
MVI 0(R2),C'0'
LH R3,LENSHIFT
LA R3,1(R3)
STH R3,LENSHIFT shift=shift!!'0'
MVC TEMP,Z
LA R2,TEMP
AH R2,LENZ
MVC 0(2,R2),=C' '
B LOOPI next i
ELOOPI EQU *
MVI TEMP,C' '
LA R2,Z
AH R2,LENZ
LH R3,=AL2(LL)
SH R3,LENZ
LA R4,TEMP
LH R5,=H'1'
ICM R5,8,=C' '
MVCL R2,R4 z=clean(z)
BR R14 end function longmult
L DS H NI DS H NJ DS H NK DS H NZ DS H NT DS H CARRY DS H NKR DS H CI DS CL1 CJ DS CL1 CZ DS CL1 CT DS CL1 CK DS CL1 ZJ DS CL1 LENXX DS H XX DS CL94 LENYY DS H YY DS CL94 LENX DS H X DS CL94 LENY DS H Y DS CL94 LENZ DS H Z DS CL94 LENT DS H T DS CL94 LENW DS H W DS CL94 LENSHIFT DS H SHIFT DS CL94 LENTEMP DS H TEMP DS CL94 LL EQU 94
YREGS
END LONGINT</lang>
- Output:
18446744073709551616 340282366920938463463374607431768211456
Ada
Using properly range-checked integers
(The source text for these examples can also be found on Bitbucket.)
First we specify the required operations and declare our number type as an array of digits (in base 2^16): <lang ada>package Long_Multiplication is
type Number (<>) is private;
Zero : constant Number; One : constant Number;
function Value (Item : in String) return Number; function Image (Item : in Number) return String;
overriding function "=" (Left, Right : in Number) return Boolean;
function "+" (Left, Right : in Number) return Number; function "*" (Left, Right : in Number) return Number;
function Trim (Item : in Number) return Number;
private
Bits : constant := 16; Base : constant := 2 ** Bits;
type Accumulated_Value is range 0 .. (Base - 1) * Base; subtype Digit is Accumulated_Value range 0 .. Base - 1;
type Number is array (Natural range <>) of Digit; for Number'Component_Size use Bits; -- or pragma Pack (Number);
Zero : constant Number := (1 .. 0 => 0); One : constant Number := (0 => 1);
procedure Divide (Dividend : in Number;
Divisor : in Digit;
Result : out Number;
Remainder : out Digit);
end Long_Multiplication;</lang> Some of the operations declared above are useful helper operations for the conversion of numbers to and from base 10 digit strings.
Then we implement the operations: <lang ada>package body Long_Multiplication is
function Value (Item : in String) return Number is
subtype Base_Ten_Digit is Digit range 0 .. 9;
Ten : constant Number := (0 => 10);
begin
case Item'Length is
when 0 =>
raise Constraint_Error;
when 1 =>
return (0 => Base_Ten_Digit'Value (Item));
when others =>
return (0 => Base_Ten_Digit'Value (Item (Item'Last .. Item'Last)))
+ Ten * Value (Item (Item'First .. Item'Last - 1));
end case;
end Value;
function Image (Item : in Number) return String is
Base_Ten : constant array (Digit range 0 .. 9) of String (1 .. 1) :=
("0", "1", "2", "3", "4", "5", "6", "7", "8", "9");
Result : Number (0 .. Item'Last);
Remainder : Digit;
begin
if Item = Zero then
return "0";
else
Divide (Dividend => Item,
Divisor => 10,
Result => Result,
Remainder => Remainder);
if Result = Zero then
return Base_Ten (Remainder);
else
return Image (Trim (Result)) & Base_Ten (Remainder);
end if;
end if;
end Image;
overriding
function "=" (Left, Right : in Number) return Boolean is
begin
for Position in Integer'Min (Left'First, Right'First) ..
Integer'Max (Left'Last, Right'Last) loop
if Position in Left'Range and Position in Right'Range then
if Left (Position) /= Right (Position) then
return False;
end if;
elsif Position in Left'Range then
if Left (Position) /= 0 then
return False;
end if;
elsif Position in Right'Range then
if Right (Position) /= 0 then
return False;
end if;
else
raise Program_Error;
end if;
end loop;
return True; end "=";
function "+" (Left, Right : in Number) return Number is
Result : Number (Integer'Min (Left'First, Right'First) ..
Integer'Max (Left'Last , Right'Last) + 1);
Accumulator : Accumulated_Value := 0;
Used : Integer := Integer'First;
begin
for Position in Result'Range loop
if Position in Left'Range then
Accumulator := Accumulator + Left (Position);
end if;
if Position in Right'Range then
Accumulator := Accumulator + Right (Position);
end if;
Result (Position) := Accumulator mod Base;
Accumulator := Accumulator / Base;
if Result (Position) /= 0 then
Used := Position;
end if;
end loop;
if Accumulator = 0 then
return Result (Result'First .. Used);
else
raise Constraint_Error;
end if;
end "+";
function "*" (Left, Right : in Number) return Number is
Accumulator : Accumulated_Value;
Result : Number (Left'First + Right'First ..
Left'Last + Right'Last + 1) := (others => 0);
Used : Integer := Integer'First;
begin
for L in Left'Range loop
for R in Right'Range loop
Accumulator := Left (L) * Right (R);
for Position in L + R .. Result'Last loop
exit when Accumulator = 0;
Accumulator := Accumulator + Result (Position);
Result (Position) := Accumulator mod Base;
Accumulator := Accumulator / Base;
Used := Position;
end loop;
end loop;
end loop;
return Result (Result'First .. Used); end "*";
procedure Divide (Dividend : in Number;
Divisor : in Digit;
Result : out Number;
Remainder : out Digit) is
Accumulator : Accumulated_Value := 0;
begin
Result := (others => 0);
for Position in reverse Dividend'Range loop
Accumulator := Accumulator * Base + Dividend (Position);
Result (Position) := Accumulator / Divisor;
Accumulator := Accumulator mod Divisor;
end loop;
Remainder := Accumulator; end Divide;
function Trim (Item : in Number) return Number is
begin
for Position in reverse Item'Range loop
if Item (Position) /= 0 then
return Item (Item'First .. Position);
end if;
end loop;
return Zero; end Trim;
end Long_Multiplication;</lang>
And finally we have the requested test application: <lang ada>with Ada.Text_IO; with Long_Multiplication;
procedure Test_Long_Multiplication is
use Ada.Text_IO, Long_Multiplication;
N : Number := Value ("18446744073709551616");
M : Number := N * N;
begin
Put_Line (Image (N) & " * " & Image (N) & " = " & Image (M));
end Test_Long_Multiplication;</lang>
- Output:
18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456
Using modular types
The following implementation uses representation of a long number by an array of 32-bit elements: <lang ada>type Long_Number is array (Natural range <>) of Unsigned_32;
function "*" (Left, Right : Long_Number) return Long_Number is
Result : Long_Number (0..Left'Length + Right'Length - 1) := (others => 0); Accum : Unsigned_64;
begin
for I in Left'Range loop
for J in Right'Range loop
Accum := Unsigned_64 (Left (I)) * Unsigned_64 (Right (J));
for K in I + J..Result'Last loop
exit when Accum = 0;
Accum := Accum + Unsigned_64 (Result (K));
Result (K) := Unsigned_32 (Accum and 16#FFFF_FFFF#);
Accum := Accum / 2**32;
end loop;
end loop;
end loop;
for Index in reverse Result'Range loop -- Normalization
if Result (Index) /= 0 then
return Result (0..Index);
end if;
end loop;
return (0 => 0);
end "*";</lang>
The task requires conversion into decimal base. For this we also need division to short number with a remainder. Here it is: <lang ada>procedure Div
( Dividend : in out Long_Number;
Last : in out Natural;
Remainder : out Unsigned_32;
Divisor : Unsigned_32
) is
Div : constant Unsigned_64 := Unsigned_64 (Divisor);
Accum : Unsigned_64 := 0;
Size : Natural := 0;
begin
for Index in reverse Dividend'First..Last loop
Accum := Accum * 2**32 + Unsigned_64 (Dividend (Index));
Dividend (Index) := Unsigned_32 (Accum / Div);
if Size = 0 and then Dividend (Index) /= 0 then
Size := Index;
end if;
Accum := Accum mod Div;
end loop;
Remainder := Unsigned_32 (Accum);
Last := Size;
end Div;</lang>
With the above the test program: <lang ada>with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO; use Ada.Text_IO; with Interfaces; use Interfaces;
procedure Long_Multiplication is
-- Insert definitions above here
procedure Put (Value : Long_Number) is
X : Long_Number := Value;
Last : Natural := X'Last;
Digit : Unsigned_32;
Result : Unbounded_String;
begin
loop
Div (X, Last, Digit, 10);
Append (Result, Character'Val (Digit + Character'Pos ('0')));
exit when Last = 0 and then X (0) = 0;
end loop;
for Index in reverse 1..Length (Result) loop
Put (Element (Result, Index));
end loop;
end Put;
X : Long_Number := (0 => 0, 1 => 0, 2 => 1) * (0 => 0, 1 => 0, 2 => 1);
begin
Put (X);
end Long_Multiplication;</lang>
Sample output:
340282366920938463463374607431768211456
Aime
<lang aime>data b, c, v; integer d, e, i, j, s;
b = 1.argv; b.dump(','); v = 2.argv; v.dump(',');
c.run(~b + ~v + 1, 0);
for (i, d in b) {
b[i] = d - '0';
}
for (j, d of v) {
d = v[j] - '0';
s = 0;
for (i, e of b) {
s += e * d + c[i + j];
c[i + j] = s % 10;
s /= 10;
}
while (s) {
s += c[i + j];
c[i + j] = s % 10;
s /= 10;
i -= 1;
}
}
c.delete(-1); c.bf_drop0("");
for (i, d in c) {
c[i] = d + '0';
}
o_form("~\n", c);</lang>
ALGOL 68
The long multiplication for the golden ratio has been included as half the digits cancel and end up as being zero. This is useful for testing.
Built in or standard distribution routines
ALGOL 68G allows any precision for long long int to be defined when the program is run, e.g. 200 digits. <lang algol68>PRAGMAT precision=200 PRAGMAT MODE INTEGER = LONG LONG INT;
LONG INT default integer width := 69; INT width = 69+2;
INT fix w = 1, fix h = 1; # round up #
LONG LONG INT golden ratio w := ENTIER ((long long sqrt(5)-1) / 2 * LENG LENG 10 ** default integer width + fix w),
golden ratio h := ENTIER ((long long sqrt(5)+1) / 2 * LENG LENG 10 ** default integer width + fix h);
test: (
print(( "The approximate golden ratios, width: ", whole(golden ratio w,width), new line, " length: ", whole(golden ratio h,width), new line, " product is exactly: ", whole(golden ratio w*golden ratio h,width*2), new line));
INTEGER two to the power of 64 = LONG 2 ** 64;
INTEGER neg two to the power of 64 = -(LONG 2 ** 64);
print(("2 ** 64 * -(2 ** 64) = ", whole(two to the power of 64*neg two to the power of 64,width), new line))
)</lang> Output:
The approximate golden ratios, width: +618033988749894848204586834365638117720309179805762862135448622705261
length: +1618033988749894848204586834365638117720309179805762862135448622705261
product is exactly: +1000000000000000000000000000000000000000000000000000000000000000000001201173450350400438606015942314498798603569682901026716145698077078121
2 ** 64 * -(2 ** 64) = -340282366920938463463374607431768211456
Implementation example
<lang algol68>MODE DIGIT = INT; MODE INTEGER = FLEX[0]DIGIT; # an arbitary number of digits #
- "digits" are stored in digit base ten, but 10000 & 2**n (inc hex) can be used #
INT digit base = 1000;
- if possible, then print the digit with one character #
STRING hex digit repr = "0123456789abcdefghijklmnopqrstuvwxyz"[AT 0]; INT digit base digit width = ( digit base <= UPB hex digit repr + 1 | 1 | 1 + ENTIER log(digit base-1) );
INT next digit = -1; # reverse order so digits appear in "normal" order when printed #
PROC raise value error = ([]STRING args)VOID:
( print(("Value Error: ", args, new line)); stop );
PROC raise not implemented error = ([]STRING args)VOID:
( print(("Not implemented Error: ", args, new line)); stop );
PROC raise integer not implemented error = (STRING message)INTEGER:
( raise not implemented error(("INTEGER ", message)); SKIP );
INT half max int = max int OVER 2; IF digit base > half max int THEN raise value error("INTEGER addition may fail") FI;
INT sqrt max int = ENTIER sqrt(max int); IF digit base > sqrt max int THEN raise value error("INTEGER multiplication may fail") FI;
- initialise/cast a INTEGER from a LONG LONG INT #
OP INTEGERINIT = (LONG LONG INT number)INTEGER:(
[1 + ENTIER (SHORTEN SHORTEN long long log(ABS number) / log(digit base))]DIGIT out; LONG LONG INT carry := number; FOR digit out FROM UPB out BY next digit TO LWB out DO LONG LONG INT prev carry := carry; carry %:= digit base; # avoid MOD as it doesn't under handle -ve numbers # out[digit out] := SHORTEN SHORTEN (prev carry - carry * digit base) OD; out
);
- initialise/cast a INTEGER from an LONG INT #
OP INTEGERINIT = (LONG INT number)INTEGER: INTEGERINIT LENG number;
- initialise/cast a INTEGER from an INT #
OP INTEGERINIT = (INT number)INTEGER: INTEGERINIT LENG LENG number;
- remove leading zero "digits" #
OP NORMALISE = ([]DIGIT number)INTEGER: (
INT leading zeros := LWB number - 1; FOR digit number FROM LWB number TO UPB number WHILE number[digit number] = 0 DO leading zeros := digit number OD; IF leading zeros = UPB number THEN 0 ELSE number[leading zeros+1:] FI
);
Define a standard representation for the INTEGER mode. Note: this is rather crude because for a large "digit base" the number is represented as blocks of decimals. It works nicely for powers of ten (10,100,1000,...), but for most larger bases (greater then 35) the repr will be a surprise.
OP REPR = (DIGIT d)STRING:
IF digit base > UPB hex digit repr THEN
STRING out := whole(ABS d, -digit base digit width);
- Replace spaces with zeros #
FOR digit out FROM LWB out TO UPB out DO
IF out[digit out] = " " THEN out[digit out] := "0" FI
OD;
out
ELSE # small enough to represent as ASCII (hex) characters #
hex digit repr[ABS d]
FI;
OP REPR = (INTEGER number)STRING:(
STRING sep = ( digit base digit width > 1 | "," | "" );
INT width := digit base digit width + UPB sep;
[width * UPB number - UPB sep]CHAR out;
INT leading zeros := LWB out - 1;
FOR digit TO UPB number DO
INT start := digit * width - width + 1;
out[start:start+digit base digit width-1] := REPR number[digit];
IF digit base digit width /= 1 & digit /= UPB number THEN
out[start+digit base digit width] := ","
FI
OD;
- eliminate leading zeros #
FOR digit out FROM LWB out TO UPB out WHILE out[digit out] = "0" OR out[digit out] = sep DO leading zeros := digit out OD;
CHAR sign = ( number[1]<0 | "-" | "+" );
- finally return the semi-normalised result #
IF leading zeros = UPB out THEN "0" ELSE sign + out[leading zeros+1:] FI
);</lang>
<lang algol68>################################################################
- Finally Define the required INTEGER multiplication OPerator. #
OP * = (INTEGER a, b)INTEGER:(
- initialise out to all zeros #
[UPB a + UPB b]INT ab; FOR place ab TO UPB ab DO ab[place ab]:=0 OD;
FOR place a FROM UPB a BY next digit TO LWB a DO DIGIT carry := 0;
- calculate each digit (whilst removing the carry) #
FOR place b FROM UPB b BY next digit TO LWB b DO
# n.b. result may be 2 digits #
INT result := ab[place a + place b] + a[place a]*b[place b] + carry;
carry := result % digit base; # avoid MOD as it doesn't under handle -ve numbers #
ab[place a + place b] := result - carry * digit base
OD;
ab[place a + LWB b + next digit] +:= carry
OD; NORMALISE ab
);</lang>
<lang algol68># The following standard operators could (potentially) also be defined # OP - = (INTEGER a)INTEGER: raise integer not implemented error("monadic minus"),
ABS = (INTEGER a)INTEGER: raise integer not implemented error("ABS"),
ODD = (INTEGER a)INTEGER: raise integer not implemented error("ODD"),
BIN = (INTEGER a)INTEGER: raise integer not implemented error("BIN");
OP + = (INTEGER a, b)INTEGER: raise integer not implemented error("addition"),
- = (INTEGER a, b)INTEGER: raise integer not implemented error("subtraction"),
/ = (INTEGER a, b)REAL: ( VOID(raise integer not implemented error("floating point division")); SKIP),
% = (INTEGER a, b)INTEGER: raise integer not implemented error("fixed point division"),
%* = (INTEGER a, b)INTEGER: raise integer not implemented error("modulo division"),
** = (INTEGER a, b)INTEGER: raise integer not implemented error("to the power of");
LONG INT default integer width := long long int width - 2;
INT fix w = -1177584, fix h = -3915074; # floating point error, probably GMP/hardware specific #
INTEGER golden ratio w := INTEGERINIT ENTIER ((long long sqrt(5)-1) / 2 * LENG LENG 10 ** default integer width + fix w),
golden ratio h := INTEGERINIT ENTIER ((long long sqrt(5)+1) / 2 * LENG LENG 10 ** default integer width + fix h);
test: (
print(( "The approximate golden ratios, width: ", REPR golden ratio w, new line, " length: ", REPR golden ratio h, new line, " product is exactly: ", REPR (golden ratio w * golden ratio h), new line));
INTEGER two to the power of 64 = INTEGERINIT(LONG 2 ** 64);
INTEGER neg two to the power of 64 = INTEGERINIT(-(LONG 2 ** 64));
print(("2 ** 64 * -(2 ** 64) = ", REPR (two to the power of 64 * neg two to the power of 64), new line))
)</lang>
Output:
The approximate golden ratios, width: +618,033,988,749,894,848,204,586,834,365,638,117,720,309,179,805,762,862,135,448,622,705,261
length: +1,618,033,988,749,894,848,204,586,834,365,638,117,720,309,179,805,762,862,135,448,622,705,261
product is exactly: +1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,001,201,173,450,350,400,438,606,015,942,314,498,798,603,569,682,901,026,716,145,698,077,078,121
2 ** 64 * -(2 ** 64) = -340,282,366,920,938,463,463,374,607,431,768,211,456
Other libraries or implementation specific extensions
As of February 2009 no open source libraries to do this task have been located.
ALGOL W
<lang algolw>begin
% long multiplication of large integers %
% large integers are represented by arrays of integers whose absolute %
% values are in 0 .. ELEMENT_MAX - 1 %
% negative large integers should have negative values in all non-zero %
% elements %
% the least significant digits of the large integer are in element 1 %
integer ELEMENT_DIGITS; % number of digits in an element of a large %
% integer %
integer ELEMENT_MAX; % max absolute value of an element of a large %
% integer - must be 10^( ELEMENT_DIGITS + 1 ) %
integer ELEMENT_COUNT; % number of elements in each large integer %
% implements long multiplication, c is set to a * b %
% c can be the same array as a or b %
% n is the number of elements in the large integers a, b and c %
procedure longMultiply( integer array a, b, c ( * )
; integer value n
) ;
begin
% multiplies the large integer in b by the integer a, the result %
% is added to c, starting from offset %
% overflow is ignored %
procedure multiplyElement( integer value a
; integer array b, c ( * )
; integer value offset, n
) ;
begin
integer carry, cPos;
carry := 0;
cPos := offset;
for bPos := 1 until highestNonZeroElementPosition( b, ( n + 1 ) - offset ) do begin
integer cElement;
cElement := c( cPos ) + ( a * b( bPos ) ) + carry;
if abs cElement < ELEMENT_MAX then carry := 0
else begin
% have digits to carry %
carry := cElement div ELEMENT_MAX;
cElement := ( abs cElement ) rem ELEMENT_MAX;
if carry < 0 then cElement := - cElement
end if_no_carry_ ;
c( cPos ) := cElement;
cPos := cPos + 1
end for_aPos ;
if cPos <= n then c( cPos ) := carry
end multiplyElement ;
integer array mResult ( 1 :: n );
% the result will be computed in mResult, allowing a or b to be c %
for rPos := 1 until n do mResult( rPos ) := 0;
% multiply and add each element to the result %
for aPos := 1 until highestNonZeroElementPosition( a, n ) do begin
if a( aPos ) not = 0 then multiplyElement( a( aPos ), b, mResult, aPos, n )
end for_aPos ;
% return the result in c %
for rPos := 1 until n do c( rPos ) := mResult( rPos )
end longMultiply ;
% writes the decimal value of a large integer a with n elements %
procedure writeonLargeInteger( integer array a ( * )
; integer value n
) ;
begin
integer aMax;
aMax := highestNonZeroElementPosition( a, n );
if aMax < 1 then writeon( "0" )
else begin
% the large integer is non-zero %
writeon( i_w := 1, s_w := 0, a( aMax ) ); % highest element %
% handle the remaining elements - show leading zeros %
for aPos := aMax - 1 step -1 until 1 do begin
integer v;
integer array digits ( 1 :: ELEMENT_DIGITS );
v := abs a( aPos );
for dPos := ELEMENT_DIGITS step -1 until 1 do begin
digits( dPos ) := v rem 10;
v := v div 10
end for_dPos;
for dPos := 1 until ELEMENT_DIGITS do writeon( i_w := 1, s_w := 0, digits( dPos ) )
end for_aPos
end if_aMax_lt_1_
end writeonLargeInteger ;
% returns the position of the highest non-zero element of the large %
% integer a with n elements %
integer procedure highestNonZeroElementPosition( integer array a ( * )
; integer value n
) ;
begin
integer aMax;
aMax := n;
while aMax > 0 and a( aMax ) = 0 do aMax := aMax - 1;
aMax
end highestNonZeroElementPosition ;
% allow each element to contain 4 decimal digits, so element by element %
% multiplication won't overflow 32-bits %
ELEMENT_DIGITS := 4;
ELEMENT_MAX := 10000;
ELEMENT_COUNT := 12; % allows up to 48 digits - enough for the task %
begin
integer array twoTo64, twoTo128 ( 1 :: ELEMENT_COUNT );
integer pwr;
% construct 2^64 in twoTo64 %
for tPos := 2 until ELEMENT_COUNT do twoTo64( tPos ) := 0;
twoTo64( 1 ) := 2;
pwr := 1;
while pwr < 64 do begin
longMultiply( twoTo64, twoTo64, twoTo64, ELEMENT_COUNT );
pwr := pwr * 2
end while_pwr_lt_64 ;
% construct 2^128 %
longMultiply( twoTo64, twoTo64, twoTo128, ELEMENT_COUNT );
write( "2^128: " );
writeonLargeInteger( twoTo128, ELEMENT_COUNT )
end
end.</lang>
- Output:
2^128: 340282366920938463463374607431768211456
AutoHotkey
ahk discussion <lang autohotkey>MsgBox % x := mul(256,256) MsgBox % x := mul(x,x) MsgBox % x := mul(x,x) ; 18446744073709551616 MsgBox % x := mul(x,x) ; 340282366920938463463374607431768211456
mul(b,c) { ; <- b*c
VarSetCapacity(a, n:=StrLen(b)+StrLen(c), 48), NumPut(0,a,n,"char")
Loop % StrLen(c) {
i := StrLen(c)+1-A_Index, cy := 0
Loop % StrLen(b) {
j := StrLen(b)+1-A_Index,
t := SubStr(a,i+j,1) + SubStr(b,j,1) * SubStr(c,i,1) + cy
cy := t // 10
NumPut(mod(t,10)+48,a,i+j-1,"char")
}
NumPut(cy+48,a,i+j-2,"char")
}
Return cy ? a : SubStr(a,2)
}</lang>
AWK
<lang awk>BEGIN {
DEBUG = 0
n = 2^64
nn = sprintf("%.0f", n)
printf "2^64 * 2^64 = %.0f\n", multiply(nn, nn)
printf "2^64 * 2^64 = %.0f\n", n*n
exit
}
function multiply(x, y, len_x,len_y,ax,ay,j,m,c,i,k,d,v,res,mul,result) {
len_x = split_reverse(x, ax)
len_y = split_reverse(y, ay)
print_array(ax)
print_array(ay)
for (j=1; j<=len_y; j++) {
m = ay[j]
c = 0
i = j - 1
for (k=1; k<=len_x; k++) {
d = ax[k]
i++
v = res[i]
if (v == "") {
append_array(res, 0)
v = 0
}
mul = v + c + d*m
c = int(mul / 10)
v = mul % 10
res[i] = v
}
append_array(res, c)
}
print_array(res)
result = reverse_join(res)
sub(/^0+/, "", result)
return result
}
function split_reverse(x, a, a_x) {
split(x, a_x, "") return reverse_array(a_x, a)
}
function reverse_array(a,b, len,i) {
len = length_array(a)
for (i in a) {
b[1+len-i] = a[i]
}
return len
}
function length_array(a, len,i) {
len = 0 for (i in a) len++ return len
}
function append_array(a, value, len) {
len = length_array(a) a[++len] = value
}
function reverse_join(a, len,str,i) {
len = length_array(a)
str = ""
for (i=len; i>=1; i--) {
str = str a[i]
}
return str
}
function print_array(a, len,i) {
if (DEBUG) {
len = length_array(a)
print "length=" len
for (i=1; i<=len; i++) {
printf("%s ", i%10)
}
print ""
for (i=1; i<=len; i++) {
#print i " " a[i]
printf("%s ", a[i])
}
print ""
print "===="
}
}</lang> outputs:
2^64 * 2^64 = 340282366920938463463374607431768211456 2^64 * 2^64 = 340282366920938463463374607431768211456
BASIC
Version 1
<lang qbasic>'PROGRAM : BIG MULTIPLICATION VER #1 'LRCVS 01.01.2010 'THIS PROGRAM SIMPLY MAKES A MULTIPLICATION 'WITH ALL THE PARTIAL PRODUCTS. '............................................................
DECLARE SUB A.INICIO (A$, B$) DECLARE SUB B.STORE (CAD$, N$) DECLARE SUB C.PIZARRA () DECLARE SUB D.ENCABEZADOS (A$, B$) DECLARE SUB E.MULTIPLICACION (A$, B$) DECLARE SUB G.SUMA () DECLARE FUNCTION F.INVCAD$ (CAD$)
RANDOMIZE TIMER CALL A.INICIO(A$, B$) CALL B.STORE(A$, "A") CALL B.STORE(B$, "B") CALL C.PIZARRA CALL D.ENCABEZADOS(A$, B$) CALL E.MULTIPLICACION(A$, B$) CALL G.SUMA
SUB A.INICIO (A$, B$)
CLS
'Note: Number of digits > 1000
INPUT "NUMBER OF DIGITS "; S
CLS
A$ = ""
B$ = ""
FOR N = 1 TO S
A$ = A$ + LTRIM$(STR$(INT(RND * 9)))
NEXT N
FOR N = 1 TO S
B$ = B$ + LTRIM$(STR$(INT(RND * 9)))
NEXT N
END SUB
SUB B.STORE (CAD$, N$)
OPEN "O", #1, N$
FOR M = LEN(CAD$) TO 1 STEP -1
WRITE #1, MID$(CAD$, M, 1)
NEXT M
CLOSE (1)
END SUB
SUB C.PIZARRA
OPEN "A", #3, "R" WRITE #3, "" CLOSE (3) KILL "R"
END SUB
SUB D.ENCABEZADOS (A$, B$)
LT = LEN(A$) + LEN(B$) + 1 L$ = STRING$(LT, " ") OPEN "A", #3, "R" MID$(L$, LT - LEN(A$) + 1) = A$ WRITE #3, L$ CLOSE (3) L$ = STRING$(LT, " ") OPEN "A", #3, "R" MID$(L$, LT - LEN(B$) - 1) = "X " + B$ WRITE #3, L$ CLOSE (3)
END SUB
SUB E.MULTIPLICACION (A$, B$)
LT = LEN(A$) + LEN(B$) + 1
L$ = STRING$(LT, " ")
C$ = ""
D$ = ""
E$ = ""
CT1 = 1
ACUM = 0
OPEN "I", #2, "B"
WHILE EOF(2) <> -1
INPUT #2, B$
OPEN "I", #1, "A"
WHILE EOF(1) <> -1
INPUT #1, A$
RP = (VAL(A$) * VAL(B$)) + ACUM
C$ = LTRIM$(STR$(RP))
IF EOF(1) <> -1 THEN D$ = D$ + RIGHT$(C$, 1)
IF EOF(1) = -1 THEN D$ = D$ + F.INVCAD$(C$)
E$ = LEFT$(C$, LEN(C$) - 1)
ACUM = VAL(E$)
WEND
CLOSE (1)
MID$(L$, LT - CT1 - LEN(D$) + 2) = F.INVCAD$(D$)
OPEN "A", #3, "R"
WRITE #3, L$
CLOSE (3)
L$ = STRING$(LT, " ")
ACUM = 0
C$ = ""
D$ = ""
E$ = ""
CT1 = CT1 + 1
WEND
CLOSE (2)
END SUB
FUNCTION F.INVCAD$ (CAD$)
LCAD = LEN(CAD$)
CADTEM$ = ""
FOR CAD = LCAD TO 1 STEP -1
CADTEM$ = CADTEM$ + MID$(CAD$, CAD, 1)
NEXT CAD
F.INVCAD$ = CADTEM$
END FUNCTION
SUB G.SUMA
CF = 0
OPEN "I", #3, "R"
WHILE EOF(3) <> -1
INPUT #3, R$
CF = CF + 1
AN = LEN(R$)
WEND
CF = CF - 2
CLOSE (3)
W$ = ""
ST = 0
ACUS = 0
FOR P = 1 TO AN
K = 0
OPEN "I", #3, "R"
WHILE EOF(3) <> -1
INPUT #3, R$
K = K + 1
IF K > 2 THEN ST = ST + VAL(MID$(R$, AN - P + 1, 1))
IF K > 2 THEN M$ = LTRIM$(STR$(ST + ACUS))
WEND
'COLOR 10: LOCATE CF + 3, AN - P + 1: PRINT RIGHT$(M$, 1); : COLOR 7
W$ = W$ + RIGHT$(M$, 1)
ACUS = VAL(LEFT$(M$, LEN(M$) - 1))
CLOSE (3)
ST = 0
NEXT P
OPEN "A", #3, "R" WRITE #3, " " + RIGHT$(F.INVCAD(W$), AN - 1) CLOSE (3) CLS PRINT "THE SOLUTION IN THE FILE: R"
END SUB</lang>
Version 2
<lang qbasic>'PROGRAM: BIG MULTIPLICATION VER # 2 'LRCVS 01/01/2010 'THIS PROGRAM SIMPLY MAKES A BIG MULTIPLICATION 'WITHOUT THE PARTIAL PRODUCTS. 'HERE SEE ONLY THE SOLUTION. '............................................................... CLS PRINT "WAIT"
NA = 2000 'NUMBER OF ELEMENTS OF THE MULTIPLY. NB = 2000 'NUMBER OF ELEMENTS OF THE MULTIPLIER. 'Solution = 4000 Exacts digits
'...................................................... OPEN "X" + ".MLT" FOR BINARY AS #1 CLOSE (1) KILL "*.MLT" '..................................................... 'CREATING THE MULTIPLY >>> A 'CREATING THE MULTIPLIER >>> B FOR N = 1 TO 2 IF N = 1 THEN F$ = "A" + ".MLT": NN = NA IF N = 2 THEN F$ = "B" + ".MLT": NN = NB
OPEN F$ FOR BINARY AS #1
FOR N2 = 1 TO NN
RANDOMIZE TIMER
X$ = LTRIM$(STR$(INT(RND * 10)))
SEEK #1, N2: PUT #1, N2, X$
NEXT N2
SEEK #1, N2
CLOSE (1)
NEXT N '..................................................... OPEN "A" + ".MLT" FOR BINARY AS #1 FOR K = 0 TO 9 NUM$ = "": Z$ = "": ACU = 0: GG = NA C$ = LTRIM$(STR$(K))
OPEN C$ + ".MLT" FOR BINARY AS #2
'OPEN "A" + ".MLT" FOR BINARY AS #1
FOR N = 1 TO NA
SEEK #1, GG: GET #1, GG, X$
NUM$ = X$
Z$ = LTRIM$(STR$(ACU + (VAL(X$) * VAL(C$))))
L = LEN(Z$)
ACU = 0
IF L = 1 THEN NUM$ = Z$: PUT #2, N, NUM$
IF L > 1 THEN ACU = VAL(LEFT$(Z$, LEN(Z$) - 1)): NUM$ = RIGHT$(Z$, 1): PUT #2, N, NUM$
SEEK #2, N: PUT #2, N, NUM$
GG = GG - 1
NEXT N
IF L > 1 THEN ACU = VAL(LEFT$(Z$, LEN(Z$) - 1)): NUM$ = LTRIM$(STR$(ACU)): XX$ = XX$ + NUM$: PUT #2, N, NUM$
'CLOSE (1)
CLOSE (2)
NEXT K CLOSE (1) '...................................................... ACU = 0 LT5 = 1 LT6 = LT5 OPEN "B" + ".MLT" FOR BINARY AS #1
OPEN "D" + ".MLT" FOR BINARY AS #3
FOR JB = NB TO 1 STEP -1
SEEK #1, JB
GET #1, JB, X$
OPEN X$ + ".MLT" FOR BINARY AS #2: LF = LOF(2): CLOSE (2)
OPEN X$ + ".MLT" FOR BINARY AS #2
FOR KB = 1 TO LF
SEEK #2, KB
GET #2, , NUM$
SEEK #3, LT5
GET #3, LT5, PR$
T$ = ""
T$ = LTRIM$(STR$(ACU + VAL(NUM$) + VAL(PR$)))
PR$ = RIGHT$(T$, 1)
ACU = 0
IF LEN(T$) > 1 THEN ACU = VAL(LEFT$(T$, LEN(T$) - 1))
SEEK #3, LT5: PUT #3, LT5, PR$
LT5 = LT5 + 1
NEXT KB
IF ACU <> 0 THEN PR$ = LTRIM$(STR$(ACU)): PUT #3, LT5, PR$
CLOSE (2)
LT6 = LT6 + 1
LT5 = LT6
ACU = 0
NEXT JB
CLOSE (3)
CLOSE (1) OPEN "D" + ".MLT" FOR BINARY AS #3: LD = LOF(3): CLOSE (3) ER = 1 OPEN "D" + ".MLT" FOR BINARY AS #3
OPEN "R" + ".MLT" FOR BINARY AS #4
FOR N = LD TO 1 STEP -1
SEEK #3, N: GET #3, N, PR$
SEEK #4, ER: PUT #4, ER, PR$
ER = ER + 1
NEXT N
CLOSE (4)
CLOSE (3) KILL "D.MLT" FOR N = 0 TO 9
C$ = LTRIM$(STR$(N)) KILL C$ + ".MLT"
NEXT N PRINT "END" PRINT "THE SOLUTION IN THE FILE: R.MLT"</lang>
Applesoft BASIC
<lang ApplesoftBasic> 100 A$ = "18446744073709551616"
110 B$ = A$ 120 GOSUB 400 130 PRINT E$ 140 END
400 REM MULTIPLY A$ * B$ 410 C$ = "":D$ = "0" 420 FOR I = LEN (B$) TO 1 STEP - 1 430 C = 0:B = VAL ( MID$ (B$,I,1)) 440 FOR J = LEN (A$) TO 1 STEP - 1 450 V = B * VAL ( MID$ (A$,J,1)) + C 460 C = INT (V / 10):V = V - C * 10 470 C$ = STR$ (V) + C$ 480 NEXT J 490 IF C THEN C$ = STR$ (C) + C$ 510 GOSUB 600"ADD C$ + D$ 520 D$ = E$:C$ = "0":J = LEN (B$) - I 530 IF J THEN J = J - 1:C$ = C$ + "0": GOTO 530 550 NEXT I 560 RETURN
600 REM ADD C$ + D$ 610 E = LEN (D$):E$ = "":C = 0 620 FOR J = LEN (C$) TO 1 STEP - 1 630 IF E THEN D = VAL ( MID$ (D$,E,1)) 640 V = VAL ( MID$ (C$,J,1)) + D + C 650 C = V > 9:V = V - 10 * C 660 E$ = STR$ (V) + E$ 670 IF E THEN E = E - 1:D = 0 680 NEXT J 700 IF E THEN V = VAL ( MID$ (D$,E,1)) + C:C = V > 9:V = V - 10 * C:E$ = STR$ (V) + E$:E = E - 1: GOTO 700 720 RETURN</lang>
Batch File
Based on the JavaScript iterative code. <lang dos>::Long Multiplication Task from Rosetta Code
- Batch File Implementation
@echo off call :longmul 18446744073709551616 18446744073709551616 answer echo(%answer% exit /b 0
rem The Hellish Procedure rem Syntax: call :longmul <n1> <n2> <variable to store product>
- longmul
setlocal enabledelayedexpansion
rem Define variables set "num1=%1" set "num2=%2" set "limit1=-1" set "limit2=-1" set "length=0" set "prod="
rem Reverse the digits of each factor
for %%A in (1,2) do (
for /l %%B in (0,1,9) do set "num%%A=!num%%A:%%B=%%B !"
for %%C in (!num%%A!) do ( set /a limit%%A+=1 & set "rev%%A=%%C!rev%%A!" )
)
rem Do the multiplication
for /l %%A in (0,1,%limit1%) do (
for /l %%B in (0,1,%limit2%) do (
set /a iter=%%A+%%B
set /a iternext=iter+1
set /a iternext2=iter+2
set /a prev=digit!iter!
set /a digit!iter!=!rev1:~%%A,1!*!rev2:~%%B,1!
rem The next line updates the length of "digits"
if !iternext! gtr !length! set length=!iternext!
if !iter! lss !length! set /a digit!iter!+=prev
set /a currdigit=digit!iter!
if !currDigit! gtr 9 (
set /a prev=digit!iternext!
set /a digit!iternext!=currdigit/10
set /a digit!iter!=currdigit%%10
rem The next line updates the length of "digits"
if !iternext2! gtr !length! set length=!iternext2!
if !iternext! lss !length! set /a digit!iternext!+=prev
)
)
)
rem Finalize product reversing the digits for /l %%F in (0,1,%length%) do set "prod=!digit%%F!!prod!" endlocal & set "%3=%prod%"
goto :eof</lang>
- Output:
340282366920938463463374607431768211456
BBC BASIC
Library method: <lang bbcbasic> INSTALL @lib$+"BB4WMAPMLIB"
MAPM_DllPath$ = @lib$+"BB4WMAPM.DLL"
PROCMAPM_Init
twoto64$ = "18446744073709551616"
PRINT "2^64 * 2^64 = " ; FNMAPM_Multiply(twoto64$, twoto64$)</lang>
Explicit method: <lang bbcbasic> twoto64$ = "18446744073709551616"
PRINT "2^64 * 2^64 = " ; FNlongmult(twoto64$, twoto64$)
END
DEF FNlongmult(num1$, num2$)
LOCAL C%, I%, J%, S%, num1&(), num2&(), num3&()
S% = LEN(num1$)+LEN(num2$)
DIM num1&(S%), num2&(S%), num3&(S%)
IF LEN(num1$) > LEN(num2$) SWAP num1$,num2$
$$^num1&(1) = num1$
num1&() AND= 15
FOR I% = LEN(num1$) TO 1 STEP -1
$$^num2&(I%) = num2$
num2&() AND= 15
num3&() += num2&() * num1&(I%)
IF I% MOD 3 = 1 THEN
C% = 0
FOR J% = S%-1 TO I%-1 STEP -1
C% += num3&(J%)
num3&(J%) = C% MOD 10
C% DIV= 10
NEXT
ENDIF
NEXT I%
num3&() += &30
num3&(S%) = 0
IF num3&(0) = &30 THEN = $$^num3&(1)
= $$^num3&(0)</lang>
C
Doing it as if by hand. <lang c>#include <stdio.h>
- include <string.h>
/* c = a * b. Caller is responsible for memory.
c must not be the same as either a or b. */
void longmulti(const char *a, const char *b, char *c) { int i = 0, j = 0, k = 0, n, carry; int la, lb;
/* either is zero, return "0" */ if (!strcmp(a, "0") || !strcmp(b, "0")) { c[0] = '0', c[1] = '\0'; return; }
/* see if either a or b is negative */ if (a[0] == '-') { i = 1; k = !k; } if (b[0] == '-') { j = 1; k = !k; }
/* if yes, prepend minus sign if needed and skip the sign */ if (i || j) { if (k) c[0] = '-'; longmulti(a + i, b + j, c + k); return; }
la = strlen(a); lb = strlen(b); memset(c, '0', la + lb); c[la + lb] = '\0';
- define I(a) (a - '0')
for (i = la - 1; i >= 0; i--) { for (j = lb - 1, k = i + j + 1, carry = 0; j >= 0; j--, k--) { n = I(a[i]) * I(b[j]) + I(c[k]) + carry; carry = n / 10; c[k] = (n % 10) + '0'; } c[k] += carry; }
- undef I
if (c[0] == '0') memmove(c, c + 1, la + lb);
return; }
int main() { char c[1024]; longmulti("-18446744073709551616", "-18446744073709551616", c); printf("%s\n", c);
return 0; }</lang>output<lang>340282366920938463463374607431768211456</lang>
C#
If you strip out the BigInteger checking, it will work with lesser versions.
This uses the decimal type, (which has a MaxValue of 79,228,162,514,264,337,593,543,950,335). By limiting it to 10^28, it allows 28 decimal digits for the hi part, and 28 decimal digits for the lo part, 56 decimal digits total. A side computation of BigInteger assures that the results are accurate.
<lang csharp>using System; using static System.Console; using BI = System.Numerics.BigInteger;
class Program {
static decimal mx = 1E28M, hm = 1E14M, a;
// allows for 56 digit representation, using 28 decimal digits from each decimal
struct bi { public decimal hi, lo; }
// sets up for squaring process
static bi set4sq(decimal a) { bi r; r.hi = Math.Floor(a / hm); r.lo = a % hm; return r; }
// outputs bi structure as string, optionally inserting commas
static string toStr(bi a, bool comma = false) {
string r = a.hi == 0 ? string.Format("{0:0}", a.lo) :
string.Format("{0:0}{1:" + new string('0', 28) + "}", a.hi, a.lo);
if (!comma) return r; string rc = "";
for (int i = r.Length - 3; i > 0; i -= 3) rc = "," + r.Substring(i, 3) + rc;
return r.Substring(0, ((r.Length + 2) % 3) + 1) + rc; }
// needed because Math.Pow() returns a double
static decimal Pow_dec(decimal bas, uint exp) {
if (exp == 0) return 1M; decimal tmp = Pow_dec(bas, exp >> 1); tmp *= tmp;
if ((exp & 1) == 0) return tmp; return tmp * bas; }
static void Main(string[] args) {
for (uint p = 64; p < 95; p += 30) { // show prescribed output and maximum power of 2 output
bi x = set4sq(a = Pow_dec(2M, p)), y; // setup for squaring process
WriteLine("The square of (2^{0}): {1,38:n0}", p, a); BI BS = BI.Pow((BI)a, 2);
y.lo = x.lo * x.lo; y.hi = x.hi * x.hi; // square lo and hi parts
a = x.hi * x.lo * 2M; // calculate midterm
y.hi += Math.Floor(a / hm); // increment hi part w/ high part of midterm
y.lo += (a % hm) * hm; // increment lo part w/ low part of midterm
while (y.lo > mx) { y.lo -= mx; y.hi++; } // check for overflow, adjust both parts as needed
WriteLine(" is {0,75} (which {1} match the BigInteger computation)\n", toStr(y, true),
BS.ToString() == toStr(y) ? "does" : "fails to"); } }
}</lang>
- Output:
The square of (2^64): 18,446,744,073,709,551,616 is 340,282,366,920,938,463,463,374,607,431,768,211,456 (which does match the BigInteger computation) The square of (2^94): 19,807,040,628,566,084,398,385,987,584 is 392,318,858,461,667,547,739,736,838,950,479,151,006,397,215,279,002,157,056 (which does match the BigInteger computation)
C++
Version 1
<lang cpp>
- include <iostream>
- include <sstream>
//-------------------------------------------------------------------------------------------------- typedef long long bigInt; //-------------------------------------------------------------------------------------------------- using namespace std; //-------------------------------------------------------------------------------------------------- class number { public:
number() { s = "0"; neg = false; }
number( bigInt a ) { set( a ); }
number( string a ) { set( a ); }
void set( bigInt a ) { neg = false; if( a < 0 ) { a = -a; neg = true; } ostringstream o; o << a; s = o.str(); clearStr(); }
void set( string a ) { neg = false; s = a; if( s.length() > 1 && s[0] == '-' ) { neg = true; } clearStr(); }
number operator * ( const number& b ) { return this->mul( b ); }
number& operator *= ( const number& b ) { *this = *this * b; return *this; }
number& operator = ( const number& b ) { s = b.s; return *this; }
friend ostream& operator << ( ostream& out, const number& a ) { if( a.neg ) out << "-"; out << a.s; return out; }
friend istream& operator >> ( istream& in, number& a ){ string b; in >> b; a.set( b ); return in; }
private:
number mul( const number& b )
{
number a; bool neg = false; string r, bs = b.s; r.resize( 2 * max( b.s.length(), s.length() ), '0' ); int xx, ss, rr, t, c, stp = 0; string::reverse_iterator xi = bs.rbegin(), si, ri; for( ; xi != bs.rend(); xi++ ) { c = 0; ri = r.rbegin() + stp; for( si = s.rbegin(); si != s.rend(); si++ ) { xx = ( *xi ) - 48; ss = ( *si ) - 48; rr = ( *ri ) - 48; ss = ss * xx + rr + c; t = ss % 10; c = ( ss - t ) / 10; ( *ri++ ) = t + 48; } if( c > 0 ) ( *ri ) = c + 48; stp++; } trimLeft( r ); t = b.neg ? 1 : 0; t += neg ? 1 : 0; if( t & 1 ) a.s = "-" + r; else a.s = r; return a;
}
void trimLeft( string& r )
{
if( r.length() < 2 ) return; for( string::iterator x = r.begin(); x != ( r.end() - 1 ); ) { if( ( *x ) != '0' ) return; x = r.erase( x ); }
}
void clearStr()
{
for( string::iterator x = s.begin(); x != s.end(); ) { if( ( *x ) < '0' || ( *x ) > '9' ) x = s.erase( x ); else x++; }
} string s; bool neg;
}; //-------------------------------------------------------------------------------------------------- int main( int argc, char* argv[] ) {
number a, b; a.set( "18446744073709551616" ); b.set( "18446744073709551616" ); cout << a * b << endl << endl;
cout << "Factor 1 = "; cin >> a; cout << "Factor 2 = "; cin >> b; cout << "Product: = " << a * b << endl << endl; return system( "pause" );
} //-------------------------------------------------------------------------------------------------- </lang>
- Output:
340282366920938463463374607431768211456 Factor 1 = 9876548974569852365985574874787454878778975948 Factor 2 = 8954564845421878741168741154541897945138974567 Product: = 88440198241770705041777453160463400993104404280916080859287340887463980926235972531076714516
Version 2
<lang cpp>
- include <iostream>
- include <vector>
using namespace std;
typedef unsigned long native_t;
struct ZPlus_ // unsigned int, represented as digits base 10 { vector<native_t> digits_; // least significant first; value is sum(digits_[i] * 10^i)
ZPlus_(native_t n) : digits_(1, n) { while(Sweep()); }
bool Sweep() // clean up digits so they are in [0,9] { bool changed = false; int carry = 0; for (auto pd = digits_.begin(); pd != digits_.end(); ++pd) { *pd += carry; carry = *pd / 10; *pd -= 10 * carry; changed = changed || carry > 0; } if (carry) digits_.push_back(carry); return changed || carry > 9; } };
ZPlus_ operator*(const ZPlus_& lhs, const ZPlus_& rhs) { ZPlus_ retval(0); // hold enough space retval.digits_.resize(lhs.digits_.size() + rhs.digits_.size(), 0ul); // accumulate one-digit multiples for (size_t ir = 0; ir < rhs.digits_.size(); ++ir) for (size_t il = 0; il < lhs.digits_.size(); ++il) retval.digits_[ir + il] += rhs.digits_[ir] * lhs.digits_[il]; // sweep clean and drop zeroes while(retval.Sweep()); while (!retval.digits_.empty() && !retval.digits_.back()) retval.digits_.pop_back(); return retval; }
ostream& operator<<(ostream& dst, const ZPlus_& n) { for (auto pd = n.digits_.rbegin(); pd != n.digits_.rend(); ++pd) dst << *pd; return dst; }
int main(int argc, char* argv[]) { int p2 = 1; ZPlus_ n(2ul); for (int ii = 0; ii < 7; ++ii) { p2 *= 2; n = n * n; cout << "2^" << p2 << " = " << n << "\n"; } return 0; }</lang>
2^2 = 4 2^4 = 16 2^8 = 256 2^16 = 65536 2^32 = 4294967296 2^64 = 18446744073709551616 2^128 = 340282366920938463463374607431768211456
Ceylon
<lang Ceylon>"run() is the main function of this module."
shared void run() {
function multiply(String|Integer|Integer[] top, String|Integer|Integer[] bottom, Integer base = 10) {
function fromString(String s) =>
s
.filter(not(','.equals))
.map((char) => Integer.parse(char.string))
.narrow<Integer>()
.sequence()
.reversed;
function toString(Integer[] ints) =>
""
.join(ints.interpose(',', 3))
.reversed
.trimLeading((char) => char in "0,");
function fromInteger(Integer int) => fromString(int.string);
function convertArg(String|Integer|Integer[] arg) =>
switch(arg)
case (is String) fromString(arg)
case (is Integer) fromInteger(arg)
case (is Integer[]) arg;
value a = convertArg(top);
value b = convertArg(bottom);
value p = a.size;
value q = b.size;
value product = Array.ofSize(p + q, 0);
for (bIndex->bDigit in b.indexed) {
variable value carry = 0;
for (aIndex->aDigit in a.indexed) {
assert (exists prodDigit = product[aIndex + bIndex]);
value temp = prodDigit + carry + aDigit * bDigit;
carry = temp / base;
product[aIndex + bIndex] = temp % base;
}
assert (exists lastDigit = product[bIndex + p]);
product[bIndex + p] = lastDigit + carry;
}
return toString(product.sequence()); }
value twoToThe64th = "18,446,744,073,709,551,616"; value expectedResult = "340,282,366,920,938,463,463,374,607,431,768,211,456"; value result = multiply(twoToThe64th, twoToThe64th);
print("The expected result is ``expectedResult``");
print("The actual result is ``result``");
print("Do they match? ``expectedResult == result then "Yes!" else "No!"``");
}</lang>
COBOL
<lang COBOL>
identification division.
program-id. long-mul.
data division.
replace ==ij-lim== by ==7== ==ir-lim== by ==14==.
working-storage section.
1 input-string pic x(26) value "18,446,744,073,709,551,616".
1 a-table.
2 a pic 999 occurs ij-lim.
1 b-table.
2 b pic 999 occurs ij-lim.
1 ir-table value all "0".
2 occurs ij-lim.
3 ir pic 999 occurs ir-lim.
1 s-table value all "0".
2 s pic 999 occurs ir-lim.
1 display.
2 temp-result pic 9(6) value 0.
2 carry pic 999 value 0.
2 remain pic 999 value 0.
1 binary.
2 i pic 9(4) value 0.
2 j pic 9(4) value 0.
2 k pic 9(4) value 0.
procedure division.
begin.
move 1 to j
perform varying i from 1 by 1 until i > ij-lim
unstring input-string delimited ","
into a (i) with pointer j
end-perform
move a-table to b-table
perform intermediate-calc
perform sum-ir
perform display-result
stop run
.
intermediate-calc.
perform varying i from ij-lim by -1 until i < 1
move 0 to carry
perform varying j from ij-lim by -1 until j < 1
compute temp-result = a (i) * b (j) + carry
divide temp-result by 1000 giving carry
remainder remain
compute k = i + j
move remain to ir (i k)
end-perform
subtract 1 from k
move carry to ir (i k)
end-perform
.
sum-ir.
move 0 to carry
perform varying k from ir-lim by -1 until k < 1
move carry to temp-result
perform varying i from ij-lim by -1 until i < 1
compute temp-result = temp-result + ir (i k)
end-perform
divide temp-result by 1000 giving carry
remainder remain
move remain to s (k)
end-perform
.
display-result.
display " " input-string
display " * " input-string
display " = " with no advancing
perform varying k from 1 by 1
until k > ir-lim or s (k) not = 0
end-perform
if s (k) < 100
move 1 to i
inspect s (k) tallying i for leading "0"
display s (k) (i:) "," with no advancing
add 1 to k
end-if
perform varying k from k by 1 until k > ir-lim
display s (k) with no advancing
if k < ir-lim
display "," with no advancing
end-if
end-perform
display space
.
end program long-mul.
</lang>
18,446,744,073,709,551,616 * 18,446,744,073,709,551,616 = 340,282,366,920,938,463,463,374,607,431,768,211,456
CoffeeScript
<lang coffeescript>
- This very limited BCD-based collection of functions
- allows for long multiplication. It works for positive
- numbers only. The assumed data structure is as follows:
- BcdInteger.from_integer(4321) == [1, 2, 3, 4]
BcdInteger =
from_string: (s) ->
arr = []
for c in s
arr.unshift parseInt(c)
arr
from_integer: (n) ->
result = []
while n > 0
result.push n % 10
n = Math.floor n / 10
result
to_string: (arr) ->
s =
for elem in arr
s = elem.toString() + s
s
sum: (arr1, arr2) ->
if arr1.length < arr2.length
return BcdInteger.sum(arr2, arr1)
carry = 0
result= []
for d1, pos in arr1
d = d1 + (arr2[pos] || 0) + carry
result.push d % 10
carry = Math.floor d / 10
if carry
result.push 1
result
multiply_by_power_of_ten: (arr, power_of_ten) ->
result = (0 for i in [0...power_of_ten])
result.concat arr
product_by_integer: (arr, n) ->
result = []
for digit, i in arr
prod = BcdInteger.from_integer n * digit
prod = BcdInteger.multiply_by_power_of_ten prod, i
result = BcdInteger.sum result, prod
result
product: (arr1, arr2) ->
result = []
for digit, i in arr1
prod = BcdInteger.product_by_integer arr2, digit
prod = BcdInteger.multiply_by_power_of_ten prod, i
result = BcdInteger.sum result, prod
result
x = BcdInteger.from_integer 1 for i in [1..64]
x = BcdInteger.product_by_integer x, 2
console.log BcdInteger.to_string x # 18446744073709551616 square = BcdInteger.product x, x console.log BcdInteger.to_string square # 340282366920938463463374607431768211456 </lang>
Common Lisp
<lang lisp>(defun number->digits (number)
(do ((digits '())) ((zerop number) digits)
(multiple-value-bind (quotient remainder) (floor number 10)
(setf number quotient)
(push remainder digits))))
(defun digits->number (digits)
(reduce #'(lambda (n d) (+ (* 10 n) d)) digits :initial-value 0))
(defun long-multiply (a b)
(labels ((first-digit (list)
"0 if list is empty, else first element of list."
(if (endp list) 0
(first list)))
(long-add (digitses &optional (carry 0) (sum '()))
"Do long addition on the list of lists of digits. Each
list of digits in digitses should begin with the least
significant digit. This is the opposite of the digit
list returned by number->digits which places the most
significant digit first. The digits returned by
long-add do have the most significant bit first."
(if (every 'endp digitses)
(nconc (number->digits carry) sum)
(let ((column-sum (reduce '+ (mapcar #'first-digit digitses)
:initial-value carry)))
(multiple-value-bind (carry column-digit)
(floor column-sum 10)
(long-add (mapcar 'rest digitses)
carry (list* column-digit sum)))))))
;; get the digits of a and b (least significant bit first), and
;; compute the zero padded rows. Then, add these rows (using
;; long-add) and convert the digits back to a number.
(do ((a (nreverse (number->digits a)))
(b (nreverse (number->digits b)))
(prefix '() (list* 0 prefix))
(rows '()))
((endp b) (digits->number (long-add rows)))
(let* ((bi (pop b))
(row (mapcar #'(lambda (ai) (* ai bi)) a)))
(push (append prefix row) rows)))))</lang>
> (long-multiply (expt 2 64) (expt 2 64)) 340282366920938463463374607431768211456
Crystal
<lang ruby>require "big"
a = 2.to_big_i ** 64
puts "#{a} * #{a} = #{a*a}" </lang>
- Output:
18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456
D
Using the standard library: <lang d>void main() {
import std.stdio, std.bigint;
writeln(2.BigInt ^^ 64 * 2.BigInt ^^ 64);
}</lang>
- Output:
340282366920938463463374607431768211456
Long multiplication, same output:
<lang d>import std.stdio, std.algorithm, std.range, std.ascii, std.string;
auto longMult(in string x1, in string x2) pure nothrow @safe {
auto digits1 = x1.representation.retro.map!q{a - '0'};
immutable digits2 = x2.representation.retro.map!q{a - '0'}.array;
uint[] res;
foreach (immutable i, immutable d1; digits1.enumerate) {
foreach (immutable j, immutable d2; digits2) {
immutable k = i + j;
if (res.length <= k)
res.length++;
res[k] += d1 * d2;
if (res[k] > 9) {
if (res.length <= k + 1)
res.length++;
res[k + 1] = res[k] / 10 + res[k + 1];
res[k] -= res[k] / 10 * 10;
}
}
}
//return res.retro.map!digits; return res.retro.map!(d => digits[d]);
}
void main() {
immutable two64 = "18446744073709551616"; longMult(two64, two64).writeln;
}</lang>
Dc
Since Dc has arbitrary precision built-in, the task is no different than a normal multiplication: <lang Dc>2 64^ 2 64^ *p</lang>
EchoLisp
We implement long multiplication by multiplying polynomials, knowing that the number 1234 is the polynomial x^3 +2x^2 +3x +4 at x=10. As we assume no bigint library is present, long-mul operates on strings. <lang lisp> (lib 'math) ;; for poly multiplication
- convert string of decimal digits to polynomial
- "1234" → x^3 +2x^2 +3x +4
- least-significant digit first
(define (string->long N) (reverse (map string->number (string->list N))))
- convert polynomial to string
(define (long->string N) (if (pair? N)
(string-append (number->string (first N)) (long->string (rest N))) ""))
- convert poly coefficients to base 10
(define (poly->10 P (carry 0)) (append (for/list ((coeff P)) (set! coeff (+ carry coeff )) (set! carry (quotient coeff 10)) ;; new carry (modulo coeff 10)) (if(zero? carry) null (list carry)))) ;; remove leading 0 if any
- long multiplication
- convert input - strings of decimal digits - to polynomials
- perform poly multiplication in base 10
- convert result to string of decimal digits
(define (long-mul A B )
(long->string (reverse (poly->10 (poly-mul (string->long A) (string->long B))))))
(define two-64 "18446744073709551616") (long-mul two-64 two-64)
→ "340282366920938463463374607431768211456"
- check it
(lib 'bigint) Lib: bigint.lib loaded. (expt 2 128)
→ 340282366920938463463374607431768211456
</lang>
Euphoria
<lang euphoria>constant base = 1000000000
function atom_to_long(atom a)
sequence s
s = {}
while a>0 do
s = append(s,remainder(a,base))
a = floor(a/base)
end while
return s
end function
function long_mult(object a, object b)
sequence c
if atom(a) then
a = atom_to_long(a)
end if
if atom(b) then
b = atom_to_long(b)
end if
c = repeat(0,length(a)+length(b))
for i = 1 to length(a) do
c[i .. i+length(b)-1] += a[i]*b
end for
for i = 1 to length(c) do
if c[i] > base then
c[i+1] += floor(c[i]/base) -- carry
c[i] = remainder(c[i],base)
end if
end for
if c[$] = 0 then
c = c[1..$-1]
end if
return c
end function
function long_to_str(sequence a)
sequence s
s = sprintf("%d",a[$])
for i = length(a)-1 to 1 by -1 do
s &= sprintf("%09d",a[i])
end for
return s
end function
sequence a, b, c
a = atom_to_long(power(2,32)) printf(1,"a is %s\n",{long_to_str(a)})
b = long_mult(a,a) printf(1,"a*a is %s\n",{long_to_str(b)})
c = long_mult(b,b) printf(1,"a*a*a*a is %s\n",{long_to_str(c)})</lang>
Output:
a is 4294967296 a*a is 18446744073709551616 a*a*a*a is 340282366920938463488374607424768211456
F#
<lang F#>> let X = 2I ** 64 * 2I ** 64 ;;
val X : System.Numerics.BigInteger = 340282366920938463463374607431768211456 </lang>
Factor
<lang factor>USING: kernel math sequences ;
- longmult-seq ( xs ys -- zs )
[ * ] cartesian-map dup length iota [ 0 <repetition> ] map [ prepend ] 2map [ ] [ [ 0 suffix ] dip [ + ] 2map ] map-reduce ;
- integer->digits ( x -- xs ) { } swap [ dup 0 > ] [ 10 /mod swap [ prefix ] dip ] while drop ;
- digits->integer ( xs -- x ) 0 [ swap 10 * + ] reduce ;
- longmult ( x y -- z ) [ integer->digits ] bi@ longmult-seq digits->integer ;</lang>
<lang factor>( scratchpad ) 2 64 ^ dup longmult . 340282366920938463463374607431768211456 ( scratchpad ) 2 64 ^ dup * . 340282366920938463463374607431768211456</lang>
Fortran
<lang fortran>module LongMoltiplication
implicit none
type longnum
integer, dimension(:), pointer :: num
end type longnum
interface operator (*)
module procedure longmolt_ll
end interface
contains
subroutine longmolt_s2l(istring, num) character(len=*), intent(in) :: istring type(longnum), intent(out) :: num integer :: i, l
l = len(istring)
allocate(num%num(l))
forall(i=1:l) num%num(l-i+1) = iachar(istring(i:i)) - 48
end subroutine longmolt_s2l
! this one performs the moltiplication function longmolt_ll(a, b) result(c) type(longnum) :: c type(longnum), intent(in) :: a, b integer, dimension(:,:), allocatable :: t integer :: ntlen, i, j
ntlen = size(a%num) + size(b%num) + 1 allocate(c%num(ntlen)) c%num = 0
allocate(t(size(b%num), ntlen)) t = 0 forall(i=1:size(b%num), j=1:size(a%num)) t(i, j+i-1) = b%num(i) * a%num(j)
do j=2, ntlen
forall(i=1:size(b%num)) t(i, j) = t(i, j) + t(i, j-1)/10
end do
forall(j=1:ntlen) c%num(j) = sum(mod(t(:,j), 10))
do j=2, ntlen
c%num(j) = c%num(j) + c%num(j-1)/10
end do
c%num = mod(c%num, 10) deallocate(t) end function longmolt_ll
subroutine longmolt_print(num) type(longnum), intent(in) :: num
integer :: i, j
do j=size(num%num), 2, -1
if ( num%num(j) /= 0 ) exit
end do
do i=j, 1, -1
write(*,"(I1)", advance="no") num%num(i)
end do
end subroutine longmolt_print
end module LongMoltiplication</lang>
<lang fortran>program Test
use LongMoltiplication
type(longnum) :: a, b, r
call longmolt_s2l("18446744073709551616", a)
call longmolt_s2l("18446744073709551616", b)
r = a * b call longmolt_print(r) write(*,*)
end program Test</lang>
FreeBASIC
<lang freebasic>' version 08-01-2017 ' compile with: fbc -s console
Const As UInteger base_ = 1000000000 ' base 1,000,000,000
Function multiply(a1 As String, b1 As String) As String
Dim As String a = a1, b = b1 Trim(a) : Trim(b) ' remove spaces If Len(a) = 0 Or Len(b) = 0 Then Return "0"
If Len(a) + Len(b) > 10000 Then
Print "number(s) are to big"
Sleep 5000,1
Return ""
End If
If Len(a) < Len(b) Then
Swap a, b
End If
Dim As ULongInt product Dim As UInteger carry, i, m, shift Dim As UInteger la = Len(a), lb = Len(b) Dim As UInteger la9 = la \ 9 + IIf((la Mod 9) = 0, 0, 1) Dim As UInteger lb9 = lb \ 9 + IIf((lb Mod 9) = 0, 0, 1) Dim As UInteger arr_a(la9), answer((la9 + lb9) + 2) Dim As Integer last = la9
' make length a, b a multipy of 9 a = Right((String(9, "0") + a), la9 * 9) b = Right((String(9, "0") + b), lb9 * 9)
For i = 1 To la9
arr_a(la9 - i +1) = Val(Mid(a, i * 9 -8, 9))
Next
Do
carry = 0
m = Val(Mid(b, lb9 * 9 -8, 9))
For i = 1 To la9
product = CULngInt(arr_a(i)) * m + answer(i + shift) + carry
carry = product \ base_
answer(i + shift) = product - carry * base_
Next
If carry <> 0 Then
last = la9 + shift +1
answer(last) = carry
End If
lb9 = lb9 -1
shift = shift +1
Loop Until lb9 = 0
Dim As String tmp = Str(answer(last))
last = last -1
While last > 0
tmp = tmp + Right(String(9,"0") + Str(answer(last)), 9)
last = last -1
Wend
Return tmp
End Function
' ------=< MAIN >=------
Dim As String a = "2", b = "2", answer Dim As UInteger i = 1, j
For j = 1 To 7
answer = multiply(a, b) a = answer b = answer i = i + i Print using "2 ^ ### = "; i; Print answer
Next
Print Print "-------------------------------------------------" Print
a = "2" : b = "1" : answer = "" For j = 1 To 128
answer = multiply(a, b) b = answer
Next Print "2 ^ 128 = "; answer
' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
2 ^ 2 = 4 2 ^ 4 = 16 2 ^ 8 = 256 2 ^ 16 = 65536 2 ^ 32 = 4294967296 2 ^ 64 = 18446744073709551616 2 ^ 128 = 340282366920938463463374607431768211456 ------------------------------------------------- 2 ^ 128 = 340282366920938463463374607431768211456
Go
<lang go>// Long multiplication per WP article referenced by task description. // That is, multiplicand is multiplied by single digits of multiplier // to form intermediate results. Intermediate results are accumulated // for the product. Used here is the abacus method mentioned by the // article, of summing intermediate results as they are produced, // rather than all at once at the end. // // Limitations: Negative numbers not supported, superfluous leading zeros // not generally removed. package main
import "fmt"
// argument validation func d(b byte) byte {
if b < '0' || b > '9' {
panic("digit 0-9 expected")
}
return b - '0'
}
// add two numbers as strings func add(x, y string) string {
if len(y) > len(x) {
x, y = y, x
}
b := make([]byte, len(x)+1)
var c byte
for i := 1; i <= len(x); i++ {
if i <= len(y) {
c += d(y[len(y)-i])
}
s := d(x[len(x)-i]) + c
c = s / 10
b[len(b)-i] = (s % 10) + '0'
}
if c == 0 {
return string(b[1:])
}
b[0] = c + '0'
return string(b)
}
// multipy a number by a single digit func mulDigit(x string, y byte) string {
if y == '0' {
return "0"
}
y = d(y)
b := make([]byte, len(x)+1)
var c byte
for i := 1; i <= len(x); i++ {
s := d(x[len(x)-i])*y + c
c = s / 10
b[len(b)-i] = (s % 10) + '0'
}
if c == 0 {
return string(b[1:])
}
b[0] = c + '0'
return string(b)
}
// multiply two numbers as strings func mul(x, y string) string {
result := mulDigit(x, y[len(y)-1])
for i, zeros := 2, ""; i <= len(y); i++ {
zeros += "0"
result = add(result, mulDigit(x, y[len(y)-i])+zeros)
}
return result
}
// requested output const n = "18446744073709551616"
func main() {
fmt.Println(mul(n, n))
}</lang> Output:
340282366920938463463374607431768211456
Haskell
<lang haskell>import Data.List (transpose, inits) import Data.Char (digitToInt)
longmult :: Integer -> Integer -> Integer longmult x y = foldl1 ((+) . (10 *)) (polymul (digits x) (digits y))
polymul :: [Integer] -> [Integer] -> [Integer] polymul xs ys =
sum <$>
transpose
(zipWith
(<>)
(inits $ repeat 0)
((\f x -> fmap ((<$> x) . f)) (*) xs ys))
digits :: Integer -> [Integer] digits = fmap (fromIntegral . digitToInt) . show
main :: IO () main = print $ (2 ^ 64) `longmult` (2 ^ 64)</lang>
- Output:
340282366920938463463374607431768211456
Icon and Unicon
Large integers are native to Icon and Unicon. Neither libraries nor special programming is required. <lang Icon>procedure main()
write(2^64*2^64)
end</lang>
- Output:
340282366920938463463374607431768211456
J
Solution: <lang j> digits =: ,.&.":
polymult =: +//.@(*/) buildDecimal=: 10x&#.
longmult=: buildDecimal@polymult&digits</lang>
Example: <lang j> longmult~ 2x^64 340282366920938463463374607431768211456</lang>
Alternatives:
longmult could have been defined concisely:
<lang j>longmult=: 10x&#.@(+//.@(*/)&(,.&.":))</lang>
Or, of course, the task may be accomplished without the verb definitions:
<lang j> 10x&#.@(+//.@(*/)&(,.&.":))~2x^64
340282366920938463463374607431768211456</lang>
Or using the code (+ 10x&*)/@|. instead of #.:
<lang j> (+ 10x&*)/@|.@(+//.@(*/)&(,.&.":))~2x^64
340282366920938463463374607431768211456</lang>
Or you could use the built-in language support for arbitrary precision multiplication:
<lang j> (2x^64)*(2x^64)
340282366920938463463374607431768211456</lang>
Explaining the component verbs:
digitstranslates a number to a corresponding list of digits;
<lang j> ,.&.": 123 1 2 3</lang>
polymult(multiplies polynomials): ref. [1]
<lang j> 1 2 3 (+//.@(*/)) 1 2 3 1 4 10 12 9</lang>
buildDecimal(translates a list of decimal digits - possibly including "carry" - to the corresponding extended precision number):
<lang j> (+ 10x&*)/|. 1 4 10 12 9 15129</lang>
Java
Decimal version
This version of the code keeps the data in base ten. By doing this, we can avoid converting the whole number to binary and we can keep things simple, but the runtime will be suboptimal.
<lang java>public class LongMult {
private static byte[] stringToDigits(String num) { byte[] result = new byte[num.length()]; for (int i = 0; i < num.length(); i++) { char c = num.charAt(i); if (c < '0' || c > '9') { throw new IllegalArgumentException("Invalid digit " + c + " found at position " + i); } result[num.length() - 1 - i] = (byte) (c - '0'); } return result; }
public static String longMult(String num1, String num2) { byte[] left = stringToDigits(num1); byte[] right = stringToDigits(num2); byte[] result = new byte[left.length + right.length]; for (int rightPos = 0; rightPos < right.length; rightPos++) { byte rightDigit = right[rightPos]; byte temp = 0; for (int leftPos = 0; leftPos < left.length; leftPos++) { temp += result[leftPos + rightPos]; temp += rightDigit * left[leftPos]; result[leftPos + rightPos] = (byte) (temp % 10); temp /= 10; } int destPos = rightPos + left.length; while (temp != 0) { temp += result[destPos] & 0xFFFFFFFFL; result[destPos] = (byte) (temp % 10); temp /= 10; destPos++; } } StringBuilder stringResultBuilder = new StringBuilder(result.length); for (int i = result.length - 1; i >= 0; i--) { byte digit = result[i]; if (digit != 0 || stringResultBuilder.length() > 0) { stringResultBuilder.append((char) (digit + '0')); } } return stringResultBuilder.toString(); }
public static void main(String[] args) { System.out.println(longMult("18446744073709551616", "18446744073709551616")); } } </lang>
Binary version
This version tries to be as efficient as possible, so it converts numbers into binary before doing any calculations. The complexity is higher because of the need to convert to and from base ten, which requires the implementation of some additional arithmetic operations beyond long multiplication itself.
<lang java>import java.util.Arrays;
public class LongMultBinary {
/** * A very basic arbitrary-precision integer class. It only handles * non-negative numbers and doesn't implement any arithmetic not necessary * for the task at hand. */ public static class MyLongNum implements Cloneable {
/* * The actual bits of the integer, with the least significant place * first. The biggest native integer type of Java is the 64-bit long, * but since we need to be able to store the result of two digits * multiplied, we have to use the second biggest native type, the 32-bit * int. All numeric types are signed in Java, but we don't want to waste * the sign bit, so we need to take extra care while doing arithmetic to * ensure unsigned semantics. */ private int[] digits;
/* * The number of digits actually used in the digits array. Since arrays * cannot be resized in Java, we are better off remembering the logical * size ourselves, instead of reallocating and copying every time we need to shrink. */ private int digitsUsed;
@Override public MyLongNum clone() { try { MyLongNum clone = (MyLongNum) super.clone(); clone.digits = clone.digits.clone(); return clone; } catch (CloneNotSupportedException e) { throw new Error("Object.clone() threw exception", e); } }
private void resize(int newLength) { if (digits.length < newLength) { digits = Arrays.copyOf(digits, newLength); } }
private void adjustDigitsUsed() { while (digitsUsed > 0 && digits[digitsUsed - 1] == 0) { digitsUsed--; } }
/** * "Short" multiplication by one digit. Used to convert strings to long numbers. */ public void multiply(int multiplier) { if (multiplier < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } resize(digitsUsed + 1); long temp = 0; for (int i = 0; i < digitsUsed; i++) { temp += (digits[i] & 0xFFFFFFFFL) * multiplier; digits[i] = (int) temp; // store the low 32 bits temp >>>= 32; } digits[digitsUsed] = (int) temp; digitsUsed++; adjustDigitsUsed(); }
/** * "Short" addition (adding a one-digit number). Used to convert strings to long numbers. */ public void add(int addend) { if (addend < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } long temp = addend; for (int i = 0; i < digitsUsed && temp != 0; i++) { temp += (digits[i] & 0xFFFFFFFFL); digits[i] = (int) temp; // store the low 32 bits temp >>>= 32; } if (temp != 0) { resize(digitsUsed + 1); digits[digitsUsed] = (int) temp; digitsUsed++; } }
/** * "Short" division (dividing by a one-digit number). Used to convert numbers to strings. * @param divisor The digit to divide by. * @return The remainder of the division. */ public int divide(int divisor) { if (divisor < 0) { throw new IllegalArgumentException( "Signed arithmetic isn't supported"); } int remainder = 0; for (int i = digitsUsed - 1; i >= 0; i--) { long twoDigits = (((long) remainder << 32) | (digits[i] & 0xFFFFFFFFL)); remainder = (int) (twoDigits % divisor); digits[i] = (int) (twoDigits / divisor); } adjustDigitsUsed(); return remainder; }
public MyLongNum(String value) { // each of our 32-bit digits can store at least 9 decimal digit's worth this.digits = new int[value.length() / 9 + 1]; this.digitsUsed = 0; // To lower the number of bignum operations, handle nine digits at a time. for (int i = 0; i < value.length(); i+=9) { String chunk = value.substring(i, Math.min(i+9, value.length())); int multiplier = 1; int addend = 0; for (int j=0; j<chunk.length(); j++) { char c = chunk.charAt(j); if (c < '0' || c > '9') { throw new IllegalArgumentException("Invalid digit " + c + " found in input"); } multiplier *= 10; addend *= 10; addend += c - '0'; } multiply(multiplier); add(addend); } }
@Override public String toString() { if (digitsUsed == 0) { return "0"; } MyLongNum dummy = this.clone(); StringBuilder resultBuilder = new StringBuilder(digitsUsed * 9); while (dummy.digitsUsed > 0) { // To limit the number of bignum divisions, handle nine digits at a time. int decimalDigits = dummy.divide(1000000000); for (int i=0; i<9; i++) { resultBuilder.append((char) (decimalDigits % 10 + '0')); decimalDigits /= 10; } } // Trim any leading zeros we may have created. while (resultBuilder.charAt(resultBuilder.length()-1) == '0') { resultBuilder.deleteCharAt(resultBuilder.length()-1); } return resultBuilder.reverse().toString(); }
/** * Long multiplication. */ public void multiply(MyLongNum multiplier) { MyLongNum left, right; // Make sure the shorter number is on the right-hand side to make things a bit more efficient. if (this.digitsUsed > multiplier.digitsUsed) { left = this; right = multiplier; } else { left = multiplier; right = this; } int[] newDigits = new int[left.digitsUsed + right.digitsUsed]; for (int rightPos = 0; rightPos < right.digitsUsed; rightPos++) { long rightDigit = right.digits[rightPos] & 0xFFFFFFFFL; long temp = 0; for (int leftPos = 0; leftPos < left.digitsUsed; leftPos++) { temp += (newDigits[leftPos + rightPos] & 0xFFFFFFFFL); temp += rightDigit * (left.digits[leftPos] & 0xFFFFFFFFL); newDigits[leftPos + rightPos] = (int) temp; temp >>>= 32; } // Roll forward any carry we may have. int destPos = rightPos + digitsUsed; while (temp != 0) { temp += (newDigits[destPos] & 0xFFFFFFFFL); newDigits[destPos] = (int) temp; temp >>>= 32; destPos++; } } this.digits = newDigits; this.digitsUsed = newDigits.length; adjustDigitsUsed(); } }
public static void main(String[] args) { MyLongNum one = new MyLongNum("18446744073709551616"); MyLongNum two = one.clone(); one.multiply(two); System.out.println(one); }
} </lang>
JavaScript
Iterative
With integer expression inputs at this scale, JavaScript still gives a slightly lossy result, despite the subsequent digit by digit string concatenation approach.
The problem is that the JavaScript Math.pow expressions become lossy at around 2^54, and Math.pow(2, 64) evaluates to a rounded:
18446744073709552000 rather than the full 18446744073709551616
This means that to handle larger inputs, the multiplication function needs to have string parameters:
<lang javascript>function mult(strNum1,strNum2){
var a1 = strNum1.split("").reverse();
var a2 = strNum2.toString().split("").reverse();
var aResult = new Array;
for ( var iterNum1 = 0; iterNum1 < a1.length; iterNum1++ ) {
for ( var iterNum2 = 0; iterNum2 < a2.length; iterNum2++ ) {
var idxIter = iterNum1 + iterNum2; // Get the current array position.
aResult[idxIter] = a1[iterNum1] * a2[iterNum2] + ( idxIter >= aResult.length ? 0 : aResult[idxIter] );
if ( aResult[idxIter] > 9 ) { // Carrying
aResult[idxIter + 1] = Math.floor( aResult[idxIter] / 10 ) + ( idxIter + 1 >= aResult.length ? 0 : aResult[idxIter + 1] );
aResult[idxIter] %= 10;
}
}
}
return aResult.reverse().join("");
}
mult('18446744073709551616', '18446744073709551616')</lang>
- Output:
340282366920938463463374607431768211456
Functional (ES 5)
The function below accepts integer string or native integer arguments, but as JavaScript (unlike Haskell and Python, for example), lacks an arbitrary precision integer type, larger inputs to this function (beyond the scale of c. 2^54) need to take the form of integer strings, to avoid rounding.
For the same reason, the output always takes the form of an arbitrary precision integer string, rather than a native integer data type. (See the largeIntegerString() helper function below)
<lang JavaScript>(function () {
'use strict';
// Javascript lacks an unbounded integer type // so this multiplication function takes and returns // long integer strings rather than any kind of native integer
// longMult :: (String | Integer) -> (String | Integer) -> String
function longMult(num1, num2) {
return largeIntegerString(
digitProducts(digits(num1), digits(num2))
);
}
// digitProducts :: [Int] -> [Int] -> [Int]
function digitProducts(xs, ys) {
return multTable(xs, ys)
.map(function (zs, i) {
return Array.apply(null, Array(i))
.map(function () {
return 0;
})
.concat(zs);
})
.reduce(function (a, x) {
if (a) {
var lng = a.length;
return x.map(function (y, i) {
return y + (i < lng ? a[i] : 0);
})
} else return x;
})
}
// largeIntegerString :: [Int] -> String
function largeIntegerString(lstColumnValues) {
var dctProduct = lstColumnValues
.reduceRight(function (a, x) {
var intSum = x + a.carried,
intDigit = intSum % 10;
return {
digits: intDigit
.toString() + a.digits,
carried: (intSum - intDigit) / 10
};
}, {
digits: ,
carried: 0
});
return (dctProduct.carried > 0 ? (
dctProduct.carried.toString()
) : ) + dctProduct.digits;
}
// multTables :: [Int] -> [Int] -> Int function multTable(xs, ys) { return ys.map(function (y) { return xs.map(function (x) { return x * y; }) }); }
// digits :: (Integer | String) -> [Integer]
function digits(n) {
return (typeof n === 'string' ? n : n.toString())
.split()
.map(function (x) {
return parseInt(x, 10);
});
}
// TEST showing that larged bounded integer inputs give only rounded results // whereas integer string inputs allow for full precision on this scale (2^128)
return {
fromIntegerStrings: longMult(
'18446744073709551616',
'18446744073709551616'
),
fromBoundedIntegers: longMult(
18446744073709551616,
18446744073709551616
)
};
})();</lang>
- Output:
{"fromIntegerStrings":"340282366920938463463374607431768211456",
"fromBoundedIntegers":"340282366920938477630474056040704000000"}
jq
Since the task description mentions 2^64, the following includes "long_power(i)" for computing n^i. <lang jq># multiply two decimal strings, which may be signed (+ or -) def long_multiply(num1; num2):
def stripsign:
.[0:1] as $a
| if $a == "-" then [ -1, .[1:]]
elif $a == "+" then [ 1, .[1:]]
else [1, .]
end;
def adjustsign(sign):
if sign == 1 then . else "-" + . end;
# mult/2 assumes neither argument has a sign
def mult(num1;num2):
(num1 | explode | map(.-48) | reverse) as $a1
| (num2 | explode | map(.-48) | reverse) as $a2
| reduce range(0; num1|length) as $i1
([]; # result
reduce range(0; num2|length) as $i2 (.;
($i1 + $i2) as $ix
| ( $a1[$i1] * $a2[$i2] +
(if $ix >= length then 0
else .[$ix]
end) ) as $r
| if $r > 9 # carrying
then
.[$ix + 1] = ($r / 10 | floor) +
(if $ix + 1 >= length then 0
else .[$ix + 1]
end)
| .[$ix] = $r - ( $r / 10 | floor ) * 10
else
.[$ix] = $r
end
)
)
| reverse | map(.+48) | implode;
(num1|stripsign) as $a1
| (num2|stripsign) as $a2
| if $a1[1] == "0" or $a2[1] == "0" then "0"
elif $a1[1] == "1" then $a2[1]|adjustsign( $a1[0] * $a2[0] )
elif $a2[1] == "1" then $a1[1]|adjustsign( $a1[0] * $a2[0] )
else mult($a1[1]; $a2[1]) | adjustsign( $a1[0] * $a2[0] )
end;</lang>
<lang jq># Emit (input)^i where input and i are non-negative decimal integers,
- represented as numbers and/or strings.
def long_power(i):
def power(i):
tostring as $self
| (i|tostring) as $i
| if $i == "0" then "1"
elif $i == "1" then $self
elif $self == "0" then "0"
else reduce range(1;i) as $_ ( $self; long_multiply(.; $self) )
end;
(i|tonumber) as $i | if $i < 4 then power($i) else ($i|sqrt|floor) as $j | ($i - $j*$j) as $k | long_multiply( power($j) | power($j) ; power($k) ) end ;</lang>
Example: <lang jq> 2 | long_power(64) | long_multiply(.;.)</lang>
- Output:
$ jq -n -f Long_multiplication.jq "340282366920938463463374607431768211456"
Julia
Module: <lang julia>module LongMultiplication
using Compat
function addwithcarry!(r, addend, addendpos)
while true
pad = max(0, addendpos - lastindex(r))
append!(r, fill(0, pad))
addendrst = addend + r[addendpos]
addend, r[addendpos] = divrem(addendrst, 10)
iszero(addend) && break
addendpos += 1
end
return r
end
function longmult(mult1::AbstractVector{T}, mult2::AbstractVector{T}) where T <: Integer
r = T[]
for (offset1, digit1) in enumerate(mult1), (offset2, digit2) in zip(eachindex(mult2) + offset1 - 1, mult2)
single_multrst = digits(digit1 * digit2)
for (addoffset, rstdigit) in zip(eachindex(single_multrst) + offset2 - 1, single_multrst)
addwithcarry!(r, rstdigit, addoffset)
end
end
return r
end
function longmult(a::T, b::T)::T where T <: Integer
mult1 = digits(a) mult2 = digits(b) r = longmult(mult1, mult2) return sum(d * T(10) ^ (e - 1) for (e, d) in enumerate(r))
end
function longmult(a::AbstractString, b::AbstractString)
if !ismatch(r"^\d+", a) || !ismatch(r"^\d+", b)
throw(ArgumentError("string must contain only digits"))
end
mult1 = reverse(collect(Char, a) .- '0')
mult2 = reverse(collect(Char, b) .- '0')
r = longmult(mult1, mult2)
return reverse(join(r))
end
end # module LongMultiplication</lang>
Main: <lang julia>@show LongMultiplication.longmult(big(2) ^ 64, big(2) ^ 64) @show LongMultiplication.longmult("18446744073709551616", "18446744073709551616")</lang>
- Output:
LongMultiplication.longmult(big(2) ^ 64, big(2) ^ 64) = 340282366920938463463374607431768211456
LongMultiplication.longmult("18446744073709551616", "18446744073709551616") = "340282366920938463463374607431768211456"
Kotlin
<lang scala>fun String.toDigits() = mapIndexed { i, c ->
if (!c.isDigit())
throw IllegalArgumentException("Invalid digit $c found at position $i")
c - '0'
}.reversed()
operator fun String.times(n: String): String {
val left = toDigits() val right = n.toDigits() val result = IntArray(left.size + right.size)
right.mapIndexed { rightPos, rightDigit ->
var tmp = 0
left.indices.forEach { leftPos ->
tmp += result[leftPos + rightPos] + rightDigit * left[leftPos]
result[leftPos + rightPos] = tmp % 10
tmp /= 10
}
var destPos = rightPos + left.size
while (tmp != 0) {
tmp += (result[destPos].toLong() and 0xFFFFFFFFL).toInt()
result[destPos] = tmp % 10
tmp /= 10
destPos++
}
}
return result.foldRight(StringBuilder(result.size), { digit, sb ->
if (digit != 0 || sb.length > 0) sb.append('0' + digit)
sb
}).toString()
}
fun main(args: Array<out String>) {
println("18446744073709551616" * "18446744073709551616")
}</lang>
Lambdatalk
<lang scheme> Natural positive numbers are defined as strings, for instance 123 -> "123". {lambda talk} has a small set of primitives working on strings, [equal?, empty?, chars, charAt, substring]
1) helper functions
{def lastchar
{lambda {:w}
{charAt {- {chars :w} 1} :w}
}} {def butlast
{lambda {:w}
{substring 0 {- {chars :w} 1} :w}
}} {def zeros
{lambda {:n}
{if {< :n 1}
then
else 0{zeros {- :n 1}}
}}}
2) add function
{def add
{def add.r
{lambda {:a :b :c :d}
{if {equal? :a #}
then {if {equal? :d 1} then 1 else}{butlast :c}
else {let { {:a :a} {:b :b} {:c :c}
{:d {+ :d {lastchar :a} {lastchar :b} }} }
{add.r {butlast :a} {butlast :b} {lastchar :d}:c
{if {equal? {chars :d} 1} then 0 else 1}}
}}}}
{lambda {:a :b}
{{lambda {:a :b :n}
{add.r #{zeros {- :n {chars :a}}}:a
#{zeros {- :n {chars :b}}}:b # 0}
} :a :b {max {chars :a} {chars :b}}}
}}
3) mul function
{def mul
{def muln
{lambda {:a :b :n}
{if {< :n 1}
then :b
else {muln :a {add :a :b} {- :n 1}}
}}}
{def mul.r
{lambda {:a :b :c :n}
{if {equal? :b #}
then :c
else {mul.r :a {butlast :b}
{add {muln :a 0 {lastchar :b}}{zeros :n} :c} {+ :n 1}}
}}}
{lambda {:a :b}
{mul.r :a #:b 0 0}
}}
4) applying to the task
Due to JS numbers limits, we compute first 2^32 using the JS pow function, then 2^64 and 2^128 using the mul function.
2^32 = '{def p32 {pow 2 32}} -> '{p32} = 4294967296 2^64 = '{def p64 {mul {p32} {p32}}} -> '{p64} = 18446744073709551616 2^128 = '{def p128 {mul {p64} {p64}}} -> '{p128} = 340282366920938463463374607431768211456
5) a more effective implementation
Lambdatalk can be helped by the lib_BN javascript library from Jonas Raoni Soares Silva and stored in a wiki page called by a {require lib_BN} command, computing becomes fast:
2^32 = {def p32 {BN.pow 2 32}} -> {p32} = 4294967296 2^64 = {def p64 {BN.* {p32} {p32}}} -> {p64} = 18446744073709551616 2^128 = {def p128 {BN.* {p64} {p64}}} -> {p128} = 340282366920938463463374607431768211456
This can be tested in http://lambdaway.free.fr/lambdaspeech/?view=numbers8 </lang>
Liberty BASIC
<lang lb> '[RC] long multiplication
'now, count 2^64 print "2^64" a$="1" for i = 1 to 64
a$ = multByD$(a$, 2)
next print a$ print "(check with native LB)" print 2^64 print "(looks OK)"
'now let's do b$*a$ stuff print print "2^64*2^64" print longMult$(a$, a$) print "(check with native LB)" print 2^64*2^64 print "(looks OK)"
end '--------------------------------------- function longMult$(a$, b$)
signA = 1 if left$(a$,1) = "-" then a$ = mid$(a$,2): signA = -1 signB = 1 if left$(b$,1) = "-" then b$ = mid$(b$,2): signB = -1
c$ = ""
t$ = ""
shift$ = ""
for i = len(a$) to 1 step -1
d = val(mid$(a$,i,1))
t$ = multByD$(b$, d)
c$ = addLong$(c$, t$+shift$)
shift$ = shift$ +"0"
'print d, t$, c$
next
if signA*signB<0 then c$ = "-" + c$
'print c$
longMult$ = c$
end function
function multByD$(a$, d) 'multiply a$ by digit d c$ = "" carry = 0 for i = len(a$) to 1 step -1
a = val(mid$(a$,i,1))
c = a*d+carry
carry = int(c/10)
c = c mod 10
'print a, c
c$ = str$(c)+c$
next
if carry>0 then c$ = str$(carry)+c$ 'print c$ multByD$ = c$
end function
function addLong$(a$, b$) 'add a$ + b$, for now only positive
l = max(len(a$), len(b$))
a$=pad$(a$,l)
b$=pad$(b$,l)
c$ = "" 'result
carry = 0
for i = l to 1 step -1
a = val(mid$(a$,i,1))
b = val(mid$(b$,i,1))
c = a+b+carry
carry = int(c/10)
c = c mod 10
'print a, b, c
c$ = str$(c)+c$
next
if carry>0 then c$ = str$(carry)+c$
'print c$
addLong$ = c$
end function
function pad$(a$,n) 'pad from right with 0 to length n
pad$ = a$
while len(pad$)<n
pad$ = "0"+pad$
wend
end function
</lang>
Lobster
Translation of Java binary version, but with base 1000000000
<lang Lobster>import std
// Very basic arbitrary-precision integers // - only non-negative numbers // - doesn't implement any arithmetic not necessary for the task at hand...
let base = 1000000000
class Bign:
digits: [int] // little endian, of base base digitsUsed: int
def clone():
return Bign { digits: copy(digits), digitsUsed: digitsUsed }
def resize(newLength):
while digits.length < newLength:
digits.push(0)
def adjustDigitsUsed():
while digitsUsed > 0 and digits[digitsUsed - 1] == 0:
digitsUsed -= 1
// multiplication by one digit; used to convert string to Bign
def muldigit(multiplier : int):
if (multiplier < 0):
return // "Signed arithmetic isn't supported"
resize(digitsUsed + 1)
var temp = 0
for(digitsUsed) i:
temp += digits[i] * multiplier
digits[i] = temp % base
temp /= base
digits[digitsUsed] = temp
digitsUsed += 1
adjustDigitsUsed()
// addition of one digit; used to convert string to Bign
def adddigit(addend: int):
if (addend < 0):
return // "Signed arithmetic isn't supported"
var temp = addend
var i = 0
while i < digitsUsed and temp != 0:
temp += digits[i]
digits[i] = temp % base
temp /= base
i += 1
if temp != 0:
resize(digitsUsed + 1)
digits[digitsUsed] = temp
digitsUsed += 1
def bign2str():
var i = digitsUsed
if i == 0:
return "0"
i -= 1
var s = string(digits[i])
while i > 0:
i -= 1
s += number_to_string(digits[i], 10, 9)
return s
def str2bign(value):
// each of our Bign digits can store 9 decimal digits
let this = Bign { digits: map(value.length() / 9 + 1): 0, digitsUsed: 0 }
// handle nine digits at a time
var i = 0
while i < value.length:
var multiplier = 1
var addend = 0
for(min(9, value.length() - i)) j:
let c = value[i+j]
//if (c < '0' or c > '9') -- what!?
multiplier *= 10
addend *= 10
addend += c - '0'
this.muldigit(multiplier)
this.adddigit(addend)
i += 9
return this
// Long multiplication
def bign_multiply(this, multiplier):
// Make sure the shorter number is on the right side to make things a bit more efficient
let left = if (this.digitsUsed > multiplier.digitsUsed): this else: multiplier
let right = if (this.digitsUsed > multiplier.digitsUsed): multiplier else: this
let newDigits = map(left.digitsUsed + right.digitsUsed): 0
for(right.digitsUsed) rightPos:
let rightDigit = right.digits[rightPos]
var temp = 0
for(left.digitsUsed) leftPos:
temp += newDigits[leftPos + rightPos]
temp += rightDigit * left.digits[leftPos]
newDigits[leftPos + rightPos] = temp % base
temp /= base
// Roll forward any carry we may have
let destPos = rightPos + left.digitsUsed
while temp != 0:
temp += newDigits[destPos]
newDigits[destPos] = temp % base
temp /= base
destPos +- 1
let bign = Bign { digits: newDigits, digitsUsed: newDigits.length }
bign.adjustDigitsUsed()
return bign
let one = str2bign("18446744073709551616") let two = one.clone() var pro = one.bign_multiply(two) print(bign2str(pro)) </lang>
- Output:
340282366920938463463374607431768211456
Maple
<lang Maple> longmult := proc(a::integer,b::integer)
local A,B,m,n,i,j; # Note, return a*b; works in Maple for any sized integer A := convert(a,base,10); B := convert(b,base,10); m := numelems(A); n := numelems(B); add( add( A[i]*B[j]*10^(j-1), j=1..n )*10^(i-1), i=1..m );
end;
> longmult( 2^64, 2^64 );
340282366920938463463374607431768211456
</lang>
Mathematica
We define the long multiplication function: <lang Mathematica> LongMultiplication[a_,b_]:=Module[{d1,d2},
d1=IntegerDigits[a]//Reverse; d2=IntegerDigits[b]//Reverse; Sum[d1id2j*10^(i+j-2),{i,1,Length[d1]},{j,1,Length[d2]}] ]</lang>
Example: <lang Mathematica> n1 = 2^64;
n2 = 2^64; LongMultiplication[n1, n2]</lang>
gives back: <lang Mathematica> 340282366920938463463374607431768211456</lang>
To check the speed difference between built-in multiplication (which is already arbitrary precision) we multiply two big numbers (2^8000 has 2409 digits!) and divide their timings: <lang Mathematica> n1=2^8000;
n2=2^8000; Timing[LongMultiplication[n1,n2]]1 Timing[n1 n2]1 Floor[%%/%]</lang>
gives back: <lang Mathematica> 72.9686
7.*10^-6 10424088</lang>
So our custom function takes about 73 second, the built-in function a couple of millionths of a second, so the long multiplication is about 10.5 million times slower! Mathematica uses Karatsuba multiplication for large integers, which is several magnitudes faster for really big numbers. Making it able to multiply in about a second; the final result has 9542426 digits; result omitted for obvious reasons.
NetRexx
A reworking of the example at Rexx Version 2. <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
numeric digits 100
runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method multiply(multiplier, multiplicand) public static
result =
mpa = s2a(multiplier)
mpb = s2a(multiplicand)
r_ = 0
rim = 1
loop bi = 1 to mpb[0]
loop ai = 1 to mpa[0]
ri = ai + bi -1
p_ = mpa[ai] * mpb[bi]
loop i_ = ri by 1 until p_ = 0
s_ = r_[i_] + p_
r_[i_] = s_ // 10
p_ = s_ % 10
end i_
rim = rim.max(i_)
end ai
end bi
r_[0] = rim
result = a2s(r_)
result = result.strip('l', 0)
if result = then result = 0
return result
-- ............................................................................. -- copy characters of a numeric string into a corresponding array -- digits are numbered 1 to n from right to left method s2a(numbr) private static
result = 0 lstr = numbr.length() loop z_ = 1 to lstr result[z_] = numbr.substr(lstr - z_ + 1, 1) end z_ result[0] = lstr return result
-- ............................................................................. -- turn the array of digits into a numeric string method a2s(numbr) private static
result = loop z_ = numbr[0] to 1 by -1 result = result || numbr[z_] end z_ return result
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static
mms = [ -
123', '123, -
012', '12, -
123456789012' , '44444444444, -
2 ** 64' , '2**64, -
0' ,0 ' -
]
ok = 0
errors = 0
loop mm over mms parse mm multiplier . ',' multiplicand . builtIn = multiplier * multiplicand calculated = multiply(multiplier, multiplicand) say 'Calculate' multiplier + 0 'x' multiplicand + 0 say 'Built in:' builtIn say 'Derived: ' calculated say if builtIn = calculated then ok = ok + 1 else errors = errors + 1 end mm say ok 'ok' say errors 'not ok'
return
</lang>
- Output:
Calculate 123 x 123 Built in: 15129 Derived: 15129 Calculate 12 x 12 Built in: 144 Derived: 144 Calculate 123456789012 x 44444444444 Built in: 5486968400478463649328 Derived: 5486968400478463649328 Calculate 18446744073709551616 x 18446744073709551616 Built in: 340282366920938463463374607431768211456 Derived: 340282366920938463463374607431768211456 Calculate 0 x 0 Built in: 0 Derived: 0 5 ok 0 not ok
Nim
<lang nim>import strutils
proc ti(a): int = ord(a) - ord('0')
proc longmulti(a, b: string): string =
var i, j, n, carry, la, lb = 0 k = false
# either is zero, return "0" if a == "0" or b == "0": return "0"
# see if either a or b is negative if a[0] == '-': i = 1; k = not k if b[0] == '-': j = 1; k = not k
# if yes, prepend minus sign if needed and skip the sign if i > 0 or j > 0: result = if k: "-" else: "" result.add longmulti(a[i..a.high], b[j..b.high]) return
result = repeatChar(a.len + b.len, '0')
for i in countdown(a.high, 0):
var carry = 0
var k = i + b.len
for j in countdown(b.high, 0):
let n = ti(a[i]) * ti(b[j]) + ti(result[k]) + carry
carry = n div 10
result[k] = chr(n mod 10 + ord('0'))
dec k
result[k] = chr(ord(result[k]) + carry)
if result[0] == '0': result[0..result.high-1] = result[1..result.high]
echo longmulti("-18446744073709551616", "-18446744073709551616")</lang> Output:
3402823669209384634633746074317682114566
Oforth
Oforth handles arbitrary precision integers, so there is no need to implement long multiplication :
- Output:
2 64 pow dup * println 340282366920938463463374607431768211456
But, if long multiplication was to be implemented :
A natural is implemented as a list of integers with base 1000000000 (in order to print them easier)
Just multiplication is implemented here.
<lang Oforth>Number Class new: Natural(v)
Natural method: initialize := v ; Natural method: _v @v ;
Natural classMethod: newValues super new ; Natural classMethod: newFrom asList self newValues ;
Natural method: *(n) | v i j l x k |
n _v ->v
ListBuffer initValue(@v size v size + 1+, 0) ->l
v size loop: i [
i v at dup ->x 0 ifEq: [ continue ]
0 @v size loop: j [
i j + 1- ->k
j @v at x * + l at(k) + 1000000000 /mod k rot l put
]
k 1+ swap l put
]
while(l last 0 == l size 0 <> and) [ l removeLast drop ]
l dup freeze Natural newValues ;
Natural method: << | i |
@v last << @v size 1 - loop: i [ @v at(@v size i -) <<wjp(0, JUSTIFY_RIGHT, 8) ] ;</lang>
- Output:
>Natural newFrom(2 16 pow) .s [1] (Natural) 65536 ok >dup * .s [1] (Natural) 4294967296 ok >dup * .s [1] (Natural) 18446744073709551616 ok >dup * .s [1] (Natural) 340282366920938463463374607431768211456 ok >_v .s [1] (List) [768211456, 374607431, 938463463, 282366920, 340] ok
Ol
Ol already supports long numbers "out-of-the-box".
<lang scheme> (define x (* 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2)) ; 2^64
(print (* x x)) </lang>
340282366920938463463374607431768211456
PARI/GP
<lang parigp>long(a,b)={
a=eval(Vec(a));
b=eval(Vec(b));
my(c=vector(#a+#b),carry=0);
for(i=1,#a,
for(j=1,#b,
c[i+j]+=a[i]*b[j]
)
);
forstep(i=#c,1,-1,
c[i] += carry;
carry = c[i] \ 10;
c[i] = c[i] % 10
);
for(i=1,#c,
if(c[i], return(concat(apply(s->Str(s),vector(#c+1-i,j,c[i+j-1])))))
);
"0"
}; long("18446744073709551616","18446744073709551616")</lang> Output:
%1 = "340282366920938463463374607431768211456"
Pascal
Extracted from a programme to calculate and factor the number (two versions) in Frederick Pohl's book The Gold at the Starbow's End, and compute Godel encodings of text. Compiles with the Free Pascal Compiler. The original would compile with Turbo Pascal (and used pointers to allow access to the "heap" storage scheme) except that does not allow functions to return a "big number" data aggregate, and it is so much nicer to be able to write X:=BigMult(A,B); The original has a special "square" calculation but this task is to exhibit long multiplication. However, raising to a power by iteration is painful, so a special routine for that. <lang Pascal> Program TwoUp; Uses DOS, crt; {Concocted by R.N.McLean (whom God preserve), Victoria university, NZ.}
Procedure Croak(gasp: string); Begin Writeln; Write(Gasp); HALT; End;
const BigBase = 10; {The base of big arithmetic.}
const BigEnuff = 333; {The most storage possible is 65532 bytes with Turbo Pascal.}
type BigNumberIndexer = word; {To access 0:BigEnuff BigNumberDigit data.}
type BigNumberDigit = byte; {The data.}
type BigNumberDigit2 = word; {Capable of digit*digit + carry. Like, 255*255 = 65025}
type BigNumber = {All sorts of arrangements are possible.}
Record {Could include a sign indication.}
TopDigit: BigNumberDigit; {Finger the high-order digit.}
digit: array[0..BigEnuff] of byte; {The digits: note the "downto" in BigShow.}
end; {Could add fractional digits too. Endless, endless.}
Procedure BigShow(var a: BigNumber); {Print the number.}
var i: integer; {A stepper.}
Begin
for i:=a.TopDigit downto 0 do {Thus high-order to low, as is the custom.}
if BigBase = 10 then write(a.digit[i]) {Constant following by the Turbo Pascal compiler}
else if BigBase = 100 then Write(a.digit[i] div 10,a.digit[i] mod 10) {Means that there will be no tests.}
else write(a.digit[i],','); {And dead code will be omitted.}
End;
Procedure BigZero(var A: BigNumber); {A:=0;}
Begin;
A.TopDigit:=0;
A.Digit[0]:=0;
End;
Procedure BigOne(var A: BigNumber); {A:=1;}
Begin;
A.TopDigit:=0;
A.Digit[0]:=1;
End;
Function BigInt(n: longint): BigNumber; {A:=N;}
var l: BigNumberIndexer;
Begin
l:=0;
if n < 0 then croak('Negative integers are not yet considered.');
repeat {At least one digit is to be placed.}
if l > BigEnuff then Croak('BigInt overflowed!'); {Oh dear.}
BigInt.Digit[l]:=N mod BigBase; {The low-order digit.}
n:=n div BigBase; {Shift down a digit.}
l:=l + 1; {Count in anticipation.}
until N = 0; {Still some number left?}
BigInt.TopDigit:=l - 1; {Went one too far.}
End;
Function BigMult(a,b: BigNumber): BigNumber; {x:=BigMult(a,b);}
{Suppose the digits of A are a5,a4,a3,a2,a1,a0...
To multiply A and B.
a5 a4 a3 a2 a1 a0: six digits, d1
x b4 b3 b2 b1 b0: five digits, d2
---------------------------
a5b0 a4b0 a3b0 a2b0 a1b0 a0b0
a5b1 a4b1 a3b1 a2b1 a1b1 a0b1
a5b2 a4b2 a3b2 a2b2 a1b2 a0b2
a5b3 a4b3 a3b3 a2b3 a1b3 a0b3
a5b4 a4b4 a3b4 a2b4 a1b4 a0b4
-------------------------------------------------------
carry 9 8 7 6 5 4 3 2 1 0: at least nine digits,
------------------------------------------------------- = d1 + d2 - 1
But the indices are also the powers, so the highest power is 9 = 5 + 4,
and a possible tenth for any carry.}
var X: BigNumber; {Scratchpad, so b:=BigMult(a,b); doesn't overwrite b as it goes...}
var d: BigNumberDigit; {A digit.}
var c: BigNumberDigit; {A carry.}
var dd: BigNumberDigit2; {A digit product.}
var i,j,l: BigNumberIndexer; {Steppers.}
Begin
if ((A.TopDigit = 0) and (A.Digit[0] = 0))
or((B.TopDigit = 0) and (B.Digit[0] = 0)) then begin BigZero(BigMult); exit; end;
l:=A.TopDigit + B.TopDigit; {Minimal digit requirement. (Counting is from zero)}
if l > BigEnuff then Croak('BigMult will overflow.');
for i:=l downto 0 do X.Digit[i]:=0; {Clear for action.}
for i:=0 to A.TopDigit do {Arbitrarily, choose A on the one hand.}
begin {Though there could be a better choice.}
d:=A.Digit[i]; {Select the digit.}
if d <> 0 then {What the hell. One in BigBase chance.}
begin {But not this time.}
l:=i; {Locate the power of BigBase.}
c:=0; {Start this digit's multiply pass.}
for j:=0 to B.TopDigit do {Stepping along B's digits.}
begin {One by one.}
dd:=BigNumberDigit2(B.Digit[j])*d + X.Digit[l] + c; {The deed.}
X.Digit[l]:=dd mod BigBase; {Place the new digit.}
c:=dd div BigBase; {And extract the carry.}
l:=l + 1; {Ready for the next power up.}
end; {Advance to it.}
if c > 0 then {The multiply done, place the carry.}
begin {Ah. We *will* use the next power up.}
if l > BigEnuff then Croak('BigMultX has overflowed.'); {Oh dear.}
X.Digit[l]:=c; {Thus as if BigMult..Digit[l] was zeroed.}
l:=l + 1; {Preserve the one-too-far for the last case}
end; {So much for a carry at the end of a pass.}
end; {So much for a non-zero digit.}
end; {On to another digit to multiply with.}
X.TopDigit:=l - 1; {Remember the one-too-far.}
BigMult:=X; {Deliver, possibly scragging A or B, or, both!}
End; {of BigMult.}
Procedure BigPower(var X: BigNumber; P: longint); {Replaces X by X**P}
var A,W: BigNumber; {Scratchpads}
label up;
Begin {Each squaring doubles the power, melding nicely with binary reduction.}
if P <= 0 then Croak('Negative powers are not accommodated!');
BigOne(A); {x**0 = 1}
W:=X; {Holds X**1, 2, 4, 8, etc.}
up:if P mod 2 = 1 then A:=BigMult(A,W); {Bit on, so include this order.}
P:=P div 2; {Halve the power contrariwise to W's doubling.}
if P > 0 then {Still some power to come?}
begin {Yes.}
W:=BigMult(W,W); {Step up to the next bit's power.}
goto up; {And see if it is "on".}
end; {Odd layout avoids multiply testing P > 0.}
X:=A; {The result.}
End;
var X: BigNumber;
var p: longint;
BEGIN
ClrScr;
WriteLn('To calculate x = 2**64, then x*x via multi-digit long multiplication.');
p:=64; {As per the specification.}
X:=BigInt(2); {Start with 2.}
BigPower(X,p); {First stage: 2**64}
Write ('x = 2**',p,' = '); BigShow(X);
WriteLn;
X:=BigMult(X,X); {Second stage.}
Write ('x*x = ');BigShow(X); {Can't have Write('x*x = ',BigShow(BigMult(X,X))), after all. Oh well.}
END.
</lang>
Output:
To calculate x = 2**64, then x*x via multi-digit long multiplication. x = 2**64 = 18446744073709551616 x*x = 340282366920938463463374607431768211456
Perl
<lang perl>#!/usr/bin/perl -w use strict;
- This should probably be done in a loop rather than be recursive.
sub add_with_carry {
my $resultref = shift; my $addend = shift; my $addendpos = shift;
push @$resultref, (0) while (scalar @$resultref < $addendpos + 1); my $addend_result = $addend + $resultref->[$addendpos]; my @addend_digits = reverse split //, $addend_result; $resultref->[$addendpos] = shift @addend_digits;
my $carry_digit = shift @addend_digits; &add_with_carry($resultref, $carry_digit, $addendpos + 1) if( defined $carry_digit )
}
sub longhand_multiplication {
my @multiplicand = reverse split //, shift;
my @multiplier = reverse split //, shift;
my @result = ();
my $multiplicand_offset = 0;
foreach my $multiplicand_digit (@multiplicand)
{
my $multiplier_offset = $multiplicand_offset;
foreach my $multiplier_digit (@multiplier)
{
my $multiplication_result = $multiplicand_digit * $multiplier_digit;
my @result_digit_addend_list = reverse split //, $multiplication_result;
my $addend_offset = $multiplier_offset;
foreach my $result_digit_addend (@result_digit_addend_list)
{
&add_with_carry(\@result, $result_digit_addend, $addend_offset++)
}
++$multiplier_offset; }
++$multiplicand_offset; }
@result = reverse @result;
return join , @result;
}
my $sixtyfour = "18446744073709551616";
my $onetwentyeight = &longhand_multiplication($sixtyfour, $sixtyfour); print "$onetwentyeight\n";</lang>
Phix
base 10^9
Simple longhand multiplication. To keep things as simple as possible, this does not handle negative numbers.
If bcd1 is a number split into digits 0..9, bcd9 is a number split into "digits" 000,000,000..999,999,999, which fit in an integer.
They are held lsb-style mainly so that trimming a trailing 0 does not alter their value.
<lang Phix>constant base = 1_000_000_000
function bcd9_mult(sequence a, sequence b)
sequence c = repeat(0,length(a)+length(b))
for i=1 to length(a) do
integer j = i+length(b)-1
c[i..j] = sq_add(c[i..j],sq_mul(a[i],b))
end for
for i=1 to length(c) do
atom ci = c[i]
if ci>base then
c[i+1] += floor(ci/base) -- carry
c[i] = remainder(ci,base)
end if
end for
if c[$]=0 then
c = c[1..$-1]
end if
return c
end function
function atom_to_bcd9(atom a) sequence s = {}
while a>0 do
s = append(s,remainder(a,base))
a = floor(a/base)
end while
return s
end function
function bcd9_to_str(sequence a) string s = sprintf("%d",a[$])
for i=length(a)-1 to 1 by -1 do
s &= sprintf("%09d",a[i])
end for
-- (might want to trim leading 0s here)
return s
end function
sequence a, b, c
a = atom_to_bcd9(power(2,32)) printf(1,"a is %s\n",{bcd9_to_str(a)})
b = bcd9_mult(a,a) printf(1,"a*a is %s\n",{bcd9_to_str(b)})
c = bcd9_mult(b,b) printf(1,"a*a*a*a is %s\n",{bcd9_to_str(c)})</lang>
- Output:
a is 4294967296 a*a is 18446744073709551616 a*a*a*a is 340282366920938463488374607488768211456
string
<lang Phix>function mul(string a, b)
bool bSign = false
if a[1]='-' then {bSign,a} = {not bSign, a[2..$]} end if
if b[1]='-' then {bSign,b} = {not bSign, b[2..$]} end if
string res = repeat('0',length(a)+length(b))
--
-- Note that i,j,k are used as negative indexes, working
-- from the right hand least significant digit leftwards.
--
for i=1 to length(a) do
integer j=1, k=i, c=0
while j<=length(b) or c do
c += res[-k]-'0'
if j<=length(b) then
c += (a[-i]-'0')*(b[-j]-'0')
j += 1
end if
res[-k] = remainder(c,10)+'0'
c = floor(c/10)
k += 1
end while
end for
res = trim_head(res,"0")
if bSign then res = '-'&res end if
return res
end function ?mul("18446744073709551616","18446744073709551616")</lang>
- Output:
"340282366920938463488374607488768211456"
builtin
(same output as immediately above) <lang Phix>include mpfr.e mpz a = mpz_init("18446744073709551616") -- or: --mpz a = mpz_init(); mpz_ui_pow_ui(a,2,64) mpz_mul(a,a,a) ?mpz_get_str(a)</lang>
PHP
<lang PHP><?php function longMult($a, $b) {
$as = (string) $a;
$bs = (string) $b;
for($pi = 0, $ai = strlen($as) - 1; $ai >= 0; $pi++, $ai--)
{
for($p = 0; $p < $pi; $p++)
{
$regi[$ai][] = 0;
}
for($bi = strlen($bs) - 1; $bi >= 0; $bi--)
{
$regi[$ai][] = $as[$ai] * $bs[$bi];
}
}
return $regi;
}
function longAdd($arr) {
$outer = count($arr);
$inner = count($arr[$outer-1]) + $outer;
for($i = 0; $i <= $inner; $i++)
{
for($o = 0; $o < $outer; $o++)
{
$val = isset($arr[$o][$i]) ? $arr[$o][$i] : 0;
@$sum[$i] += $val;
}
}
return $sum;
}
function carry($arr) {
for($i = 0; $i < count($arr); $i++)
{
$s = (string) $arr[$i];
switch(strlen($s))
{
case 2:
$arr[$i] = $s{1};
@$arr[$i+1] += $s{0};
break;
case 3:
$arr[$i] = $s{2};
@$arr[$i+1] += $s{0}.$s{1};
break;
}
}
return ltrim(implode(,array_reverse($arr)),'0');
}
function lm($a,$b) {
return carry(longAdd(longMult($a,$b)));
}
if(lm('18446744073709551616','18446744073709551616') == '340282366920938463463374607431768211456')
{
echo 'pass!';
}; // 2^64 * 2^64
</Lang>
PicoLisp
<lang PicoLisp> (de multi (A B)
(setq A (format A) B (reverse (chop B)))
(let Result 0
(for (I . X) B
(setq Result (+ Result (* (format X) A (** 10 (dec I)))))) ) )
</lang>
PL/I
<lang PL/I>/* Multiply a by b, giving c. */ multiply: procedure (a, b, c);
declare (a, b, c) (*) fixed decimal (1); declare (d, e, f) (hbound(a,1)) fixed decimal (1); declare pr (-hbound(a,1) : hbound(a,1)) fixed decimal (1); declare p fixed decimal (2), (carry, s) fixed decimal (1); declare neg bit (1) aligned; declare (i, j, n, offset) fixed binary (31);
n = hbound(a,1);
d = a;
e = b;
s = a(1) + b(1);
neg = (s = 9);
if a(1) = 9 then call complement (d);
if b(1) = 9 then call complement (e);
pr = 0;
offset = 0; carry = 0;
do i = n to 1 by -1;
do j = n to 1 by -1;
p = d(i) * e(j) + pr(j-offset) + carry;
if p > 9 then do; carry = p/10; p = mod(p, 10); end; else carry = 0;
pr(j-offset) = p;
end;
offset = offset + 1;
end;
do i = hbound(a,1) to 1 by -1;
c(i) = pr(i);
end;
do i = -hbound(a,1) to 1;
if pr(i) ^= 0 then signal fixedoverflow;
end;
if neg then call complement (c);
end multiply;
complement: procedure (a);
declare a(*) fixed decimal (1); declare i fixed binary (31), carry fixed decimal (1); declare s fixed decimal (2);
carry = 1;
do i = hbound(a,1) to 1 by -1;
s = 9 - a(i) + carry;
if s > 9 then do; s = s - 10; carry = 1; end; else carry = 0;
a(i) = s;
end;
end complement;</lang> Calling sequence: <lang PL/I> a = 0; b = 0; c = 0;
a(60) = 1;
do i = 1 to 64; /* Generate 2**64 */
call add (a, a, b);
put skip;
call output (b);
a = b;
end;
call multiply (a, b, c);
put skip;
call output (c);</lang>
Final output:
18446744073709551616 340282366920938463463374607431768211456
PL/M
Uses bytes instead of integers to hold the digits.
Tested using a PLM286 to C converter and a simple I/O library. <lang plm>LONGMULT: DO;
/* long multiplication of large integers */
/* large integers are represented by arrays of bytes whose values are */
/* a single decimal digit of the number */
/* the least significant digits of the large integer are in element 1 */
/* element 0 should contain 0 if the number is positive, 1 if negative */
/* External I/O routines */
WRITE$STRING: PROCEDURE( S ) EXTERNAL; DECLARE S POINTER; END WRITE$STRING;
WRITE$CHAR: PROCEDURE( C ) EXTERNAL; DECLARE C BYTE; END WRITE$CHAR;
WRITE$NL: PROCEDURE EXTERNAL; END WRITE$NL;
/* End external routines */
DECLARE DIGIT$BASE LITERALLY '10';
DECLARE DIGIT$COUNT LITERALLY '49'; /* allows up to 48 digits */
/* - enough for the task */
DECLARE LARGE$INTEGER LITERALLY '(49)BYTE';
/* returns the position of the highest non-zero digit of the large */
/* integer a with n digits */
HIGHEST$NONZERO$DIGIT$POSITION: PROCEDURE( A$PTR, N ) BYTE;
DECLARE A$PTR POINTER;
DECLARE N BYTE;
DECLARE A BASED A$PTR LARGE$INTEGER;
DECLARE AMAX BYTE;
AMAX = N;
DO WHILE AMAX >= 1 AND A( AMAX ) = 0; AMAX = AMAX - 1; END;
RETURN AMAX;
END HIGHEST$NONZERO$DIGIT$POSITION ;
/* multiplies the large integer in b by the integer a, the result */
/* is added to c, starting from start */
/* overflow is ignored */
MULTIPLY$ELEMENT: PROCEDURE( A, B$PTR, C$PTR, START );
DECLARE ( B$PTR, C$PTR ) POINTER;
DECLARE ( A, START ) BYTE;
DECLARE ( B BASED B$PTR, C BASED C$PTR ) LARGE$INTEGER;
DECLARE ( CDIGIT, CARRY, BPOS, CPOS ) BYTE;
CARRY = 0;
CPOS = START;
DO BPOS = 1 TO HIGHEST$NONZERO$DIGIT$POSITION( B$PTR, DIGIT$COUNT - START );
CDIGIT = C( CPOS ) + ( A * B( BPOS ) ) + CARRY;
IF CDIGIT < DIGIT$BASE THEN CARRY = 0;
ELSE DO;
/* have digits to carry */
CARRY = CDIGIT / DIGIT$BASE;
CDIGIT = CDIGIT MOD DIGIT$BASE;
END;
C( CPOS ) = CDIGIT;
CPOS = CPOS + 1;
END;
IF CPOS <= ( DIGIT$COUNT - 1 ) THEN C( CPOS ) = CARRY;
END MULTIPLY$ELEMENT ;
/* implements long multiplication, c is set to a * b */
/* c can be the same array as a or b */
LONG$MULTIPLY: PROCEDURE( A$PTR, B$PTR, C$PTR );
DECLARE ( A$PTR, B$PTR, C$PTR ) POINTER;
DECLARE ( A BASED A$PTR
, B BASED B$PTR
, C BASED C$PTR
) LARGE$INTEGER;
DECLARE MRESULT LARGE$INTEGER;
DECLARE RPOS BYTE;
/* the result will be computed in mResult, allowing a or b to be c */
DO RPOS = 1 TO DIGIT$COUNT - 1; MRESULT( RPOS ) = 0; END;
/* multiply and add each digit to the result */
DO RPOS = 1 TO HIGHEST$NONZERO$DIGIT$POSITION( A$PTR, DIGIT$COUNT - 1 );
IF A( RPOS ) <> 0 THEN DO;
CALL MULTIPLY$ELEMENT( A( RPOS ), B$PTR, @MRESULT, RPOS );
END;
END;
/* return the result in c */
DO RPOS = 1 TO DIGIT$COUNT - 1; C( RPOS ) = MRESULT( RPOS ); END;
C( 0 ) = A( 0 ) XOR B( 0 ); /* handle the sign */
END;
/* writes the decimal value of a large integer a with n digits */
WRITE$LARGE$INTEGER: PROCEDURE( A$PTR );
DECLARE A$PTR POINTER;
DECLARE A BASED A$PTR LARGE$INTEGER;
DECLARE ( AMAX, APOS ) BYTE;
AMAX = HIGHEST$NONZERO$DIGIT$POSITION( A$PTR, DIGIT$COUNT - 1 );
IF AMAX < 1 THEN CALL WRITE$CHAR( '0' );
ELSE DO;
/* the large integer is non-zero */
IF A( 0 ) <> 0 THEN CALL WRITE$CHAR( '-' );
APOS = AMAX;
CALL WRITE$CHAR( A( APOS ) + '0' ); /* highest digit */
DO WHILE APOS > 1;
APOS = APOS - 1;
CALL WRITE$CHAR( A( APOS ) + '0' );
END;
END;
END WRITE$LARGE$INTEGER ;
/* calculate and output 2^128 */
MAIN: PROCEDURE;
DECLARE ( TWOTO64, TWOTO128 ) LARGE$INTEGER;
DECLARE ( PWR, TPOS ) BYTE;
/* construct 2^64 in twoTo64 */
DO TPOS = 0 TO DIGIT$COUNT - 1; TWOTO64( TPOS ) = 0; END;
TWOTO64( 1 ) = 2;
PWR = 1;
DO WHILE PWR < 64;
CALL LONG$MULTIPLY( @TWOTO64, @TWOTO64, @TWOTO64 );
PWR = PWR + PWR;
END;
/* construct 2^128 */
CALL LONG$MULTIPLY( @TWOTO64, @TWOTO64, @TWOTO128 );
CALL WRITE$STRING( @( '2^128: ', 0 ) );
CALL WRITE$LARGE$INTEGER( @TWOTO128 );
CALL WRITE$NL();
END MAIN;
END LONGMULT;</lang>
- Output:
2^128: 340282366920938463463374607431768211456
PowerShell
Implementation
<lang PowerShell>
- LongAddition only supports Unsigned Integers represented as Strings/Character Arrays
Function LongAddition ( [Char[]] $lhs, [Char[]] $rhs ) { $lhsl = $lhs.length $rhsl = $rhs.length if(($lhsl -gt 0) -and ($rhsl -gt 0)) { $maxplace = [Math]::Max($rhsl,$lhsl)+1 1..$maxplace | ForEach-Object { $carry = 0 $result = "" } { $add1 = 0 $add2 = 0 if( $_ -le $lhsl ) { $add1 = [int]$lhs[ -$_ ] - 48 } if( $_ -le $rhsl ) { $add2 = [int]$rhs[ -$_ ] - 48 } $iresult = $add1 + $add2 + $carry if( ( $_ -lt $maxplace ) -or ( $iresult -gt 0 ) ) { $result = "{0}{1}" -f ( $iresult % 10 ),$result } $carry = [Math]::Floor( $iresult / 10 ) } { $result } } elseif($lhsl -gt 0) { [String]::Join( , $lhs ) } elseif($rhsl -gt 0) { [String]::Join( , $rhs ) } else { "0" } }
- LongMultiplication only supports Unsigned Integers represented as Strings/Character Arrays
Function LongMultiplication ( [Char[]] $lhs, [Char[]] $rhs ) { $lhsl = $lhs.length $rhsl = $rhs.length if(($lhsl -gt 0) -and ($rhsl -gt 0)) { 1..$lhsl | ForEach-Object { $carry0 = "" $result0 = "" } { $i = -$_ $add1 = ( 1..$rhsl | ForEach-Object { $carry1 = 0 $result1 = "" } { $j = -$_ $mult1 = [int]$lhs[ $i ] - 48 $mult2 = [int]$rhs[ $j ] - 48 $iresult1 = $mult1 * $mult2 + $carry1 $result1 = "{0}{1}" -f ( $iresult1 % 10 ), $result1 $carry1 = [Math]::Floor( $iresult1 / 10 ) } { if( $carry1 -gt 0 ) { $result1 = "{0}{1}" -f $carry1, $result1 } $result1 } ) $iresult0 = ( LongAddition $add1 $carry0 ) $iresultl = $iresult0.length $result0 = "{0}{1}" -f $iresult0[-1],$result0 if( $iresultl -gt 1 ) { $carry0 = [String]::Join( , $iresult0[ -$iresultl..-2 ] ) } else { $carry0 = "" } } { if( $carry0 -ne "" ) { $result0 = "{0}{1}" -f $carry0, $result0 } $result0 } } else { "0" } }
LongMultiplication "18446744073709551616" "18446744073709551616"</lang>
Library Method
<lang PowerShell> [BigInt]$n = [Math]::Pow(2,64) [BigInt]::Multiply($n,$n) </lang> Output:
340282366920938463463374607431768211456
Prolog
Arbitrary precision arithmetic is native in most Prolog implementations. <lang Prolog> ?- X is 2**64 * 2**64. X = 340282366920938463463374607431768211456.</lang>
PureBasic
Explicit Implementation
<lang purebasic>Structure decDigitFmt ;decimal digit format
Array Digit.b(0) ;contains each digit of number, right-most digit is index 0
digitCount.i ;zero based
sign.i ; {x < 0} = -1, {x = 0} = 0, {x > 0} = 1
EndStructure
Global zero_decDigitFmt.decDigitFmt ;represents zero in the decimal digit format
- converts string representation of integer into the digit format, number can include signus but no imbedded spaces
Procedure stringToDecDigitFmt(numString.s, *x.decDigitFmt)
Protected *c.Character, digitIdx, digitCount
If numString And *x
*c.Character = @numString
Repeat
Select *c\c
Case '0' To '9', '-', '+'
*c + SizeOf(Character)
Default
numString = Left(numString, *c - @numString)
Break
EndSelect
ForEver
*c = @numString
Select *c\c
Case '-'
*x\sign = -1
*c + SizeOf(Character)
Case '+'
*x\sign = 1
*c + SizeOf(Character)
Case '0' To '9'
*x\sign = 1
EndSelect
numString = LTrim(PeekS(*c), "0") ;remove leading zeroes
If numString = "" ;is true if equal to zero or if only a signus is present
CopyStructure(@zero_decDigitFmt, *x, decDigitFmt)
ProcedureReturn
EndIf
*c = @numString
digitCount = Len(PeekS(*c)) - 1
Dim *x\Digit(digitCount)
*x\digitCount = digitCount
digitIdx = 0
While *c\c
If *c\c >= '0' And *c\c <= '9'
*x\Digit(digitCount - digitIdx) = *c\c - '0'
digitIdx + 1
*c + SizeOf(Character)
Else
Break
EndIf
Wend
EndIf
EndProcedure
- converts digit format representation of integer into string representation
Procedure.s decDigitFmtToString(*x.decDigitFmt)
Protected i, number.s
If *x
If *x\sign = 0
number = "0"
Else
For i = *x\digitCount To 0 Step -1
number + Str(*x\Digit(i))
Next
number = LTrim(number, "0")
If *x\sign = -1
number = "-" + number
EndIf
EndIf
EndIf
ProcedureReturn number
EndProcedure
- handles only positive numbers and zero, negative numbers left as an exercise for the reader ;)
Procedure add_decDigitFmt(*a.decDigitFmt, *b.decDigitFmt, *sum.decDigitFmt, digitPos = 0) ;*sum contains the result of (*a ) * 10^digitPos + (*b)
Protected carry, i, newDigitCount, workingSum, a_dup.decDigitFmt
If *a And *b And *sum
If *a = *sum: CopyStructure(*a, @a_dup, decDigitFmt): *a = @a_dup: EndIf ;handle special case of *sum + *b = *sum
If *b <> *sum: CopyStructure(*b, *sum, decDigitFmt): EndIf ;handle general case of *a + *b = *sum and special case of *a + *sum = *sum
;calculate number of digits needed for sum and resize array of digits if necessary
newDigitCount = *a\digitCount + digitPos
If newDigitCount >= *sum\digitCount
If *sum\digitCount = newDigitCount And *sum\Digit(*sum\digitCount) <> 0
newDigitCount + 1
EndIf
If *sum\digitCount <> newDigitCount
*sum\digitCount = newDigitCount
Redim *sum\Digit(*sum\digitCount)
EndIf
EndIf
i = 0
Repeat
If i <= *a\digitCount
workingSum = *a\Digit(i) + *sum\Digit(digitPos) + carry
Else
workingSum = *sum\Digit(digitPos) + carry
EndIf
If workingSum > 9
carry = 1
workingSum - 10
Else
carry = 0
EndIf
*sum\Digit(digitPos) = workingSum
digitPos + 1
i + 1
Until i > *a\digitCount And carry = 0
If *a\sign <> 0 Or *sum\sign <> 0
*sum\sign = 1 ;only handle positive numbers and zero for now
EndIf
EndIf
EndProcedure
Procedure multiply_decDigitFmt(*a.decDigitFmt, *b.decDigitFmt, *product.decDigitFmt) ;*product contains the result of (*a) * (*b)
Protected i, digitPos, productSignus
Protected Dim multTable.decDigitFmt(9)
Protected NewList digitProduct.decDigitFmt()
If *a And *b And *product
If *a\sign = 0 Or *b\sign = 0
CopyStructure(zero_decDigitFmt, *product, decDigitFmt)
ProcedureReturn
EndIf
If *b\digitCount > *a\digitCount: Swap *a, *b: EndIf
;build multiplication table
CopyStructure(*a, @multTable(1), decDigitFmt): multTable(1)\sign = 1 ;always positive
For i = 2 To 9
add_decDigitFmt(*a, multTable(i - 1), multTable(i))
Next
;collect individual digit products for later summation; these could also be added as we go along
For i = 0 To *b\digitCount
AddElement(digitProduct())
digitProduct() = multTable(*b\Digit(i))
Next
;determine sign of product
If *a\sign <> *b\sign
productSignus = -1
Else
productSignus = 1
EndIf
digitPos = 0
CopyStructure(zero_decDigitFmt, *product, decDigitFmt)
ForEach digitProduct()
add_decDigitFmt(digitProduct(), *product, *product, digitPos)
digitPos + 1
Next
*product\sign = productSignus ;set sign of product
EndIf
EndProcedure
- handles only positive integer exponents or an exponent of zero, does not raise an error for 0^0
Procedure exponent_decDigitFmt(*a.decDigitFmt, exponent, *product.decDigitFmt)
Protected i, a_dup.decDigitFmt
If *a And *product And exponent >= 0
If *a = *product: CopyStructure(*a, @a_dup, decDigitFmt): *a = @a_dup: EndIf
stringToDecDigitFmt("1", *product)
For i = 1 To exponent: multiply_decDigitFmt(*product, *a, *product): Next
EndIf
EndProcedure
If OpenConsole()
Define a.decDigitFmt, product.decDigitFmt
stringToDecDigitFmt("2", a)
exponent_decDigitFmt(a, 64, a) ;2^64
multiply_decDigitFmt(a, a, product)
PrintN("The result of 2^64 * 2^64 is " + decDigitFmtToString(product))
Print(#crlf$ + #crlf$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf</lang> Output:
The result of 2^64 * 2^64 is 340282366920938463463374607431768211456
Library Method
Using Decimal.pbi by Stargåte allows for calculation with long numbers, this is useful since version 4.41 of PureBasic mostly only supporter data types native to x86/x64/PPC etc processors. <lang PureBasic>XIncludeFile "decimal.pbi"
Define.Decimal *a, *b
- a=PowerDecimal(IntegerToDecimal(2),IntegerToDecimal(64))
- b=TimesDecimal(*a,*a,#NoDecimal)
Print("2^64*2^64 = "+DecimalToString(*b))</lang>
Outputs
2^64*2^64 = 340282366920938463463374607431768211456
Python
(Note that Python comes with arbitrary length integers).
<lang python>#!/usr/bin/env python print 2**64*2**64</lang>
<lang python>#!/usr/bin/env python
def add_with_carry(result, addend, addendpos):
while True:
while len(result) < addendpos + 1:
result.append(0)
addend_result = str(int(addend) + int(result[addendpos]))
addend_digits = list(addend_result)
result[addendpos] = addend_digits.pop()
if not addend_digits:
break
addend = addend_digits.pop()
addendpos += 1
def longhand_multiplication(multiplicand, multiplier):
result = []
for multiplicand_offset, multiplicand_digit in enumerate(reversed(multiplicand)):
for multiplier_offset, multiplier_digit in enumerate(reversed(multiplier), start=multiplicand_offset):
multiplication_result = str(int(multiplicand_digit) * int(multiplier_digit))
for addend_offset, result_digit_addend in enumerate(reversed(multiplication_result), start=multiplier_offset):
add_with_carry(result, result_digit_addend, addend_offset)
result.reverse()
return .join(result)
if __name__ == "__main__":
sixtyfour = "18446744073709551616"
onetwentyeight = longhand_multiplication(sixtyfour, sixtyfour) print(onetwentyeight)</lang>
Shorter version:
<lang python>Long multiplication
from functools import reduce
def longmult(x, y):
Long multiplication.
return reduce(
digitSum,
polymul(digits(x), digits(y)), 0
)
def digitSum(a, x):
Left to right decimal digit summing. return a * 10 + x
def polymul(xs, ys):
List of specific products.
return map(
lambda *vs: sum(filter(None, vs)),
*[
[0] * i + zs for i, zs in
enumerate(mult_table(xs, ys))
]
)
def mult_table(xs, ys):
Rows of all products. return [[x * y for x in xs] for y in ys]
def digits(x):
Digits of x as a list of integers. return [int(c) for c in str(x)]
if __name__ == '__main__':
print(
longmult(2 ** 64, 2 ** 64)
)</lang>
Quackery
Long multiplication as it was taught at primary school, using natural numbers (including zero) in the base 10 numeral system, with a slight variation in that it maintains a running total rather than adding up the intermediate results column-wise at the end. Starting point is "learn your tables". Presumptions are an understanding of "equal to" and "not equal to", and the ability to split a one or two digit number into tens and units.
(Splitting a number into tens and units is achieved using /mod to keep the tables compact. Numbers in the addition and multiplication tables could be represented as, for example, [ 8 1 ] instead of 81 and 10 /mod replaced with unpack or, as the nests only contains numbers, do.)
In addition to the specified task, we were always encouraged to show our workings.
<lang Quackery> [ [] swap witheach
[ char 0 -
swap join ] ] is $->long ( $ --> L )
[ number$ $->long ] is long ( n --> L )
[ reverse behead
swap witheach
[ swap 10 * + ] ] is long->num ( L --> n )
[ reverse witheach echo ] is echolong ( L --> )
[ over size
over size -
dup dip
[ 0 < if swap ]
abs times
[ 0 join ] ] is zeropad ( L L --> L L )
[ [ table
[ 0 1 2 3 4 5 6 7 8 9 ]
[ 1 2 3 4 5 6 7 8 9 10 ]
[ 2 3 4 5 6 7 8 9 10 11 ]
[ 3 4 5 6 7 8 9 10 11 12 ]
[ 4 5 6 7 8 9 10 11 12 13 ]
[ 5 6 7 8 9 10 11 12 13 14 ]
[ 6 7 8 9 10 11 12 13 14 15 ]
[ 7 8 9 10 11 12 13 14 15 16 ]
[ 8 9 10 11 12 13 14 15 16 17 ]
[ 9 10 11 12 13 14 15 16 17 18 ] ]
swap peek 10 /mod ] is add ( n n --> n n )
[ dip add add dip [ add nip ] swap ] is addc ( n n c --> n c )
[ zeropad
0 temp put
[] unrot witheach
[ dip behead
temp take
addc
temp put
rot swap join swap ]
drop
temp take dup 0 !=
iff join else drop ] is longadd ( L L --> L )
[ [ table
[ 0 0 0 0 0 0 0 0 0 0 ]
[ 0 1 2 3 4 5 6 7 8 9 ]
[ 0 2 4 6 8 10 12 14 16 18 ]
[ 0 3 6 9 12 15 18 21 24 27 ]
[ 0 4 8 12 16 20 24 28 32 36 ]
[ 0 5 10 15 20 25 30 35 40 45 ]
[ 0 6 12 18 24 30 36 42 48 54 ]
[ 0 7 14 21 28 35 42 49 56 63 ]
[ 0 8 16 24 32 40 48 56 64 72 ]
[ 0 9 18 27 36 45 54 63 72 81 ] ]
swap peek 10 /mod ] is mult ( n n --> n n )
[ dip mult add dip [ add nip ] swap ] is multc ( n n c --> n c )
[ dup 0 = iff
[ 2drop 0 long ] done
0 temp put
[] unrot swap witheach
[ over temp take
multc
temp put
swap dip join ]
drop
temp take dup 0 !=
iff join else drop ] is shortmult ( L n --> L )
[ dup 0 long != iff [ 0 swap join ] ] is timesten ( L --> L )
[ dup 0 long = iff
[ 2drop 0 long ] done
0 long unrot
witheach
[ dip dup shortmult
rot longadd swap
timesten ]
drop ] is longmult ( L L --> L )
( ------------------------ additional to task ------------------------ )
[ stack ] is linelength ( --> s )
[ linelength share times
[ char - emit ]
cr ] is separator ( --> )
[ linelength share over size - times sp echolong cr ] is showlong ( --> )
[ over size
over size + linelength put
over showlong
dup showlong
separator
dup 0 long = iff
[ 2drop 0 long ] done
0 long unrot
witheach
[ dip dup shortmult
dup showlong
rot longadd swap
timesten ]
separator
swap showlong
separator
drop linelength release ] is workings ( L L --> )
say "Using long multiplication: "
2 64 ** long dup longmult dup echolong cr
say "Using built-in arithmetic: " 2 128 ** dup echo cr cr
swap long->num = iff
say "10/10, Gold star!"
else
say "0/10, See me after class."
cr cr say "(Show your workings.)" cr cr 2 64 ** long dup workings cr</lang>
- Output:
Using long multiplication: 340282366920938463463374607431768211456
Using built-in arithmetic: 340282366920938463463374607431768211456
10/10, Gold star!
(Show your workings.)
18446744073709551616
18446744073709551616
----------------------------------------
110680464442257309696
184467440737095516160
11068046444225730969600
18446744073709551616000
922337203685477580800000
9223372036854775808000000
166020696663385964544000000
0
12912720851596686131200000000
55340232221128654848000000000
1291272085159668613120000000000
0
73786976294838206464000000000000
737869762948382064640000000000000
12912720851596686131200000000000000
110680464442257309696000000000000000
737869762948382064640000000000000000
7378697629483820646400000000000000000
147573952589676412928000000000000000000
184467440737095516160000000000000000000
----------------------------------------
340282366920938463463374607431768211456
----------------------------------------
R
Using GMP
<lang R>library(gmp) a <- as.bigz("18446744073709551616") mul.bigz(a,a)</lang>
"340282366920938463463374607431768211456"
A native implementation
This code is more verbose than necessary, for ease of understanding. <lang R>longmult <- function(xstr, ystr) {
#get the number described in each string
getnumeric <- function(xstr) as.numeric(unlist(strsplit(xstr, "")))
x <- getnumeric(xstr)
y <- getnumeric(ystr)
#multiply each pair of digits together
mat <- apply(x %o% y, 1, as.character)
#loop over columns, then rows, adding zeroes to end of each number in the matrix to get the correct positioning
ncols <- ncol(mat)
cols <- seq_len(ncols)
for(j in cols)
{
zeroes <- paste(rep("0", ncols-j), collapse="")
mat[,j] <- paste(mat[,j], zeroes, sep="")
}
nrows <- nrow(mat)
rows <- seq_len(nrows)
for(i in rows)
{
zeroes <- paste(rep("0", nrows-i), collapse="")
mat[i,] <- paste(mat[i,], zeroes, sep="")
}
#add zeroes to the start of the each number, so they are all the same length
len <- max(nchar(mat))
strcolumns <- formatC(cbind(as.vector(mat)), width=len)
strcolumns <- gsub(" ", "0", strcolumns)
#line up all the numbers below each other
strmat <- matrix(unlist(strsplit(strcolumns, "")), byrow=TRUE, ncol=len)
#convert to numeric and add them
mat2 <- apply(strmat, 2, as.numeric)
sum1 <- colSums(mat2)
#repeat the process on each of the totals, until each total is a single digit
repeat
{
ntotals <- length(sum1)
totals <- seq_len(ntotals)
for(i in totals)
{
zeroes <- paste(rep("0", ntotals-i), collapse="")
sum1[i] <- paste(sum1[i], zeroes, sep="")
}
len2 <- max(nchar(sum1))
strcolumns2 <- formatC(cbind(as.vector(sum1)), width=len2)
strcolumns2 <- gsub(" ", "0", strcolumns2)
strmat2 <- matrix(unlist(strsplit(strcolumns2, "")), byrow=TRUE, ncol=len2)
mat3 <- apply(strmat2, 2, as.numeric)
sum1 <- colSums(mat3)
if(all(sum1 < 10)) break
}
#Concatenate the digits together
ans <- paste(sum1, collapse="")
ans
}
a <- "18446744073709551616" longmult(a, a)</lang>
"340282366920938463463374607431768211456"
Racket
<lang Racket>
- lang racket
(define (mult A B)
(define nums
(let loop ([B B] [zeros '()])
(if (null? B)
'()
(cons (append zeros (let loop ([c 0] [A A])
(cond [(pair? A)
(define-values [q r]
(quotient/remainder
(+ c (* (car A) (car B)))
10))
(cons r (loop q (cdr A)))]
[(zero? c) '()]
[else (list c)])))
(loop (cdr B) (cons 0 zeros))))))
(let loop ([c 0] [nums nums])
(if (null? nums)
'()
(let-values ([(q r) (quotient/remainder (apply + c (map car nums)) 10)])
(cons r (loop q (filter pair? (map cdr nums))))))))
(define (number->list n)
(if (zero? n) '()
(let-values ([(q r) (quotient/remainder n 10)])
(cons r (number->list q)))))
(define 2^64 (number->list (expt 2 64))) (for-each display (reverse (mult 2^64 2^64))) (newline)
- for comparison
(* (expt 2 64) (expt 2 64))
- Output
- 340282366920938463463374607431768211456
- 340282366920938463463374607431768211456
</lang>
Raku
(formerly Perl 6)
For efficiency (and novelty), this program explicitly implements long multiplication, but in base 10000. That base was chosen because multiplying two 5-digit numbers can overflow a 32-bit integer, but two 4-digit numbers cannot. <lang perl6>sub num_to_groups ( $num ) { $num.flip.comb(/.**1..4/)».flip }; sub groups_to_num ( @g ) { [~] flat @g.pop, @g.reverse».fmt('%04d') };
sub long_multiply ( Str $x, Str $y ) {
my @group_sums;
for flat num_to_groups($x).pairs X num_to_groups($y).pairs -> $xp, $yp {
@group_sums[ $xp.key + $yp.key ] += $xp.value * $yp.value;
}
for @group_sums.keys -> $k {
next if @group_sums[$k] < 10000;
@group_sums[$k+1] += @group_sums[$k].Int div 10000;
@group_sums[$k] %= 10000;
}
return groups_to_num @group_sums;
}
my $str = '18446744073709551616'; long_multiply( $str, $str ).say;
- cross-check with native implementation
say +$str * +$str;</lang>
- Output:
340282366920938463463374607431768211456 340282366920938463463374607431768211456
REXX
version 1
This REXX version supports:
- leading signs
- decimal points
- automatically adjusting the number of decimal digits needed
Programming note: && is REXX's exclusive or operand. <lang rexx>/*REXX program performs long multiplication on two numbers (without the "E"). */ numeric digits 300 /*be able to handle gihugeic input #s. */ parse arg x y . /*obtain optional arguments from the CL*/ if x== | x=="," then x=2**64 /*Not specified? Then use the default.*/ if y== | y=="," then y=x /* " " " " " " */ if x<0 && y<0 then sign= '-' /*there only a single negative number? */
else sign= /*no, then result sign must be positive*/
xx=x; x=strip(x, 'T', .); x1=left(x, 1) /*remove any trailing decimal points. */ yy=y; y=strip(y, 'T', .); y1=left(y, 1) /* " " " " " */ if x1=='-' | x1=="+" then x=substr(x, 2) /*remove a leading ± sign. */ if y1=='-' | y1=="+" then y=substr(y, 2) /* " " " " " */ parse var x '.' xf; parse var y "." yf /*obtain the fractional part of X and Y*/
- =length(xf || yf) /*#: digits past the decimal points (.)*/
x=space( translate( x, , .), 0) /*remove decimal point if there is any.*/ y=space( translate( y, , .), 0) /* " " " " " " " */ Lx=length(x); Ly=length(y) /*get the lengths of the new X and Y. */ numeric digits max(digits(), Lx + Ly) /*use a new decimal digits precision.*/ $=0 /*$: is the product (so far). */
do j=Ly by -1 for Ly /*almost like REXX does it, ··· but no.*/
$=$ + ((x*substr(y, j, 1))copies(0, Ly-j) )
end /*j*/
f=length($) - # /*does product has enough decimal digs?*/ if f<0 then $=copies(0, abs(f) + 1)$ /*Negative? Add leading 0s for INSERT.*/ say 'long mult:' xx "*" yy '──►' sign || strip( insert(., $, length($) - #), 'T', .) say ' built─in:' xx "*" yy '──►' xx*yy /*stick a fork in it, we're all done. */</lang> output when using the default inputs:
long mult: 18446744073709551616 * 18446744073709551616 ──► 340282366920938463463374607431768211456 built─in: 18446744073709551616 * 18446744073709551616 ──► 340282366920938463463374607431768211456
output when using the inputs of: 123 -456789000
long mult: 123 * -456789000 ──► -56185047000 built─in: 123 * -456789000 ──► -56185047000
output when using the inputs of: -123.678 +456789000
long mult: -123.678 * +456789000 ──► -56494749942.000 built─in: -123.678 * +456789000 ──► -56494749942.000
version 2
<lang rexx>/* REXX **************************************************************
- While REXX can multiply arbitrary large integers
- here is the algorithm asked for by the task description
- 13.05.2013 Walter Pachl
- /
cnt.=0 Numeric Digits 100 Call test 123 123 Call test 12 12 Call test 123456789012 44444444444 Call test 2**64 2**64 Call test 0 0 say cnt.0ok 'ok' say cnt.0nok 'not ok' Exit test:
Parse Arg a b soll=a*b haben=multiply(a b) Say 'soll =' soll Say 'haben=' haben If haben<>soll Then cnt.0nok=cnt.0nok+1 Else cnt.0ok=cnt.0ok+1 Return
multiply: Procedure /* REXX **************************************************************
- Multiply(a b) -> a*b
- /
Parse Arg a b
Call s2a 'a'
Call s2a 'b'
r.=0
rim=1
r0=0
Do bi=1 To b.0
Do ai=1 To a.0
ri=ai+bi-1
p=a.ai*b.bi
Do i=ri by 1 Until p=0
s=r.i+p
r.i=s//10
p=s%10
End
rim=max(rim,i)
End
End
res=strip(a2s('r'),'L','0')
If res= Then
res='0'
Return res
s2a: /**********************************************************************
- copy characters of a string into a corresponding array
- digits are numbered 1 to n fron right to left
- /
Parse arg name string=value(name) lstring=length(string) do z=1 to lstring Call value name'.'z,substr(string,lstring-z+1,1) End Call value name'.0',lstring Return
a2s: /**********************************************************************
- turn the array of digits into a string
- /
call trace 'o' Parse Arg name ol= Do z=rim To 1 By -1 ol=ol||value(name'.z') End Return ol</lang>
Output:
soll = 15129 haben= 15129 soll = 144 haben= 144 soll = 5486968400478463649328 haben= 5486968400478463649328 soll = 340282366920938463463374607431768211456 haben= 340282366920938463463374607431768211456 soll = 0 haben= 0 5 ok 0 not ok
Ring
<lang ring> decimals(0) see pow(2,64)*pow(2,64) + nl </lang> Output:
340282366920938463463374607431768211456
Ruby
<lang ruby>def longmult(x,y)
result = [0]
j = 0
y.digits.each do |m|
c = 0
i = j
x.digits.each do |d|
v = result[i]
result << 0 if v.zero?
c, v = (v + c + d*m).divmod(10)
result[i] = v
i += 1
end
result[i] += c
j += 1
end
# calculate the answer from the result array of digits
result.reverse.inject(0) {|sum, n| 10*sum + n}
end
n=2**64 printf " %d * %d = %d\n", n, n, n*n printf "longmult(%d, %d) = %d\n", n, n, longmult(n,n)</lang>
18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456 longmult(18446744073709551616, 18446744073709551616) = 340282366920938463463374607431768211456
Scala
This implementation does not rely on an arbitrary precision numeric type. Instead, only single digits are ever multiplied or added, and all partial results are kept as string.
<lang scala>def addNums(x: String, y: String) = {
val padSize = x.length max y.length
val paddedX = "0" * (padSize - x.length) + x
val paddedY = "0" * (padSize - y.length) + y
val (sum, carry) = (paddedX zip paddedY).foldRight(("", 0)) {
case ((dx, dy), (acc, carry)) =>
val sum = dx.asDigit + dy.asDigit + carry
((sum % 10).toString + acc, sum / 10)
}
if (carry != 0) carry.toString + sum else sum
}
def multByDigit(num: String, digit: Int) = {
val (mult, carry) = num.foldRight(("", 0)) {
case (d, (acc, carry)) =>
val mult = d.asDigit * digit + carry
((mult % 10).toString + acc, mult / 10)
}
if (carry != 0) carry.toString + mult else mult
}
def mult(x: String, y: String) =
y.foldLeft("")((acc, digit) => addNums(acc + "0", multByDigit(x, digit.asDigit)))</lang>
Sample:
scala> mult("18446744073709551616", "18446744073709551616")
res25: java.lang.String = 340282366920938463463374607431768211456
Scala 2.8 introduces `scanLeft` and `scanRight` which can be used to simplify this further:
<lang scala>def adjustResult(result: IndexedSeq[Int]) = (
result .map(_ % 10) // remove carry from each digit .tail // drop the seed carry .reverse // put most significant digits on the left .dropWhile(_ == 0) // remove leading zeroes .mkString
)
def addNums(x: String, y: String) = {
val padSize = (x.length max y.length) + 1 // We want to keep a zero to the left, to catch the carry
val paddedX = "0" * (padSize - x.length) + x
val paddedY = "0" * (padSize - y.length) + y
adjustResult((paddedX zip paddedY).scanRight(0) {
case ((dx, dy), last) => dx.asDigit + dy.asDigit + last / 10
})
}
def multByDigit(num: String, digit: Int) = adjustResult(("0"+num).scanRight(0)(_.asDigit * digit + _ / 10))
def mult(x: String, y: String) =
y.foldLeft("")((acc, digit) => addNums(acc + "0", multByDigit(x, digit.asDigit)))
</lang>
Scheme
Since Scheme already supports arbitrary precision arithmetic, build it out of church numerals. Don't try converting these to native integers. You will die waiting for the answer.
<lang scheme>(define one (lambda (f) (lambda (x) (f x)))) (define (add a b) (lambda (f) (lambda (x) ((a f) ((b f) x))))) (define (mult a b) (lambda (f) (lambda (x) ((a (b f)) x)))) (define (expo a b) (lambda (f) (lambda (x) (((b a) f) x)))) (define two (add one one)) (define six (add two (add two two))) (define sixty-four (expo two six)) (display (mult (expo two sixty-four) (expo two sixty-four)))</lang>
- Output:
(as run on Chicken Scheme on tio)
#<procedure (? f)>
Seed7
Seed7 supports arbitrary-precision arithmetic. The library bigint.s7i defines the type bigInteger. A bigInteger is a signed integer number of unlimited size. With library support the task can be solved by using the multiplication operator *:
<lang seed7>$ include "seed7_05.s7i";
include "bigint.s7i";
const proc: main is func
begin writeln(2_**64 * 2_**64); end func;</lang>
Output:
340282366920938463463374607431768211456
This task seems to prefer an inferior implementation of a long multiplication, where long numbers are stored in decimal strings. Besides type safety there are seveal other drawbacks triggered by such a representation. E.g.: In almost all cases a representation with decimal strings leads to significant lower computing speed. The multiplication example below uses the requested inferior implementation:
<lang seed7>$ include "seed7_05.s7i";
const func string: (in string: a) * (in string: b) is func
result
var string: product is "";
local
var integer: i is 1;
var integer: j is 1;
var integer: k is 0;
var integer: carry is 0;
begin
if startsWith(a, "-") then
if startsWith(b, "-") then
product := a[2 ..] * b[2 ..];
else
product := "-" & a[2 ..] * b;
end if;
elsif startsWith(b, "-") then
product := "-" & a * b[2 ..];
else
product := "0" mult length(a) + length(b);
for i range length(a) downto 1 do
k := i + length(b);
carry := 0;
for j range length(b) downto 1 do
carry +:= (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0')) + (ord(product[k]) - ord('0'));
product @:= [k] chr(carry rem 10 + ord('0'));
carry := carry div 10;
decr(k);
end for;
product @:= [k] chr(ord(product[k]) + carry);
end for;
while startsWith(product, "0") and length(product) >= 2 do
product := product[2 ..];
end while;
end if;
end func;
const proc: main is func
begin
writeln("-18446744073709551616" * "-18446744073709551616");
end func;</lang>
The output is the same as with the superior solution.
Sidef
(Note that arbitrary precision arithmetic is native in Sidef). <lang ruby>say (2**64 * 2**64);</lang>
<lang ruby>func add_with_carry(result, addend, addendpos) {
loop {
while (result.len < addendpos+1) {
result.append(0)
}
var addend_digits = (addend.to_i + result[addendpos] -> to_s.chars)
result[addendpos] = addend_digits.pop
addend_digits.len > 0 || break
addend = addend_digits.pop
addendpos++
}
} func longhand_multiplication(multiplicand, multiplier) {
var result = [] var multiplicand_offset = 0
multiplicand.reverse.each { |multiplicand_digit|
var multiplier_offset = multiplicand_offset
multiplier.reverse.each { |multiplier_digit|
var multiplication_result = (multiplicand_digit.to_i * multiplier_digit.to_i -> to_s)
var addend_offset = multiplier_offset
multiplication_result.reverse.each { |result_digit_addend|
add_with_carry(result, result_digit_addend, addend_offset)
addend_offset++
}
multiplier_offset++
}
multiplicand_offset++
}
return result.join.reverse
} say longhand_multiplication('18446744073709551616', '18446744073709551616')</lang>
- Output:
340282366920938463463374607431768211456
Slate
<lang slate>(2 raisedTo: 64) * (2 raisedTo: 64).</lang>
Smalltalk
Note that arbitrary precision arithmetic is native in Smalltalk, and no-one would reinvent the wheel. <lang smalltalk>(2 raisedTo: 64) * (2 raisedTo: 64).</lang> or, to display it: <lang smalltalk>Transcript showCR:(2 raisedTo: 64) * (2 raisedTo: 64). "if ** is defined as alias: " Transcript showCR:(2 ** 64) * (2 ** 64).</lang>
- Output:
340282366920938463463374607431768211456
There has been some discussion, if the above is fair... but then, I guess even using 32bit arithmetic is unfair to 8bit assembly language machines ;-) or comparing languages which have a 128bit integer against others... And is it fair, to change the challenge afterwards?
Anyway, here is a version which works on 2-digit decimal machine of the 1940's... (not that I know of any Smalltalk ever ported to a Zuse 1 :-)
<lang Smalltalk>"/ mhmh hard to avoid largeInteger arithmetic, "/ as the language does not specify, how many bits are used to represent "/ SmallIntegers, and when the VM uses LargeInts. "/ Lets assume, we run on a 2-digit decimal machine (smile). "/ So lets work hard to avoid any convenient VM support, "/ by doing decimal arithmetic (running on a decimal machine from the 1940s) "/ and only allow 0..99 in a word (assuming it has a 2*2->4 digit multiply available) "/ (smile: remember the Knuth MIX machine?) "/ Long integers are represented as an array of such 2-digit words (least significant first).
"/ the code below should never ever been taken serious "/ Not even as didactic example. "/ NOONE WOULD EVER DO SUCH A STUPID THING
WORDMAX := 100. add :=
[:a :b |
Array streamContents:[:s |
|cy|
cy := 0.
1 to:(a size max:b size) do:[:wordIndex |
|sum|
wA := a at:wordIndex ifAbsent:0.
wB := b at:wordIndex ifAbsent:0.
sum := (wA + wB + cy).
cy := (sum // WORDMAX).
s nextPut:(sum % WORDMAX).
].
cy ~~ 0 ifTrue:[s nextPut:cy].
].
].
"/ test 12,34 + 1 a := #( 34 12 ). b := #( 1 ). self assert:( add value:a value:b ) = #( 35 12 ).
"/ test 99,99 + 1 a := #( 99 99 ). b := #( 1 ). self assert:( add value:a value:b ) = #( 00 00 1 ).
"/ test 99,99,99,99 + 99,99,99,99 a := #( 99 99 99 99 ). b := #( 99 99 99 99 ). self assert:( add value:a value:b ) = #( 98 99 99 99 1).
mulW :=
[:a :w |
|cy|
cy := 0.
Array streamContents:[:s |
a do:[:wordA |
|product|
product := (wordA * w) + cy.
s nextPut:(product % WORDMAX).
cy := (product // WORDMAX)
].
cy ~~ 0 ifTrue:[s nextPut:cy].
]
].
"/ test 1 * 2 a := #( 1 ). self assert:( mulW value:a value:2) = #( 2).
"/ test 2 * 99 a := #( 2 ). self assert:( mulW value:a value:99) = #( 98 1).
"/ test 99,99,99,99 * 99 a := #( 99 99 99 99 ). self assert:( mulW value:a value:99) = #( 01 99 99 99 98 ).
mul :=
[:a :b |
|sum|
sum := #( 0 ).
b doWithIndex:[:wordB :wordIndex |
partSum := mulW value:a value:wordB.
shifted := (Array new:wordIndex-1 withAll:0),partSum.
sum := add value:sum value:shifted.
].
sum.
].
"/ test 99,99,99,99 * 99 a := #( 99 99 99 99 ). b := #( 99 ). self assert:( mul value:a value:b) = #( 01 99 99 99 98 ).
raise :=
[:a :exp |
|e rslt|
rslt := #(1).
t := a.
e := exp.
[e ~~ 0] whileTrue:[
[(e bitAnd:1) == 0] whileTrue:[
e := e bitShift:-1.
t := mul value:t value:t.
].
e := e - 1.
rslt := mul value:rslt value:t.
].
rslt.
].
"/ test 2 ** 64 a := #( 2 ). self assert:( raise value:a value:64) = #( 16 16 55 09 37 07 44 67 44 18).
"/ test (2 ** 64) * (2 ** 64) a := #( 2 ). t := raise value:a value:64. rslt := mul value:t value:t. self assert:rslt = #( 56 14 21 68 17 43 07 46 37 63 34 46 38 09 92 66 23 28 40 3).
"/ the biggest plus of having a decimal machine is that it makes printing so easy... printOn :=
[:n :stream |
|first|
first := true.
n reverseDo:[:_2Digits |
first
ifTrue:[ stream nextPutAll:(_2Digits printString)]
ifFalse:[ stream nextPutAll:(_2Digits printString leftPaddedTo:2 with:$0)].
first := false.
].
].
printOn value:rslt value:Transcript.
"/ verify... printedString := String streamContents:[:s | printOn value:rslt value:s]. self assert:(printedString = (2**64) squared printString)</lang>
- Output:
3402823669293846346337467431768211456
The above code does not really integrate into the Smalltalk class library. For example, it will not allow mixed mode arithmetic between regular integers and Rosetta integers. Here is a full example in portable chunk file format which makes mixed mode arithmetic completely transparent (I implemented only addition and multiplication): <lang smalltalk>Integer
subclass: #RosettaInteger instanceVariableNames:'digitArray' classVariableNames:'WORDMAX' package:'Rosetta demos'
!
!RosettaInteger class methodsFor:'initialization'!
initialize
WORDMAX := 100.
! !
!RosettaInteger class methodsFor:'instance creation'!
newWithDigits:digitArray
"returns a new RosettaInteger with a digitArray"
^ self basicNew digits:digitArray
!
fromInteger:anInteger
"returns a new RosettaInteger with anInteger's value"
|digits gen|
gen := [:n :s |
s nextPut:(n % 100).
n > 99 ifTrue:[ gen value:(n // 100) value:s]].
digits := Array streamContents:[:s | gen value:anInteger value:s].
^ self newWithDigits:digits
! !
!RosettaInteger class methodsFor:'helpers'!
addDigits:a and:b
|add|
add :=
[:a :b |
Array streamContents:[:s |
|cy|
cy := 0.
1 to:(a size max:b size) do:[:wordIndex |
|sum|
wA := a at:wordIndex ifAbsent:0.
wB := b at:wordIndex ifAbsent:0.
sum := (wA + wB + cy).
cy := (sum // WORDMAX).
s nextPut:(sum % WORDMAX).
].
cy ~~ 0 ifTrue:[s nextPut:cy].
].
].
^ add value:a value:b
!
mulDigits:a and:b
|mulW|
mulW :=
[:a :w |
|cy|
cy := 0.
Array streamContents:[:s |
a do:[:wordA |
|product|
product := (wordA * w) + cy.
s nextPut:(product % WORDMAX).
cy := (product // WORDMAX)
].
cy ~~ 0 ifTrue:[s nextPut:cy].
]
].
mul :=
[:a :b |
|sum|
sum := #( 0 ).
b doWithIndex:[:wordB :wordIndex |
partSum := mulW value:a value:wordB.
shifted := (Array new:wordIndex-1 withAll:0),partSum.
sum := self addDigits:sum and:shifted.
].
sum.
].
^ mul value:a value:b.
! !
!RosettaInteger methodsFor:'private accessing'! digits
"return my digitArray"
^ digitArray
!
digits:digits
"set my digitArray"
digitArray := digits
! !
!RosettaInteger methodsFor:'arithmetic'! + aNumber
^ aNumber sumFromRosettaInteger:self
!
- aNumber
^ aNumber productFromRosettaInteger:self
!
raisedTo:exp
|raise|
raise :=
[:a :exp |
|e rslt|
rslt := #(1).
t := a.
e := exp.
[e ~~ 0] whileTrue:[
[(e bitAnd:1) == 0] whileTrue:[
e := e bitShift:-1.
t := self class mulDigits:t and:t.
].
e := e - 1.
rslt := self class mulDigits:rslt and:t.
].
rslt.
].
^ self class newWithDigits:(raise value:(self digits) value:exp)
!
sumFromRosettaInteger:anRInt
^ self class
newWithDigits:(self class addDigits:(anRInt digits) and:(self digits))
!
productFromRosettaInteger:anRInt
^ self class newWithDigits:(self class mulDigits:(anRInt digits) and:(self digits))
! !
!RosettaInteger methodsFor:'printing'!
printOn:aStream
|print|
print :=
[:n :stream |
|first|
first := true.
n reverseDo:[:_2Digits |
first
ifTrue:[ stream nextPutAll:(_2Digits printString)]
ifFalse:[ stream nextPutAll:(_2Digits printString leftPaddedTo:2 with:$0)].
first := false.
].
].
print value:(self digits) value:aStream
! !
!Integer methodsFor:'converting'!
asRosettaInteger
^ RosettaInteger fromInteger:self
! !
!Integer methodsFor:'double dispatching'!
sumFromRosettaInteger:anRInt
^ anRInt + (RosettaInteger fromInteger:self)
!
productFromRosettaInteger:anRInt
^ anRInt * (RosettaInteger fromInteger:self)
! !
RosettaInteger initialize !
a := 124 asRosettaInteger. e'a is: {a} ({a class})' printCR. b := 333 asRosettaInteger. e'b is: {b} ({b class})'printCR. a_plus_b := a+b. e'(a+b) is: {a_plus_b} ({(a_plus_b) class})' printCR.
c := 999 asRosettaInteger. e'c is: {c} ({c class})' printCR. c_plus_1 := c+1. e'c+1 is: {c_plus_1} ({(c_plus_1) class})' printCR.
d := 100 asRosettaInteger. e'd is: {d} ({d class})' printCR. d_squared := d squared. e'd squared is: {d_squared} ({d_squared class})' printCR.
e := 2 asRosettaInteger. e_raisedTo_64 := e raisedTo:64. e'2 raisedTo:64 is: {e_raisedTo_64} ({e_raisedTo_64 class})' printCR.
e_raisedTo_64_squared := (e raisedTo:64) squared. e'result is: {e_raisedTo_64_squared} ({e_raisedTo_64_squared class})' printCR.
Transcript show:'once again: '. result := (2 asRosettaInteger raisedTo:64) squared. Transcript showCR:result.</lang>
- Output:
a is: 124 (RosettaInteger) b is: 333 (RosettaInteger) (a+b) is: 457 (RosettaInteger) c is: 999 (RosettaInteger) c+1 is: 1000 (RosettaInteger) d is: 100 (RosettaInteger) d squared is: 10000 (RosettaInteger) 2 raisedTo:64 is: 18446744073709551616 (RosettaInteger) result is: 340282366920938463463374607431768211456 (RosettaInteger) once again: 340282366920938463463374607431768211456
Tcl
Tcl 8.5 supports arbitrary-precision integers, which improves math operations on large integers. It is easy to define our own by following rules for long multiplication; we can then check this against the built-in's result: <lang tcl>package require Tcl 8.5
proc longmult {x y} {
set digits [lreverse [split $x ""]]
set result {0}
set j -2
foreach m [lreverse [split $y ""]] {
set c 0 set i [incr j] foreach d $digits { set v [lindex $result [incr i]] if {$v eq ""} { lappend result 0 set v 0 } regexp (.)(.)$ 0[expr {$v + $c + $d*$m}] -> c v lset result $i $v } lappend result $c
}
# Reconvert digit list into a decimal number
set result [string trimleft [join [lreverse $result] ""] 0]
if {$result == ""} then {return 0} else {return $result}
}
puts [set n [expr {2**64}]] puts [longmult $n $n] puts [expr {$n * $n}]</lang> outputs
18446744073709551616 340282366920938463463374607431768211456 340282366920938463463374607431768211456
UNIX Shell
In real shell scripts, I would use either `bc` or `dc` for this:
<lang sh>multiply() { echo "$1 $2 * p" | dc; }</lang>
But you can also do it with bash's built-in arithmetic:
<lang bash>add() { # arbitrary-precision addition
local a="$1" b="$2" sum= carry=0
if (( ${#a} < ${#b} )); then
local t="$a"
a="$b" b="$t"
fi
while (( ${#a} )); do
local -i d1="${a##${a%?}}" d2="10#0${b##${b%?}}" s=carry+d1+d2
sum="${s##${s%?}}$sum"
carry="10#0${s%?}"
a="${a%?}" b="${b%?}"
done
echo "$sum"
}
multiply() { # arbitrary-precision multiplication
local a="$1" b="$2" product=0
if (( ${#a} < ${#b} )); then
local t="$a"
a="$b" b="$t"
fi
local zeroes=
while (( ${#b} )); do
local m1="$a"
local m2="${b##${b%?}}"
local partial=$zeroes
local -i carry=0
while (( ${#m1} )); do
local -i d="${m1##${m1%?}}"
m1="${m1%?}"
local -i p=d*m2+carry
partial="${p##${p%?}}$partial"
carry="10#0${p%?}"
done
partial="${carry#0}$partial"
product="$(add "$product" "$partial")"
zeroes=0$zeroes
b="${b%?}"
done
echo "$product"
}</lang>
Output is the same either way:
$ multiply 18446744073709551616 18446744073709551616 340282366920928463463374607431768211456
Ursala
Natural numbers of unlimited size are a built in type, and arithmetic operations on them are available as library functions. However, since the task calls for explicitly implementing long multiplication, here is an implementation using nothing but language primitives. The numbers are represented as lists of booleans, LSB first. The compiler already knows how to parse and display them in decimal.
<lang Ursala>successor = ~&a^?\1! ~&ah?/~&NfatPRC ~&NNXatPC
sum = ~&B^?a\~&Y@a ~&B?abh/successor@alh2fabt2RC ~&Yabh2Ofabt2RC
product = ~&alrB^& sum@NfalrtPXPRCarh2alPNQX
x = 18446744073709551616
- show+
y = %nP product@iiX x</lang> output:
340282366920938463463374607431768211456
Vedit macro language
This example multiplies the value on current line with the value on next line and stores result on the 3rd line. <lang vedit>BOL
- 11 = EOL_Pos-Cur_Pos
- 12 = EOL_Pos-1
Line(1)
- 21 = EOL_Pos-Cur_Pos
- 22 = EOL_Pos-1
EOL Ins_Newline Ins_Char('0', COUNT, #11+#21)
- 32 = Cur_Pos-1
for (#2 = 0; #2 < #21; #2++) {
Goto_Pos(#22-#2) #5 = Cur_Char - '0'
for (#1 = 0; #1 < #11; #1++) {
Goto_Pos(#12-#1) #6 = Cur_Char - '0'
#7 = #5 * #6 #3 = #1 + #2 while (#7 > 0) { Goto_Pos(#32-#3) #7 += Cur_Char - '0' Ins_Char(#7%10 + '0', OVERWRITE) #3++ #7 = #7/10
} }
} </lang> Sample input and output:
18446744073709551616 18446744073709551616 0340282366920938463463374607431768211456
Visual Basic .NET
This uses the decimal type, (which has a MaxValue of 79,228,162,514,264,337,593,543,950,335). By limiting it to 10^28, it allows 28 decimal digits for the hi part, and 28 decimal digits for the lo part, 56 decimal digits total. A side computation of BigInteger assures that the results are accurate. <lang vbnet>Imports System Imports System.Console Imports BI = System.Numerics.BigInteger
Module Module1
Dim a As Decimal, mx As Decimal = 1E28D, hm As Decimal = 1E14D
' allows for 56 digit representation, using 28 decimal digits from each decimal
Structure bd
Public hi, lo As Decimal
End Structure
' outputs bd structure as string, optionally inserting commas
Function toStr(ByVal a As bd, ByVal Optional comma As Boolean = False) As String
Dim r As String = If(a.hi = 0, String.Format("{0:0}", a.lo),
String.Format("{0:0}{1:" & New String("0"c, 28) & "}", a.hi, a.lo))
If Not comma Then Return r
Dim rc As String = ""
For i As Integer = r.Length - 3 To 0 Step -3
rc = "," & r.Substring(i, 3) & rc : Next
toStr = r.Substring(0, r.Length Mod 3) & rc
toStr = toStr.Substring(If(toStr.Chars(0) = "," , 1, 0))
End Function
' needed because Math.Pow() returns a double
Function Pow_dec(ByVal bas As Decimal, ByVal exp As UInteger) As Decimal
If exp = 0 Then Pow_dec = 1D else Pow_dec = Pow_dec(bas, exp >> 1) : _
Pow_dec *= Pow_dec : If (exp And 1) <> 0 Then Pow_dec *= bas
End Function
Sub Main(ByVal args As String())
For p As UInteger = 64 To 95 - 1 Step 30 ' show prescribed output and maximum power of 2 output
Dim y As bd, x As bd : a = Pow_dec(2D, p) ' init the bd variables, a = decimal value to be squared
WriteLine("The square of (2^{0}): {1,38:n0}", p, a)
x.hi = Math.Floor(a / hm) : x.lo = a Mod hm ' setup for the squaring process
Dim BS As BI = BI.Pow(CType(a, BI), 2) ' for the BigInteger checking of result
y.lo = x.lo * x.lo : y.hi = x.hi * x.hi ' square the lo and the hi parts
a = x.hi * x.lo * 2D ' calculate the middle term (mid-term)
y.hi += Math.Floor(a / hm) : y.lo += (a Mod hm) * hm ' increment hi and lo parts with high and low parts of the mid-term
While y.lo > mx : y.lo -= mx : y.hi += 1 : End While ' check for overflow, adjust both parts as needed
WriteLine(" is {0,75} (which {1} match the BigInteger computation)" & vbLf,
toStr(y, True), If(BS.ToString() = toStr(y), "does", "fails to"))
Next
End Sub
End Module</lang>
- Output:
Shown are the prescribed output and the maximum power of two that can be squared by this bd structure without overflowing.
The square of (2^64): 18,446,744,073,709,551,616 is 340,282,366,920,938,463,463,374,607,431,768,211,456 (which does match the BigInteger computation) The square of (2^94): 19,807,040,628,566,084,398,385,987,584 is 392,318,858,461,667,547,739,736,838,950,479,151,006,397,215,279,002,157,056 (which does match the BigInteger computation)
Wren
<lang ecmascript>import "/fmt" for Fmt
// argument validation var d = Fn.new { |b|
if (b < 48 || b > 57) Fiber.abort("digit 0-9 expected")
return b - 48
}
// converts a list of bytes to a string var b2s = Fn.new { |b| b.map { |e| String.fromByte(e) }.join() }
// add two numbers as strings var add = Fn.new { |x, y|
if (y.count > x.count) {
var t = x
x = y
y = t
}
var b = List.filled(x.count+1, 0)
var c = 0
for (i in 1..x.count) {
if (i <= y.count) c = c + d.call(y[y.count-i].bytes[0])
var s = d.call(x[x.count-i].bytes[0]) + c
c = (s/10).floor
b[b.count-i] = (s%10) + 48
}
if (c == 0) return b2s.call(b[1..-1])
b[0] = c + 48
return b2s.call(b)
}
// multiply a number by a single digit var mulDigit = Fn.new { |x, y|
if (y == 48) return "0"
y = d.call(y)
var b = List.filled(x.count+1, 0)
var c = 0
for (i in 1..x.count) {
var s = d.call(x[x.count-i].bytes[0]) * y + c
c = (s/10).floor
b[b.count-i] = (s%10) + 48
}
if (c == 0) return b2s.call(b[1..-1])
b[0] = c + 48
return b2s.call(b)
}
// multiply two numbers as strings var mul = Fn.new { |x, y|
var result = mulDigit.call(x, y[y.count-1].bytes[0])
var zeros = ""
var i = 2
while (i <= y.count) {
zeros = zeros + "0"
result = add.call(result, mulDigit.call(x, y[y.count-i].bytes[0]) + zeros)
i = i + 1
}
result = result.trimStart("0")
if (result == "") result = "0"
return result
}
var n = "18446744073709551616" Fmt.print("$,s", mul.call(n, n))</lang>
- Output:
340,282,366,920,938,463,463,374,607,431,768,211,456
XPL0
<lang XPL0>include c:\cxpl\stdlib; char Two64, Product(40); [Two64:= "18446744073709551616"; StrNMul(Two64, Two64, Product, 20); Product(39):= Product(39)!$80; \terminate string Text(0, Product+1); \skip leading zero ]</lang>
Output:
340282366920938463463374607431768211456
zkl
[gnu] BigNums are supported via an extension library <lang zkl>var BN=Import("zklBigNum"); BN(2).pow(64) * BN(2).pow(64) 340282366920938463463374607431768211456
BN(2).pow(128) : "%,d".fmt(_) 340,282,366,920,938,463,463,374,607,431,768,211,456
//42!, also BN(42).factorial()
[2..42].reduce(fcn(p,n){p*n},BN(1)) : "%,d".fmt(_) 1,405,006,117,752,879,898,543,142,606,244,511,569,936,384,000,000,000</lang>