Egyptian division: Difference between revisions
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Quotient = 17 Remainder = 2 |
Quotient = 17 Remainder = 2 |
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=={{header|Swift}}== |
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<lang swift>extension BinaryInteger { |
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@inlinable |
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public func egyptianDivide(by divisor: Self) -> (quo: Self, rem: Self) { |
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let table = |
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(0...).lazy |
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.map({i -> (Self, Self) in |
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let power = Self(2).power(Self(i)) |
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return (power, power * divisor) |
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}) |
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.prefix(while: { $0.1 <= self }) |
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.reversed() |
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let (answer, acc) = table.reduce((Self(0), Self(0)), {cur, row in |
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let ((ans, acc), (power, doubling)) = (cur, row) |
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return acc + doubling <= self ? (ans + power, doubling + acc) : cur |
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}) |
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return (answer, Self((acc - self).magnitude)) |
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} |
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@inlinable |
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public func power(_ n: Self) -> Self { |
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return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *) |
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} |
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} |
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let dividend = 580 |
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let divisor = 34 |
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let (quo, rem) = dividend.egyptianDivide(by: divisor) |
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print("\(dividend) divided by \(divisor) = \(quo) rem \(rem)") |
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</lang> |
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{{out}} |
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<pre>580 divided by 34 = 17 rem 2</pre> |
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=={{header|VBA}}== |
=={{header|VBA}}== |
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Revision as of 11:00, 13 November 2019
You are encouraged to solve this task according to the task description, using any language you may know.
Egyptian division is a method of dividing integers using addition and doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
- Start the construction of a table of two columns:
powers_of_2, anddoublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column. - Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
- Continue with successive i’th rows of 2^i and 2^i * divisor.
- Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
- We now assemble two separate sums that both start as zero, called here answer and accumulator
- Consider each row of the table, in the reverse order of its construction.
- If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
- When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2 doublings 1 34 2 68 4 136 8 272 16 544
Initialization of sums:
powers_of_2 doublings answer accumulator 1 34 2 68 4 136 8 272 16 544 0 0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2 doublings answer accumulator 1 34 2 68 4 136 8 272 16 544 16 544
powers_of_2 doublings answer accumulator 1 34 2 68 4 136 827216 544 16 544
powers_of_2 doublings answer accumulator 1 34 2 68 413616 544 827216 544
powers_of_2 doublings answer accumulator 1 34 26816 544 4136827216 544
powers_of_2 doublings answer accumulator 1 34 17 578 2684136827216 544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
- Functions should be clear interpretations of the algorithm.
- Use the function to divide 580 by 34 and show the answer here, on this page.
- References
- Related tasks
Ada
<lang Ada> with Ada.Text_IO;
procedure Egyptian_Division is
procedure Divide (a : Natural; b : Positive; q, r : out Natural) is
doublings : array (0..31) of Natural; -- The natural type holds values < 2^32 so no need going beyond
m, sum, last_index_touched : Natural := 0;
begin
for i in doublings'Range loop
m := b * 2**i;
exit when m > a ;
doublings (i) := m;
last_index_touched := i;
end loop;
q := 0;
for i in reverse doublings'First .. last_index_touched loop
m := sum + doublings (i);
if m <= a then
sum := m;
q := q + 2**i;
end if;
end loop;
r := a -sum;
end Divide;
q, r : Natural;
begin
Divide (580,34, q, r);
Ada.Text_IO.put_line ("Quotient="&q'Img & " Remainder="&r'img);
end Egyptian_Division; </lang>
- Output:
Quotient= 17 Remainder= 2
ALGOL 68
<lang algol68>BEGIN
# performs Egyptian division of dividend by divisor, setting quotient and remainder #
# this uses 32 bit numbers, so a table of 32 powers of 2 should be sufficient #
# ( divisors > 2^30 will probably overflow - this is not checked here ) #
PROC egyptian division = ( INT dividend, divisor, REF INT quotient, remainder )VOID:
BEGIN
[ 1 : 32 ]INT powers of 2, doublings;
# initialise the powers of 2 and doublings tables #
powers of 2[ 1 ] := 1;
doublings [ 1 ] := divisor;
INT table pos := 1;
WHILE table pos +:= 1;
powers of 2[ table pos ] := powers of 2[ table pos - 1 ] * 2;
doublings [ table pos ] := doublings [ table pos - 1 ] * 2;
doublings[ table pos ] <= dividend
DO
SKIP
OD;
# construct the accumulator and answer #
INT accumulator := 0, answer := 0;
WHILE table pos >=1
DO
IF ( accumulator + doublings[ table pos ] ) <= dividend
THEN
accumulator +:= doublings [ table pos ];
answer +:= powers of 2[ table pos ]
FI;
table pos -:= 1
OD;
quotient := answer;
remainder := ABS ( accumulator - dividend )
END # egyptian division # ;
# task test case # INT quotient, remainder; egyptian division( 580, 34, quotient, remainder ); print( ( "580 divided by 34 is: ", whole( quotient, 0 ), " remainder: ", whole( remainder, 0 ), newline ) )
END</lang>
- Output:
580 divided by 34 is: 17 remainder: 2
AppleScript
Unfold to derive successively doubled rows, fold to sum quotient and derive remainder <lang AppleScript>-- EGYPTIAN DIVISION ------------------------------------
-- eqyptianQuotRem :: Int -> Int -> (Int, Int) on egyptianQuotRem(m, n)
script expansion
on |λ|(ix)
set {i, x} to ix
if x > m then
Nothing()
else
Just({ix, {i + i, x + x}})
end if
end |λ|
end script
script collapse
on |λ|(ix, qr)
set {i, x} to ix
set {q, r} to qr
if x < r then
{q + i, r - x}
else
qr
end if
end |λ|
end script
return foldr(collapse, {0, m}, ¬
unfoldr(expansion, {1, n}))
end egyptianQuotRem
-- TEST -------------------------------------------------
on run
egyptianQuotRem(580, 34)
end run
-- GENERIC FUNCTIONS ------------------------------------
-- Just :: a -> Maybe a on Just(x)
{type:"Maybe", Nothing:false, Just:x}
end Just
-- Nothing :: Maybe a on Nothing()
{type:"Maybe", Nothing:true}
end Nothing
-- foldr :: (a -> b -> b) -> b -> [a] -> b on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
end foldr
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- > unfoldr (\b -> if b == 0 then Nothing else Just (b, b-1)) 10 -- > [10,9,8,7,6,5,4,3,2,1] -- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v)
set xr to {v, v} -- (value, remainder)
set xs to {}
tell mReturn(f)
repeat -- Function applied to remainder.
set mb to |λ|(item 2 of xr)
if Nothing of mb then
exit repeat
else -- New (value, remainder) tuple,
set xr to Just of mb
-- and value appended to output list.
set end of xs to item 1 of xr
end if
end repeat
end tell
return xs
end unfoldr</lang>
- Output:
{17, 2}
AutoHotkey
<lang AutoHotkey>divident := 580 divisor := 34
answer := accumulator := 0 obj := [] , div := divisor
while (div < divident) { obj[2**(A_Index-1)] := div ; obj[powers_of_2] := doublings div *= 2 ; double up }
while obj.MaxIndex() ; iterate rows "in the reverse order" { if (accumulator + obj[obj.MaxIndex()] <= divident) ; If (accumulator + current doubling) <= dividend { accumulator += obj[obj.MaxIndex()] ; add current doubling to the accumulator answer += obj.MaxIndex() ; add the powers_of_2 value to the answer. } obj.pop() ; remove current row } MsgBox % divident "/" divisor " = " answer ( divident-accumulator > 0 ? " r" divident-accumulator : "")</lang>
Outputs:
580/34 = 17 r2
BaCon
<lang c>
'---Ported from the c code example to BaCon by bigbass
'================================================================================== FUNCTION EGYPTIAN_DIVISION(long dividend, long divisor, long remainder) TYPE long '================================================================================== '--- remainder is the third parameter, pass 0 if you do not need the remainder
DECLARE powers[64] TYPE long DECLARE doublings[64] TYPE long
LOCAL i TYPE long
FOR i = 0 TO 63 STEP 1 powers[i] = 1 << i doublings[i] = divisor << i IF (doublings[i] > dividend) THEN BREAK ENDIF NEXT
LOCAL answer TYPE long LOCAL accumulator TYPE long answer = 0 accumulator = 0
WHILE i >= 0 '--- If the current value of the accumulator added to the '--- doublings cell would be less than or equal to the '--- dividend then add it to the accumulator IF (accumulator + doublings[i] <= dividend) THEN accumulator = accumulator + doublings[i] answer = answer + powers[i] ENDIF DECR i WEND
IF remainder THEN remainder = dividend - accumulator PRINT dividend ," / ", divisor, " = " , answer ," remainder " , remainder
PRINT "Decoded the answer to a standard fraction"
PRINT (remainder + 0.0 )/ (divisor + 0.0) + answer
PRINT
ELSE PRINT dividend ," / ", divisor , " = " , answer ENDIF
RETURN answer
ENDFUNCTION
'--- the large number divided by the smaller number
'--- the third argument is 1 if you want to have a remainder
'--- and 0 if you dont want to have a remainder
EGYPTIAN_DIVISION(580,34,1) EGYPTIAN_DIVISION(580,34,0)
EGYPTIAN_DIVISION(580,34,1)
</lang>
C
<lang c>
- include <stdio.h>
- include <stdlib.h>
- include <stdint.h>
- include <assert.h>
uint64_t egyptian_division(uint64_t dividend, uint64_t divisor, uint64_t *remainder) { // remainder is an out parameter, pass NULL if you do not need the remainder
static uint64_t powers[64]; static uint64_t doublings[64];
int i;
for(i = 0; i < 64; i++) { powers[i] = 1 << i; doublings[i] = divisor << i; if(doublings[i] > dividend) break; }
uint64_t answer = 0; uint64_t accumulator = 0;
for(i = i - 1; i >= 0; i--) { // If the current value of the accumulator added to the // doublings cell would be less than or equal to the // dividend then add it to the accumulator if(accumulator + doublings[i] <= dividend) { accumulator += doublings[i]; answer += powers[i]; } }
if(remainder) *remainder = dividend - accumulator; return answer; }
void go(uint64_t a, uint64_t b) { uint64_t x, y; x = egyptian_division(a, b, &y); printf("%llu / %llu = %llu remainder %llu\n", a, b, x, y); assert(a == b * x + y); }
int main(void) { go(580, 32); } </lang>
C++
<lang cpp>#include <cassert>
- include <iostream>
typedef unsigned long ulong;
/*
* Remainder is an out paramerter. Use nullptr if the remainder is not needed. */
ulong egyptian_division(ulong dividend, ulong divisor, ulong* remainder) {
constexpr int SIZE = 64; ulong powers[SIZE]; ulong doublings[SIZE]; int i = 0;
for (; i < SIZE; ++i) {
powers[i] = 1 << i;
doublings[i] = divisor << i;
if (doublings[i] > dividend) {
break;
}
}
ulong answer = 0; ulong accumulator = 0;
for (i = i - 1; i >= 0; --i) {
/*
* If the current value of the accumulator added to the
* doublings cell would be less than or equal to the
* dividend then add it to the accumulator
*/
if (accumulator + doublings[i] <= dividend) {
accumulator += doublings[i];
answer += powers[i];
}
}
if (remainder) {
*remainder = dividend - accumulator;
}
return answer;
}
void print(ulong a, ulong b) {
using namespace std;
ulong x, y; x = egyptian_division(a, b, &y);
cout << a << " / " << b << " = " << x << " remainder " << y << endl; assert(a == b * x + y);
}
int main() {
print(580, 34);
return 0;
}</lang>
- Output:
580 / 34 = 17 remainder 2
C#
<lang csharp> using System; using System.Collections;
namespace Egyptian_division { class Program { public static void Main(string[] args) { Console.Clear(); Console.WriteLine(); Console.WriteLine(" Egyptian division "); Console.WriteLine(); Console.Write(" Enter value of dividend : "); int dividend = int.Parse(Console.ReadLine());
Console.Write(" Enter value of divisor : "); int divisor = int.Parse(Console.ReadLine());
Divide(dividend, divisor);
Console.WriteLine(); Console.Write("Press any key to continue . . . "); Console.ReadKey(true);
}
static void Divide(int dividend, int divisor) { // // Local variable declaration and initialization // int result = 0; int reminder = 0;
int powers_of_two = 0; int doublings = 0;
int answer = 0; int accumulator = 0;
int two = 2; int pow = 0; int row = 0;
// // Tables declaration // ArrayList table_powers_of_two = new ArrayList(); ArrayList table_doublings = new ArrayList();
// // Fill and Show table values // Console.WriteLine(" "); Console.WriteLine(" powers_of_2 doublings "); Console.WriteLine(" ");
// Set initial values powers_of_two = 1; doublings = divisor; while( doublings <= dividend ) { // Set table value table_powers_of_two.Add( powers_of_two ); table_doublings.Add( doublings );
// Show new table row Console.WriteLine("{0,8}{1,16}",powers_of_two, doublings);
pow++;
powers_of_two = (int)Math.Pow( two, pow ); doublings = powers_of_two * divisor; } Console.WriteLine(" ");
// // Calculate division and Show table values // row = pow - 1; Console.WriteLine(" "); Console.WriteLine(" powers_of_2 doublings answer accumulator"); Console.WriteLine(" "); Console.SetCursorPosition(Console.CursorLeft, Console.CursorTop + row);
pow--; while( pow >= 0 && accumulator < dividend ) { // Get values from tables doublings = int.Parse(table_doublings[pow].ToString()); powers_of_two = int.Parse(table_powers_of_two[pow].ToString());
if(accumulator + int.Parse(table_doublings[pow].ToString()) <= dividend ) { // Set new values accumulator += doublings; answer += powers_of_two;
// Show accumulated row values in different collor Console.ForegroundColor = ConsoleColor.Green; Console.Write("{0,8}{1,16}",powers_of_two, doublings); Console.ForegroundColor = ConsoleColor.Green; Console.WriteLine("{0,10}{1,12}", answer, accumulator); Console.SetCursorPosition(Console.CursorLeft, Console.CursorTop - 2); } else { // Show not accumulated row walues Console.ForegroundColor = ConsoleColor.DarkGray; Console.Write("{0,8}{1,16}",powers_of_two, doublings); Console.ForegroundColor = ConsoleColor.Gray; Console.WriteLine("{0,10}{1,12}", answer, accumulator); Console.SetCursorPosition(Console.CursorLeft, Console.CursorTop - 2); }
pow--;
}
Console.WriteLine(); Console.SetCursorPosition(Console.CursorLeft, Console.CursorTop + row + 2); Console.ResetColor();
// Set result and reminder result = answer; if( accumulator < dividend ) { reminder = dividend - accumulator;
Console.WriteLine(" So " + dividend + " divided by " + divisor + " using the Egyptian method is \n " + result + " remainder (" + dividend + " - " + accumulator + ") or " + reminder); Console.WriteLine(); } else { reminder = 0;
Console.WriteLine(" So " + dividend + " divided by " + divisor + " using the Egyptian method is \n " + result + " remainder " + reminder); Console.WriteLine(); } } } } </lang>
- Program Input and Output
- Instead of bold and strikeout text format, numbers are represented in different color:
Egyptian division
Enter value of dividend : 580
Enter value of divisor : 34
powers_of_2 doublings
1 34
2 68
4 136
8 272
16 544
powers_of_2 doublings answer accumulator
1 34 17 578
2 68 16 544
4 136 16 544
8 272 16 544
16 544 16 544
So 580 divided by 34 using the Egyptian method is
17 remainder (580 - 578) or 2
Press any key to continue . . .
D
<lang D> import std.stdio;
version(unittest) {
// empty
} else {
int main(string[] args) {
import std.conv;
if (args.length < 3) {
stderr.writeln("Usage: ", args[0], " dividend divisor");
return 1;
}
ulong dividend = to!ulong(args[1]);
ulong divisor = to!ulong(args[2]);
ulong remainder;
auto ans = egyptian_division(dividend, divisor, remainder);
writeln(dividend, " / ", divisor, " = ", ans, " rem ", remainder);
return 0; }
}
ulong egyptian_division(ulong dividend, ulong divisor, out ulong remainder) {
enum SIZE = 64; ulong[SIZE] powers; ulong[SIZE] doublings; int i;
for (; i<SIZE; ++i) {
powers[i] = 1 << i;
doublings[i] = divisor << i;
if (doublings[i] > dividend) {
break;
}
}
ulong answer; ulong accumulator;
for (i=i-1; i>=0; --i) {
if (accumulator + doublings[i] <= dividend) {
accumulator += doublings[i];
answer += powers[i];
}
}
remainder = dividend - accumulator; return answer;
}
unittest {
ulong remainder;
assert(egyptian_division(580UL, 34UL, remainder) == 17UL); assert(remainder == 2);
} </lang>
Erlang
<lang erlang>-module(egypt).
-export([ediv/2]).
ediv(A, B) ->
{Twos, Ds} = genpowers(A, [1], [B]),
{Quot, C} = accumulate(A, Twos, Ds),
{Quot, abs(C - A)}.
genpowers(A, [_|Ts], [D|Ds]) when D > A -> {Ts, Ds}; genpowers(A, [T|_] = Twos, [D|_] = Ds) -> genpowers(A, [2*T|Twos], [D*2|Ds]).
accumulate(N, Twos, Ds) -> accumulate(N, Twos, Ds, 0, 0). accumulate(_, [], [], Q, C) -> {Q, C}; accumulate(N, [T|Ts], [D|Ds], Q, C) when (C + D) =< N -> accumulate(N, Ts, Ds, Q+T, C+D); accumulate(N, [_|Ts], [_|Ds], Q, C) -> accumulate(N, Ts, Ds, Q, C). </lang>
- Output:
1> egypt:ediv(580,34).
{17,2}
F#
<lang fsharp>// A function to perform Egyptian Division: Nigel Galloway August 11th., 2017 let egyptianDivision N G =
let rec fn n g = seq{yield (n,g); yield! fn (n+n) (g+g)}
Seq.foldBack (fun (n,i) (g,e)->if (i<=g) then ((g-i),(e+n)) else (g,e)) (fn 1 G |> Seq.takeWhile(fun (_,g)->g<=N)) (N,0)
</lang> Which may be used: <lang fsharp> let (n,g) = egyptianDivision 580 34 printfn "580 divided by 34 is %d remainder %d" g n </lang>
- Output:
580 divided by 34 is 17 remainder 2
Factor
<lang factor>USING: assocs combinators formatting kernel make math sequences ; IN: rosetta-code.egyptian-division
- table ( dividend divisor -- table )
[ [ 2dup >= ] [ dup , 2 * ] while ] { } make 2nip
dup length <iota> [ 2^ ] map zip <reversed> ;
- accum ( a b dividend -- c )
[ 2dup [ first ] bi@ + ] dip < [ [ + ] 2map ] [ drop ] if ;
- ediv ( dividend divisor -- quotient remainder )
{
[ table ]
[ 2drop { 0 0 } ]
[ drop [ accum ] curry reduce first2 swap ]
[ drop - abs ]
} 2cleave ;
580 34 ediv "580 divided by 34 is %d remainder %d\n" printf</lang>
- Output:
580 divided by 34 is 17 remainder 2
FreeBASIC
<lang freebasic>' version 09-08-2017 ' compile with: fbc -s console
Data 580, 34
Dim As UInteger dividend, divisor, answer, accumulator, i ReDim As UInteger table(1 To 32, 1 To 2)
Read dividend, divisor
i = 1 table(i, 1) = 1 : table(i, 2) = divisor
While table(i, 2) < dividend
i += 1 table(i, 1) = table(i -1, 1) * 2 table(i, 2) = table(i -1, 2) * 2
Wend
i -= 1 answer = table(i, 1) accumulator = table(i, 2)
While i > 1
i -= 1
If table(i,2)+ accumulator <= dividend Then
answer += table(i, 1)
accumulator += table(i, 2)
End If
Wend
Print Str(dividend); " divided by "; Str(divisor); " using Egytian division"; Print " returns "; Str(answer); " mod(ulus) "; Str(dividend-accumulator)
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
580 divided by 34 using Egytian division returns 17 mod(ulus) 2
Go
<lang go>package main
import "fmt"
func egyptianDivide(dividend, divisor int) (quotient, remainder int) {
if dividend < 0 || divisor <= 0 {
panic("Invalid argument(s)")
}
if dividend < divisor {
return 0, dividend
}
powersOfTwo := []int{1}
doublings := []int{divisor}
doubling := divisor
for {
doubling *= 2
if doubling > dividend {
break
}
l := len(powersOfTwo)
powersOfTwo = append(powersOfTwo, powersOfTwo[l-1]*2)
doublings = append(doublings, doubling)
}
answer := 0
accumulator := 0
for i := len(doublings) - 1; i >= 0; i-- {
if accumulator+doublings[i] <= dividend {
accumulator += doublings[i]
answer += powersOfTwo[i]
if accumulator == dividend {
break
}
}
}
return answer, dividend - accumulator
}
func main() {
dividend := 580 divisor := 34 quotient, remainder := egyptianDivide(dividend, divisor) fmt.Println(dividend, "divided by", divisor, "is", quotient, "with remainder", remainder)
}</lang>
- Output:
580 divided by 34 is 17 with remainder 2
Haskell
Deriving division from (+) and (-) by unfolding from a seed pair (1, divisor) up to a series of successively doubling pairs, and then refolding that series of 'two column rows' back down to a (quotient, remainder) pair, using (0, dividend) as the initial accumulator value. In other words, taking the divisor as a unit, and deriving the binary composition of the dividend in terms of that unit. <lang Haskell>import Data.List (unfoldr)
egyptianQuotRem :: Integer -> Integer -> (Integer, Integer) egyptianQuotRem m n =
let expansion (i, x)
| x > m = Nothing
| otherwise = Just ((i, x), (i + i, x + x))
collapse (i, x) (q, r)
| x < r = (q + i, r - x)
| otherwise = (q, r)
in foldr collapse (0, m) $ unfoldr expansion (1, n)
main :: IO () main = print $ egyptianQuotRem 580 34</lang>
- Output:
(17,2)
We can make the process of calculation more visible by adding a trace layer:
<lang Haskell>import Data.List (unfoldr) import Debug.Trace (trace)
egyptianQuotRem :: Int -> Int -> (Int, Int) egyptianQuotRem m n =
let rows =
unfoldr
(\(i, x) ->
if x > m
then Nothing
else Just ((i, x), (i + i, x + x)))
(1, n)
in trace
(unlines
[ "Number pair unfolded to series of doubling rows:"
, show rows
, "\nRows refolded down to (quot, rem):"
, show (0, m)
])
foldr
(\(i, x) (q, r) ->
if x < r
then trace
(concat
["(+", show i, ", -", show x, ") -> rem ", show (r - x)])
(q + i, r - x)
else (q, r))
(0, m)
rows
main :: IO () main = print $ egyptianQuotRem 580 34</lang>
- Output:
Number pair unfolded to series of doubling rows: [(1,34),(2,68),(4,136),(8,272),(16,544)] Rows refolded down to (quot, rem): (0,580) (+16, -544) -> rem 36 (+1, -34) -> rem 2 (17,2)
Another approach, using lazy lists and foldr:
<lang haskell>doublings = iterate (* 2)
powers = doublings 1
k n (u, v) (ans, acc) =
if v + ans <= n then (v + ans, u + acc) else (ans, acc)
egy n = snd . foldr (k n) (0, 0) . zip powers . takeWhile (<= n) . doublings
main :: IO () main = print $ egy 580 34</lang>
- Output:
17
J
Implementation:
<lang J>doublings=:_1 }. (+:@]^:(> {:)^:a: (,~ 1:)) ansacc=: 1 }. (] + [ * {.@[ >: {:@:+)/@([,.doublings) egydiv=: (0,[)+1 _1*ansacc</lang>
Task example:
<lang J> 580 doublings 34
1 34 2 68 4 136 8 272
16 544
580 ansacc 34
17 578
580 egydiv 34
17 2</lang>
Notes:
pre
When building the doublings table, we don't actually know we've exceeded our numerator until we are done. This would result in an excess row, so we have to explicitly not include that excess row in our doublings result.
Our "fold" is actually not directly on the result of doublings - for our fold, we add another column where every value is the numerator. This conveniently makes it available for comparison at every stage of the fold and seems a more concise approach than creating a closure. (We do not include this extra value in our ansacc result, of course.)
Java
<lang Java> import java.util.ArrayList; import java.util.List;
public class EgyptianDivision {
/**
* Runs the method and divides 580 by 34
*
* @param args not used
*/
public static void main(String[] args) {
divide(580, 34);
}
/**
* Divides dividend by divisor using the Egyptian Division-Algorithm and prints the
* result to the console
*
* @param dividend
* @param divisor
*/
public static void divide(int dividend, int divisor) {
List<Integer> powersOf2 = new ArrayList<>();
List<Integer> doublings = new ArrayList<>();
//populate the powersof2- and doublings-columns
int line = 0;
while ((Math.pow(2, line) * divisor) <= dividend) { //<- could also be done with a for-loop
int powerOf2 = (int) Math.pow(2, line);
powersOf2.add(powerOf2);
doublings.add(powerOf2 * divisor);
line++;
}
int answer = 0;
int accumulator = 0;
//Consider the rows in reverse order of their construction (from back to front of the List<>s)
for (int i = powersOf2.size() - 1; i >= 0; i--) {
if (accumulator + doublings.get(i) <= dividend) {
accumulator += doublings.get(i);
answer += powersOf2.get(i);
}
}
System.out.println(String.format("%d, remainder %d", answer, dividend - accumulator));
}
}
</lang>
- Output:
17, remainder 2
JavaScript
ES6
<lang JavaScript>(() => {
'use strict';
// EGYPTIAN DIVISION --------------------------------
// eqyptianQuotRem :: Int -> Int -> (Int, Int)
const eqyptianQuotRem = (m, n) => {
const expansion = ([i, x]) =>
x > m ? (
Nothing()
) : Just([
[i, x],
[i + i, x + x]
]);
const collapse = ([i, x], [q, r]) =>
x < r ? (
[q + i, r - x]
) : [q, r];
return foldr(
collapse,
[0, m],
unfoldr(expansion, [1, n])
);
};
// TEST ---------------------------------------------
// main :: IO ()
const main = () =>
showLog(
eqyptianQuotRem(580, 34)
);
// -> [17, 2]
// GENERIC FUNCTIONS --------------------------------
// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});
// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});
// flip :: (a -> b -> c) -> b -> a -> c
const flip = f =>
1 < f.length ? (
(a, b) => f(b, a)
) : (x => y => f(y)(x));
// foldr :: (a -> b -> b) -> b -> [a] -> b const foldr = (f, a, xs) => xs.reduceRight(flip(f), a);
// unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
const unfoldr = (f, v) => {
let
xr = [v, v],
xs = [];
while (true) {
const mb = f(xr[1]);
if (mb.Nothing) {
return xs
} else {
xr = mb.Just;
xs.push(xr[0])
}
}
};
// showLog :: a -> IO ()
const showLog = (...args) =>
console.log(
args
.map(JSON.stringify)
.join(' -> ')
);
// MAIN --- return main();
})();</lang>
- Output:
[17,2]
Julia
<lang julia>function egyptiandivision(dividend::Int, divisor::Int)
N = 64
powers = Vector{Int}(N)
doublings = Vector{Int}(N)
ind = 0
for i in 0:N-1
powers[i+1] = 1 << i
doublings[i+1] = divisor << i
if doublings[i+1] > dividend ind = i-1; break end
end
ans = acc = 0
for i in ind:-1:0
if acc + doublings[i+1] ≤ dividend
acc += doublings[i+1]
ans += powers[i+1]
end
end
return ans, dividend - acc
end
q, r = egyptiandivision(580, 34) println("580 ÷ 34 = $q (remains $r)")
using Base.Test
@testset "Equivalence to divrem builtin function" begin
for x in rand(1:100, 100), y in rand(1:100, 10) @test egyptiandivision(x, y) == divrem(x, y) end
end</lang>
- Output:
580 ÷ 34 = 17 (remains 2) Test Summary: | Pass Total Equivalence to divrem builtin function | 1000 1000
Kotlin
<lang scala>// version 1.1.4
data class DivMod(val quotient: Int, val remainder: Int)
fun egyptianDivide(dividend: Int, divisor: Int): DivMod {
require (dividend >= 0 && divisor > 0)
if (dividend < divisor) return DivMod(0, dividend)
val powersOfTwo = mutableListOf(1)
val doublings = mutableListOf(divisor)
var doubling = divisor
while (true) {
doubling *= 2
if (doubling > dividend) break
powersOfTwo.add(powersOfTwo[powersOfTwo.lastIndex] * 2)
doublings.add(doubling)
}
var answer = 0
var accumulator = 0
for (i in doublings.size - 1 downTo 0) {
if (accumulator + doublings[i] <= dividend) {
accumulator += doublings[i]
answer += powersOfTwo[i]
if (accumulator == dividend) break
}
}
return DivMod(answer, dividend - accumulator)
}
fun main(args: Array<String>) {
val dividend = 580
val divisor = 34
val (quotient, remainder) = egyptianDivide(dividend, divisor)
println("$dividend divided by $divisor is $quotient with remainder $remainder")
}</lang>
- Output:
580 divided by 34 is 17 with remainder 2
Lua
<lang lua>function egyptian_divmod(dividend,divisor)
local pwrs, dbls = {1}, {divisor}
while dbls[#dbls] <= dividend do
table.insert(pwrs, pwrs[#pwrs] * 2)
table.insert(dbls, pwrs[#pwrs] * divisor)
end
local ans, accum = 0, 0
for i=#pwrs-1,1,-1 do
if accum + dbls[i] <= dividend then
accum = accum + dbls[i]
ans = ans + pwrs[i]
end
end
return ans, math.abs(accum - dividend)
end
local i, j = 580, 34 local d, m = egyptian_divmod(i, j) print(i.." divided by "..j.." using the Egyptian method is "..d.." remainder "..m)</lang>
- Output:
580 divided by 34 using the Egyptian method is 17 remainder 2
Modula-2
<lang modula2>MODULE EgyptianDivision; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,ReadChar;
PROCEDURE EgyptianDivision(dividend,divisor : LONGCARD; VAR remainder : LONGCARD) : LONGCARD; CONST
SZ = 64;
VAR
powers,doublings : ARRAY[0..SZ] OF LONGCARD; answer,accumulator : LONGCARD; i : INTEGER;
BEGIN
FOR i:=0 TO SZ-1 DO
powers[i] := 1 SHL i;
doublings[i] := divisor SHL i;
IF doublings[i] > dividend THEN
BREAK
END
END;
answer := 0;
accumulator := 0;
FOR i:=i-1 TO 0 BY -1 DO
IF accumulator + doublings[i] <= dividend THEN
accumulator := accumulator + doublings[i];
answer := answer + powers[i]
END
END;
remainder := dividend - accumulator; RETURN answer
END EgyptianDivision;
VAR
buf : ARRAY[0..63] OF CHAR; div,rem : LONGCARD;
BEGIN
div := EgyptianDivision(580, 34, rem);
FormatString("580 divided by 34 is %l remainder %l\n", buf, div, rem);
WriteString(buf);
ReadChar
END EgyptianDivision.</lang>
Nim
<lang nim>import strformat
func egyptianDivision(dividend, divisor: int): tuple[quotient, remainder: int] =
if dividend < 0 or divisor <= 0:
raise newException(IOError, "Invalid argument(s)")
if dividend < divisor:
return (0, dividend)
var powersOfTwo: array[sizeof(int) * 8, int]
var doublings: array[sizeof(int) * 8, int]
for i, _ in powersOfTwo:
powersOfTwo[i] = 1 shl i
doublings[i] = divisor shl i
if doublings[i] > dividend:
break
var answer = 0
var accumulator = 0
for i in countdown(len(doublings) - 1, 0):
if accumulator + doublings[i] <= dividend:
inc accumulator, doublings[i]
inc answer, powersOfTwo[i]
if accumulator == dividend:
break
(answer, dividend - accumulator)
let dividend = 580 let divisor = 34 var (quotient, remainder) = egyptianDivision(dividend, divisor) echo fmt"{dividend} divided by {divisor} is {quotient} with remainder {remainder}"</lang>
- Output:
580 divided by 34 is 17 with remainder 2
Perl
<lang perl>sub egyptian_divmod {
my($dividend, $divisor) = @_; die "Invalid divisor" if $divisor <= 0;
my @table = ($divisor); push @table, 2*$table[-1] while $table[-1] <= $dividend;
my $accumulator = 0;
for my $k (reverse 0 .. $#table) {
next unless $dividend >= $table[$k];
$accumulator += 1 << $k;
$dividend -= $table[$k];
}
$accumulator, $dividend;
}
for ([580,34], [578,34], [7532795332300578,235117]) {
my($n,$d) = @$_; printf "Egyption divmod %s %% %s = %s remainder %s\n", $n, $d, egyptian_divmod( $n, $d )
}</lang>
- Output:
Egyption divmod 580 % 34 = 17 remainder 2 Egyption divmod 578 % 34 = 17 remainder 0 Egyption divmod 7532795332300578 % 235117 = 32038497141 remainder 81
Perl 6
Normal version
Only works with positive real numbers, not negative or complex. <lang perl6>sub egyptian-divmod (Real $dividend is copy where * >= 0, Real $divisor where * > 0) {
my $accumulator = 0;
([1, $divisor], { [.[0] + .[0], .[1] + .[1]] } … ^ *.[1] > $dividend)
.reverse.map: { $dividend -= .[1], $accumulator += .[0] if $dividend >= .[1] }
$accumulator, $dividend;
}
- TESTING
for 580,34, 578,34, 7532795332300578,235117 -> $n, $d {
printf "%s divmod %s = %s remainder %s\n",
$n, $d, |egyptian-divmod( $n, $d )
}</lang>
- Output:
580 divmod 34 = 17 remainder 2 578 divmod 34 = 17 remainder 0 7532795332300578 divmod 235117 = 32038497141 remainder 81
More "Egyptian" version
As a preceding version was determined to be "let's just say ... not Egyptian" we submit an alternate which is hopefully more "Egyptian". Now only handles positive Integers up to 10 million, mostly due to limitations on Egyptian notation for numbers.
Note: if the below is just a mass of "unknown glyph" boxes, try installing Googles free Noto Sans Egyptian Hieroglyphs font.
This is intended to be humorous and should not be regarded as good (or even sane) programming practice. That being said, 𓂽 & 𓂻 really are the ancient Egyptian symbols for addition and subtraction, and the Egyptian number notation is as accurate as possible. Everything else owes more to whimsy than rigor. <lang perl6>my (\𓄤, \𓄊, \𓎆, \𓄰) = (0, 1, 10, 10e7); sub infix:<𓂽> { $^𓃠 + $^𓃟 } sub infix:<𓂻> { $^𓃲 - $^𓆊 } sub infix:<𓈝> { $^𓃕 < $^𓃢 } sub 𓁶 (Int \𓆉) {
my \𓁢 = [« 𓏺 𓏻 𓏼 𓏽 𓏾 𓏿 𓐀 𓐁 𓐂»], [« 𓎆 𓎏 𓎐 𓎑 𓎊 𓎋 𓎌 𓎍 𓎎»],
[« 𓍢 𓍣 𓍤 𓍥 𓍦 𓍧 𓍨 𓍩 𓍪»], [« 𓆼 𓆽 𓆾 𓆿 𓇀 𓇁 𓇂 𓇃 𓇄»],
[« 𓂭 𓂮 𓂯 𓂰 𓂱 𓂲 𓂳 𓂴 𓂵»], ['𓆐' Xx ^𓎆], ['𓁨' Xx ^𓎆];
([~] 𓆉.polymod( 𓎆 xx * ).map( { 𓁢[$++;$_] } ).reverse) || '𓄤'
}
sub infix:<𓅓> (Int $𓂀 is copy where 𓄤 𓂻 𓄊 𓈝 * 𓈝 𓄰, Int \𓌳 where 𓄤 𓈝 * 𓈝 𓄰) {
my $𓎦 = 𓄤;
([𓄊,𓌳], { [.[𓄤] 𓂽 .[𓄤], .[𓄊] 𓂽 .[𓄊]] } … ^$𓂀 𓈝 *.[𓄊])
.reverse.map: { $𓂀 𓂻= .[𓄊], $𓎦 𓂽= .[𓄤] if .[𓄊] 𓈝 ($𓂀 𓂽 𓄊) }
$𓎦, $𓂀;
}
- TESTING
for 580,34, 578,34, 2300578,23517 -> \𓃾, \𓆙 {
printf "%s divmod %s = %s remainder %s =OR= %s 𓅓 %s = %s remainder %s\n",
𓃾, 𓆙, |(𓃾 𓅓 𓆙), (𓃾, 𓆙, |(𓃾 𓅓 𓆙))».&𓁶;
}</lang>
- Output:
580 divmod 34 = 17 remainder 2 =OR= 𓍦𓎍 𓅓 𓎐𓏽 = 𓎆𓐀 remainder 𓏻 578 divmod 34 = 17 remainder 0 =OR= 𓍦𓎌𓐁 𓅓 𓎐𓏽 = 𓎆𓐀 remainder 𓄤 2300578 divmod 23517 = 97 remainder 19429 =OR= 𓁨𓁨𓆐𓆐𓆐𓍦𓎌𓐁 𓅓 𓂮𓆾𓍦𓎆𓐀 = 𓎎𓐀 remainder 𓂭𓇄𓍥𓎏𓐂
Phix
<lang Phix>procedure egyptian_division(integer dividend, divisor) integer p2 = 1, dbl = divisor, ans = 0, accum = 0 sequence p2s = {}, dbls = {}, args
while dbl<=dividend do
p2s = append(p2s,p2)
dbls = append(dbls,dbl)
dbl += dbl
p2 += p2
end while
for i=length(p2s) to 1 by -1 do
if accum+dbls[i]<=dividend then
accum += dbls[i]
ans += p2s[i]
end if
end for
args = {dividend,divisor,ans,abs(accum-dividend)}
printf(1,"%d divided by %d is: %d remainder %d\n",args)
end procedure
egyptian_division(580,34)</lang>
- Output:
580 divided by 34 is: 17 remainder 2
PicoLisp
<lang PicoLisp>(seed (in "/dev/urandom" (rd 8)))
(de divmod (Dend Disor)
(cons (/ Dend Disor) (% Dend Disor)) )
(de egyptian (Dend Disor)
(let
(P 0
D Disor
S
(make
(while (>= Dend (setq @@ (+ D D)))
(yoke
(cons
(** 2 (swap 'P (inc P)))
(swap 'D @@) ) ) ) )
P (** 2 P) )
(mapc
'((L)
(and
(>= Dend (+ D (cdr L)))
(inc 'P (car L))
(inc 'D (cdr L)) ) )
S )
(cons P (abs (- Dend D))) ) )
(for N 1000
(let (A (rand 1 1000) B (rand 1 A))
(test (divmod A B) (egyptian A B)) ) )
(println (egyptian 580 34))</lang>
- Output:
(17 . 2)
Python
Idiomatic
<lang python>from itertools import product
def egyptian_divmod(dividend, divisor):
assert divisor != 0
pwrs, dbls = [1], [divisor]
while dbls[-1] <= dividend:
pwrs.append(pwrs[-1] * 2)
dbls.append(pwrs[-1] * divisor)
ans, accum = 0, 0
for pwr, dbl in zip(pwrs[-2::-1], dbls[-2::-1]):
if accum + dbl <= dividend:
accum += dbl
ans += pwr
return ans, abs(accum - dividend)
if __name__ == "__main__":
# Test it gives the same results as the divmod built-in
for i, j in product(range(13), range(1, 13)):
assert egyptian_divmod(i, j) == divmod(i, j)
# Mandated result
i, j = 580, 34
print(f'{i} divided by {j} using the Egyption method is %i remainder %i'
% egyptian_divmod(i, j))</lang>
Sample output
580 divided by 34 using the Egyption method is 17 remainder 2
Functional
Expressing the summing catamorphism in terms of functools.reduce, and the preliminary expansion (by repeated addition to self) in terms of an unfoldl function, which is dual to reduce, and constructs a list from a seed value.
Multiplication and division operators are both avoided, in the spirit of the Rhind Papyrus derivations of both (*) and (/) from plain addition and subtraction.
Also in deference to the character of the Rhind methods, the (unfoldl) unfolding of the seed values to a list of progressively doubling rows is recursively defined, and mutation operations are avoided. The efficiency of the Egyptian method's exponential expansion means that there is no need here, even with larger numbers, to compress space by using an imperative translation of the higher-order unfold function.
<lang python>Quotient and remainder of division by the Rhind papyrus method.
from functools import reduce
- eqyptianQuotRem :: Int -> Int -> (Int, Int)
def eqyptianQuotRem(m):
Quotient and remainder derived by the Eqyptian method.
def expansion(xi):
Doubled value, and next power of two - both by self addition.
x, i = xi
return Nothing() if x > m else Just(
((x + x, i + i), xi)
)
def collapse(qr, ix):
Addition of a power of two to the quotient,
and subtraction of a paired value from the remainder.
i, x = ix
q, r = qr
return (q + i, r - x) if x < r else qr
return lambda n: reduce(
collapse,
unfoldl(expansion)(
(1, n)
),
(0, m)
)
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Test
print(
eqyptianQuotRem(580)(34)
)
- GENERIC FUNCTIONS ---------------------------------------
- Just :: a -> Maybe a
def Just(x):
Constructor for an inhabited Maybe (option type) value.
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
- Nothing :: Maybe a
def Nothing():
Constructor for an empty Maybe (option type) value.
return {'type': 'Maybe', 'Nothing': True}
- unfoldl :: (b -> Maybe (b, a)) -> b -> [a]
def unfoldl(f):
Dual to reduce or foldl.
Where a fold reduces a list to a summary value,
unfoldl builds a list from a seed value.
When f returns Just(a, b), a is appended to the list,
and the residual b becomes the argument for the next
application of f.
When f returns Nothing, the completed list is returned.
def go(v):
xr = v, v
xs = []
while True:
mb = f(xr[0])
if mb.get('Nothing'):
return xs
else:
xr = mb.get('Just')
xs.insert(0, xr[1])
return xs
return lambda x: go(x)
- MAIN ----------------------------------------------------
if __name__ == '__main__':
main()</lang>
- Output:
(17, 2)
Racket
<lang racket>#lang racket
(define (quotient/remainder-egyptian dividend divisor (trace? #f))
(define table
(for*/list ((power_of_2 (sequence-map (curry expt 2) (in-naturals)))
(doubling (in-value (* divisor power_of_2)))
#:break (> doubling dividend))
(list power_of_2 doubling)))
(when trace?
(displayln "Table\npow_2\tdoubling")
(for ((row table)) (printf "~a\t~a~%" (first row) (second row))))
(define-values (answer accumulator)
(for*/fold ((answer 0) (accumulator 0))
((row (reverse table))
(acc′ (in-value (+ accumulator (second row)))))
(when trace? (printf "row:~a\tans/acc:~a ~a\t" row answer accumulator))
(cond
[(<= acc′ dividend)
(define ans′ (+ answer (first row)))
(when trace? (printf "~a <= ~a -> ans′/acc′:~a ~a~%" acc′ dividend ans′ acc′))
(values ans′ acc′)]
[else
(when trace? (printf "~a > ~a [----]~%" acc′ dividend))
(values answer accumulator)])))
(values answer (- dividend accumulator)))
(module+ test
(require rackunit) (let-values (([q r] (quotient/remainder-egyptian 580 34))) (check-equal? q 17) (check-equal? r 2))
(let-values (([q r] (quotient/remainder-egyptian 192 3))) (check-equal? q 64) (check-equal? r 0)))
(module+ main
(quotient/remainder-egyptian 580 34 #t))</lang>
- Output:
Table pow_2 doubling 1 34 2 68 4 136 8 272 16 544 row:(16 544) ans/acc:0 0 544 <= 580 -> ans′/acc′:16 544 row:(8 272) ans/acc:16 544 816 > 580 [----] row:(4 136) ans/acc:16 544 680 > 580 [----] row:(2 68) ans/acc:16 544 612 > 580 [----] row:(1 34) ans/acc:16 544 578 <= 580 -> ans′/acc′:17 578 17 2
REXX
Only addition and subtraction is used in this version of the Egyptian division method. <lang rexx>/*REXX program performs division on positive integers using the Egyptian division method*/ numeric digits 1000 /*support gihugic numbers & be gung-ho.*/ parse arg n d . /*obtain optional arguments from the CL*/ if d== | d=="," then do; n= 580; d= 34 /*Not specified? Then use the defaults*/
end
call EgyptDiv n, d /*invoke the Egyptian Division function*/ parse var result q r /*extract the quotient & the remainder.*/ say n ' divided by ' d " is " q ' with a remainder of ' r exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ EgyptDiv: procedure; parse arg num,dem /*obtain the numerator and denominator.*/
p= 1; t= dem /*initialize the double & power values.*/
do #=1 until t>num /*construct the power & doubling lists.*/
pow.#= p; p= p + p /*build power entry; bump power value.*/
dbl.#= t; t= t + t /* " doubling " ; " doubling val.*/
end /*#*/
acc=0; ans=0 /*initialize accumulator & answer to 0 */
do s=# by -1 for # /* [↓] process the table "backwards". */
sum= acc + dbl.s /*compute the sum (to be used for test)*/
if sum>num then iterate /*Is sum to big? Then ignore this step*/
acc= sum /*use the "new" sum for the accumulator*/
ans= ans + pow.s /*calculate the (newer) running answer.*/
end /*s*/
return ans num-acc /*return the answer and the remainder. */</lang>
- output when using the default inputs:
580 divided by 34 is 17 with a remainder of 2
- output when using the input of: 9876543210111222333444555666777888999 13579
9876543210111222333444555666777888999 divided by 13579 is 727339510281406755537562093436769 with a remainder of 2748
Ring
<lang ring> load "stdlib.ring"
table = newlist(32, 2) dividend = 580 divisor = 34
i = 1 table[i][1] = 1 table[i][2] = divisor
while table[i] [2] < dividend
i = i + 1
table[i][1] = table[i -1] [1] * 2
table[i][2] = table[i -1] [2] * 2
end i = i - 1 answer = table[i][1] accumulator = table[i][2]
while i > 1
i = i - 1
if table[i][2]+ accumulator <= dividend
answer = answer + table[i][1]
accumulator = accumulator + table[i][2]
ok
end
see string(dividend) + " divided by " + string(divisor) + " using egytian division" + nl see " returns " + string(answer) + " mod(ulus) " + string(dividend-accumulator) </lang> Output:
580 divided by 34 using egytian division returns 17 mod(ulus) 2
Ruby
<lang ruby>def egyptian_divmod(dividend, divisor)
table = 1, divisor table << table.last.map{|e| e*2} while table.last.first * 2 <= dividend answer, accumulator = 0, 0 table.reverse_each do |pow, double| if accumulator + double <= dividend accumulator += double answer += pow end end [answer, dividend - accumulator]
end
puts "Quotient = %s Remainder = %s" % egyptian_divmod(580, 34) </lang>
- Output:
Quotient = 17 Remainder = 2
Rust
<lang rust>fn egyptian_divide(dividend: u32, divisor: u32) -> (u32, u32) {
let dividend = dividend as u64;
let divisor = divisor as u64;
let pows = (0..32).map(|p| 1 << p);
let doublings = (0..32).map(|p| divisor << p);
let (answer, sum) = doublings
.zip(pows)
.rev()
.skip_while(|(i, _)| i > ÷nd )
.fold((0, 0), |(answer, sum), (double, power)| {
if sum + double < dividend {
(answer + power, sum + double)
} else {
(answer, sum)
}
});
(answer as u32, (dividend - sum) as u32)
}
fn main() {
let (div, rem) = egyptian_divide(580, 34);
println!("580 divided by 34 is {} remainder {}", div, rem);
}</lang>
- Output:
580 divided by 34 is 17 remainder 2
Scala
- Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>object EgyptianDivision extends App {
private def divide(dividend: Int, divisor: Int): Unit = {
val powersOf2, doublings = new collection.mutable.ListBuffer[Integer]
//populate the powersof2- and doublings-columns
var line = 0
while ((math.pow(2, line) * divisor) <= dividend) {
val powerOf2 = math.pow(2, line).toInt
powersOf2 += powerOf2
doublings += (powerOf2 * divisor)
line += 1
}
var answer, accumulator = 0
//Consider the rows in reverse order of their construction (from back to front of the List)
var i = powersOf2.size - 1
for (i <- powersOf2.size - 1 to 0 by -1)
if (accumulator + doublings(i) <= dividend) {
accumulator += doublings(i)
answer += powersOf2(i)
}
println(f"$answer%d, remainder ${dividend - accumulator}%d")
}
divide(580, 34)
}</lang>
Sidef
<lang ruby>func egyptian_divmod(dividend, divisor) {
var table = 1, divisor table << table[-1].map{|e| 2*e } while (2*table[-1][0] <= dividend) var (answer, accumulator) = (0, 0) table.reverse.each { |pair| var (pow, double) = pair... if (accumulator + double <= dividend) { accumulator += double answer += pow } } return (answer, dividend - accumulator)
}
say ("Quotient = %s Remainder = %s" % egyptian_divmod(580, 34))</lang>
- Output:
Quotient = 17 Remainder = 2
Swift
<lang swift>extension BinaryInteger {
@inlinable
public func egyptianDivide(by divisor: Self) -> (quo: Self, rem: Self) {
let table =
(0...).lazy
.map({i -> (Self, Self) in
let power = Self(2).power(Self(i))
return (power, power * divisor)
})
.prefix(while: { $0.1 <= self })
.reversed()
let (answer, acc) = table.reduce((Self(0), Self(0)), {cur, row in
let ((ans, acc), (power, doubling)) = (cur, row)
return acc + doubling <= self ? (ans + power, doubling + acc) : cur })
return (answer, Self((acc - self).magnitude)) }
@inlinable
public func power(_ n: Self) -> Self {
return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *)
}
}
let dividend = 580 let divisor = 34 let (quo, rem) = dividend.egyptianDivide(by: divisor)
print("\(dividend) divided by \(divisor) = \(quo) rem \(rem)") </lang>
- Output:
580 divided by 34 = 17 rem 2
VBA
<lang vb>Option Explicit
Private Type MyTable
powers_of_2 As Long doublings As Long
End Type
Private Type Assemble
answer As Long accumulator As Long
End Type
Private Type Division
Quotient As Long Remainder As Long
End Type
Private Type DivEgyp
Dividend As Long Divisor As Long
End Type
Private Deg As DivEgyp
Sub Main() Dim d As Division
Deg.Dividend = 580 Deg.Divisor = 34 d = Divise(CreateTable) Debug.Print "Quotient = " & d.Quotient & " Remainder = " & d.Remainder
End Sub
Private Function CreateTable() As MyTable() Dim t() As MyTable, i As Long
Do
i = i + 1
ReDim Preserve t(i)
t(i).powers_of_2 = 2 ^ (i - 1)
t(i).doublings = Deg.Divisor * t(i).powers_of_2
Loop While 2 * t(i).doublings <= Deg.Dividend
CreateTable = t
End Function
Private Function Divise(t() As MyTable) As Division Dim a As Assemble, i As Long
a.accumulator = 0
a.answer = 0
For i = UBound(t) To LBound(t) Step -1
If a.accumulator + t(i).doublings <= Deg.Dividend Then
a.accumulator = a.accumulator + t(i).doublings
a.answer = a.answer + t(i).powers_of_2
End If
Next
Divise.Quotient = a.answer
Divise.Remainder = Deg.Dividend - a.accumulator
End Function</lang>
- Output:
Quotient = 17 Remainder = 2
Visual Basic .NET
<lang vbnet>Module Module1
Function EgyptianDivision(dividend As ULong, divisor As ULong, ByRef remainder As ULong) As ULong
Const SIZE = 64
Dim powers(SIZE) As ULong
Dim doublings(SIZE) As ULong
Dim i = 0
While i < SIZE
powers(i) = 1 << i
doublings(i) = divisor << i
If doublings(i) > dividend Then
Exit While
End If
i = i + 1
End While
Dim answer As ULong = 0
Dim accumulator As ULong = 0
i = i - 1
While i >= 0
If accumulator + doublings(i) <= dividend Then
accumulator += doublings(i)
answer += powers(i)
End If
i = i - 1
End While
remainder = dividend - accumulator
Return answer
End Function
Sub Main(args As String())
If args.Length < 2 Then
Dim name = Reflection.Assembly.GetEntryAssembly().Location
Console.Error.WriteLine("Usage: {0} dividend divisor", IO.Path.GetFileNameWithoutExtension(name))
Return
End If
Dim dividend = CULng(args(0))
Dim divisor = CULng(args(1))
Dim remainder As ULong
Dim ans = EgyptianDivision(dividend, divisor, remainder)
Console.WriteLine("{0} / {1} = {2} rem {3}", dividend, divisor, ans, remainder)
End Sub
End Module</lang>
- Output:
580 / 34 = 17 rem 2
zkl
<lang zkl>fcn egyptianDivmod(dividend,divisor){
table:=[0..].pump(List, 'wrap(n){ // (2^n,divisor*2^n)
r:=T( p:=(2).pow(n), s:=divisor*p); (s<=dividend) and r or Void.Stop });
accumulator:=0;
foreach p2,d in (table.reverse()){
if(dividend>=d){ accumulator+=p2; dividend-=d; }
}
return(accumulator,dividend);
}</lang> <lang zkl>foreach dividend,divisor in (T(T(580,34), T(580,17), T(578,34), T(7532795332300578,235117))){
println("%d %% %d = %s".fmt(dividend,divisor,egyptianDivmod(dividend,divisor)));
}</lang>
- Output:
580 % 34 = L(17,2) 580 % 17 = L(34,2) 578 % 34 = L(17,0) 7532795332300578 % 235117 = L(32038497141,81)