Long multiplication
You are encouraged to solve this task according to the task description, using any language you may know.
In this task, explicitly implement long multiplication. This is one possible approach to arbitrary-precision integer algebra.
For output, display the result of 2^64 * 2^64. The decimal representation of 2^64 is:
18446744073709551616
The output of 2^64 * 2^64 is 2^128, and that is:
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
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.
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$:E = E - 1: GOTO 700 720 RETURN</lang>
BBC BASIC
Library method: <lang bbcbasic> INSTALL @lib$+"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++
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
C#
System.Numerics.BigInteger was added with C# 4.
<lang csharp>using System; using System.Numerics;
class Program {
static void Main() {
BigInteger pow2_64 = BigInteger.Pow(2, 64);
BigInteger result = BigInteger.Multiply(pow2_64, pow2_64);
Console.WriteLine(result);
}
}</lang>
Output:
340282366920938463463374607431768211456
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 (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 (digits a)))
(b (nreverse (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
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>
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>
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) import Data.Char (digitToInt) import Data.List (inits)
digits :: Integer -> [Integer] digits = map (fromIntegral . digitToInt) . show
lZZ = inits $ repeat 0
table f = map . flip (map . f)
polymul = ((map sum . transpose . zipWith (++) lZZ) .) . table (*)
longmult = (foldl1 ((+) . (10 *)) .) . (. digits) . polymul . digits
main = print $ (2 ^ 64) `longmult` (2 ^ 64)</lang>
- Output:
340282366920938463463374607431768211456
Icon and Unicon
Icon and Unicon support large integers. <lang Icon>procedure main() write(2^64*2^64) end</lang>
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 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
<lang javascript>function mult(num1,num2){ var a1 = num1.split("").reverse(); var a2 = num2.split("").reverse(); var aResult = new Array;
for ( iterNum1 = 0; iterNum1 < a1.length; iterNum1++ ) { for ( iterNum2 = 0; iterNum2 < a2.length; iterNum2++ ) { 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] -= Math.floor( aResult[idxIter] / 10 ) * 10; } } } return aResult.reverse().join(""); }</lang>
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"
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>
Maple
<lang Maple> > longmult := proc( a, b )
local la, lb, digit; la := length(a); lb := length(b); digit := (n,i)->iquo(n,10^(i-1)) mod 10; add( add( digit(a,la-i+1) * digit(b,lb-j+1) *10^(la-i+lb-j), i=1..la), j=1..lb );
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 &= ifTrue: [ continue ]
0 @v size loop: j [
i j + 1 - ->k
j @v at x * + l at(k) + 1000000000 /mod swap l put(k)
]
l put(k 1 +)
]
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, Stream.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
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>
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 ) { [~] @g.pop, @g.reverse».fmt('%04d') };
sub long_multiply ( Str $x, Str $y ) {
my @group_sums;
for num_to_groups($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[$k] %= 10000;
}
return groups_to_num @group_sums;
}
long_multiply( '18446744073709551616', '18446744073709551616' ).say;</lang>
In any case, integers are specced to be arbitrarily large, and most implementations support this already:
niecza> 18446744073709551616 * 18446744073709551616 340282366920938463463374607431768211456
PHP
<lang PHP><?php function longMult($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
PowerShell
<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>
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>#!/usr/bin/env python
def digits(x):
return [int(c) for c in str(x)]
def mult_table(xs, ys):
return [[x * y for x in xs] for y in ys]
def polymul(xs, ys):
return map(lambda *vs: sum(filter(None, vs)),
*[[0] * i + zs for i, zs in enumerate(mult_table(xs, ys))])
def longmult(x, y):
result = 0
for v in polymul(digits(x), digits(y)):
result = result * 10 + v
return result
if __name__ == "__main__":
print longmult(2**64, 2**64)</lang>
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>
REXX
version 1
This REXX version supports:
- leading signs
- decimal points
<lang rexx>/*REXX program performs long multiplication on two numbers (without 'E')*/ numeric digits 100; d='.' /*be able to handle input numbers*/ parse arg x y . /*accept the (possible) two nums.*/ if x== then x=2**64 /*Not specified? Use the default*/ if y== then y=x /* " " " " " */ if x<0 && y<0 then sgn='-' /*only one argument is negative? */
else sgn= /*no, then the sign is positive. */
xx=x; x=strip(x,'T',d) /*remove any trailing decimal pt.*/ yy=y; y=strip(y,'T',d) /* " " " " ".*/ _=left(x,1); if _=='-' | _=='+' then x=substr(x,2) /*remove leading ±*/ _=left(y,1); if _=='-' | _=='+' then y=substr(y,2) /* " " "*/
/*[↑] above code for a Regina bug*/
/*otherwise: x=abs(x) will do it*/
dp=0; Lx=length(x); Ly=length(y) /*get the lengths of new X and Y.*/ f=pos(d,x); if f\==0 then dp= Lx-f /*calculate size of dec fraction.*/ f=pos(d,y); if f\==0 then dp=dp+Ly-f /* " " " " " */ x=space(translate(x,,d),0) /*remove decimal point, if any. */ y=space(translate(y,,d),0) /* " " " " " */ Lx=length(x); Ly=length(y) /*get the lengths of new X and Y.*/ numeric digits max(digits(),Lx+Ly) p=0 /*the product so far. */
do j=Ly by -1 for Ly /*almost like REXX does it,but no*/
p=p+((x*substr(y,j,1))copies(0,Ly-j))
end /*j*/
say f=length(p)-dp /*does product has enough digits?*/ if f<0 then p=copies(0,abs(f)+1)p /*Neg? Add leading 0s for INSERT*/ say ' built-in:' xx '*' yy '=' xx*yy say 'long mult:' xx '*' yy '=' sgn ||strip(insert(d,p,length(p)-dp),'T',d)
/*stick a fork in it, we're done.*/</lang>
output when using the default inputs of: 2^64 2^64
built-in: 18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456 long mult: 18446744073709551616 * 18446744073709551616 = 340282366920938463463374607431768211456
output when using the inputs of: 123 -456789000
built-in: 123 * -456789000 = -56185047000 long mult: 123 * -456789000 = -56185047000
output when using the inputs of: -123.678 +456789000
built-in: -123.678 * +456789000 = -56494749942.000 long mult: -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
Ruby
<lang ruby>def longmult(x,y)
digits = reverse_split_number(x)
result = [0]
j = 0
reverse_split_number(y).each do |m|
c = 0
i = j
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
def reverse_split_number(m)
digits = [] while m > 0 m, v = m.divmod 10 digits << v end digits
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>
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) {
while (true) {
while (result.len < addendpos+1) {
result.append(0);
};
var addend_digits = (addend.to_i + result[addendpos].to_i -> to_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
<lang smalltalk>(2 raisedTo: 64) * (2 raisedTo: 64).</lang>
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
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>
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