Egyptian division: Difference between revisions
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580 divided by 34 is 17 with remainder 2 |
580 divided by 34 is 17 with remainder 2 |
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</pre> |
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=={{header|Lua}}== |
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{{trans|Python}} |
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<lang lua>function egyptian_divmod(dividend,divisor) |
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local pwrs, dbls = {1}, {divisor} |
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while dbls[#dbls] <= dividend do |
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table.insert(pwrs, pwrs[#pwrs] * 2) |
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table.insert(dbls, pwrs[#pwrs] * divisor) |
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end |
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local ans, accum = 0, 0 |
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for i=#pwrs-1,1,-1 do |
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if accum + dbls[i] <= dividend then |
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accum = accum + dbls[i] |
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ans = ans + pwrs[i] |
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end |
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end |
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return ans, math.abs(accum - dividend) |
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end |
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local i, j = 580, 34 |
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local d, m = egyptian_divmod(i, j) |
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print(i.." divided by "..j.." using the Egyptian method is "..d.." remainder "..m)</lang> |
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{{out}} |
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<pre>580 divided by 34 using the Egyptian method is 17 remainder 2</pre> |
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=={{header|Modula-2}}== |
=={{header|Modula-2}}== |
Revision as of 19:50, 26 January 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 rows, fold to sum quotient and derive remainder <lang AppleScript>-- EGYPTIAN DIVISION ---------------------------------------------------------
-- egyptianQuotRem :: Int -> Int -> (Int, Int) on egyptianQuotRem(m, n)
script doubledRows script double on |λ|(x) x + x end |λ| end script on |λ|(ix) if item 2 of ix > m then {nothing:true} else {just:ix, new:map(double, ix), nothing:false} end if end |λ| end script set rows to unfoldr(doubledRows, [1, n]) script quotientSum on |λ|(ix, qr) set {i, x} to ix set {q, r} to qr if x < r then {q + i, r - x} else {q, r} end if end |λ| end script foldr(quotientSum, {0, m}, rows)
end egyptianQuotRem
-- TEST ---------------------------------------------------------------------- on run
egyptianQuotRem(580, 34) --> {17, 2}
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- foldr :: (a -> b -> a) -> a -> [b] -> a
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
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- unfoldr :: (b -> Maybe (a, b)) -> b -> [a] on unfoldr(f, v)
set mf to mReturn(f) set lst to {} set recM to {nothing:false, new:v} repeat while (not (nothing of recM)) set recM to mf's |λ|(new of recM) if not nothing of recM then set end of lst to just of recM end repeat lst
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>
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);
}
/** * Dividesdividend
bydivisor
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 -----------------------------------------------------
// egyptianQuotRem :: Int -> Int -> (Int, Int) const egyptianQuotRem = (m, n) => { const rows = unfoldr( ix => { const v = ix.new return v.x > m ? { nothing: true } : { just: v, new: { i: v.i * 2, x: v.x * 2 } } }, { i: 1, x: n } ); return foldr(({ i: i, x: x }, [q, r]) => x < r ? [q + i, r - x] : [q, r], [0, m], rows); };
// GENERIC FUNCTIONS -----------------------------------------------------
// flip :: (a -> b -> c) -> b -> a -> c const flip = f => (a, b) => f.apply(null, [b, a]);
// foldr (a -> b -> b) -> b -> [a] -> b const foldr = (f, a, xs) => xs.reduceRight(flip(f), a);
// show :: a -> String const show = (...x) => JSON.stringify.apply( null, x.length > 1 ? [x[1], null, x[0]] : x );
// unfoldr :: (b -> Maybe (a, b)) -> b -> [a] const unfoldr = (mf, v) => { let xs = []; return (until( m => m.nothing, m => { const m2 = mf(m); return ( xs = m2.nothing ? xs : xs.concat(m2.just), m2 ); }, { nothing: false, just: v, new: v, } ), xs); };
// until :: (a -> Bool) -> (a -> a) -> a -> a const until = (p, f, x) => { let v = x; while (!p(v)) v = f(v); return v; };
// TEST ------------------------------------------------------------------ return show( egyptianQuotRem(580, 34) );
})();</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 $remainder")
}</lang>
- Output:
580 divided by 34 is 17 with remainder 2
Lua
<lang lua>function egyptian_divmod(dividend,divisor)
local pwrs, dbls = {1}, {divisor} while dbls[#dbls] <= dividend do table.insert(pwrs, pwrs[#pwrs] * 2) table.insert(dbls, pwrs[#pwrs] * divisor) end local ans, accum = 0, 0
for i=#pwrs-1,1,-1 do if accum + dbls[i] <= dividend then accum = accum + dbls[i] ans = ans + pwrs[i] end end
return ans, math.abs(accum - dividend)
end
local i, j = 580, 34 local d, m = egyptian_divmod(i, j) print(i.." divided by "..j.." using the Egyptian method is "..d.." remainder "..m)</lang>
- Output:
580 divided by 34 using the Egyptian method is 17 remainder 2
Modula-2
<lang modula2>MODULE EgyptianDivision; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,ReadChar;
PROCEDURE EgyptianDivision(dividend,divisor : LONGCARD; VAR remainder : LONGCARD) : LONGCARD; CONST
SZ = 64;
VAR
powers,doublings : ARRAY[0..SZ] OF LONGCARD; answer,accumulator : LONGCARD; i : INTEGER;
BEGIN
FOR i:=0 TO SZ-1 DO powers[i] := 1 SHL i; doublings[i] := divisor SHL i; IF doublings[i] > dividend THEN BREAK END END;
answer := 0; accumulator := 0; FOR i:=i-1 TO 0 BY -1 DO IF accumulator + doublings[i] <= dividend THEN accumulator := accumulator + doublings[i]; answer := answer + powers[i] END END;
remainder := dividend - accumulator; RETURN answer
END EgyptianDivision;
VAR
buf : ARRAY[0..63] OF CHAR; div,rem : LONGCARD;
BEGIN
div := EgyptianDivision(580, 34, rem); FormatString("580 divided by 34 is %l remainder %l\n", buf, div, rem); WriteString(buf);
ReadChar
END EgyptianDivision.</lang>
Perl
<lang perl>sub egyptian_divmod {
my($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
Imperative
<lang python>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)
- Test it gives the same results as the divmod built-in
from itertools import product 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 catamorphism in terms of functools.reduce, and the preliminary expansion in terms of an unfoldl function, which is dual to reduce:
<lang python>from functools import (reduce)
- eqyptianQuotRem :: Int -> Int -> (Int, Int)
def eqyptianQuotRem(m):
def expansion(xi): x, i = xi return Nothing() if x > m else Just( ((x + x, i + i), xi) )
def collapse(qr, ix): 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 ----------------------------------------------------
def main():
print( eqyptianQuotRem(580)(34) )
- GENERIC -------------------------------------------------
- Just :: a -> Maybe a
def Just(x):
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
- Nothing :: Maybe a
def Nothing():
return {'type': 'Maybe', 'Nothing': True}
- unfoldl(lambda x: Just(((x - 1), x)) if 0 != x else Nothing())(10)
- -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
- unfoldl :: (b -> Maybe (b, a)) -> b -> [a]
def unfoldl(f):
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 v: go(v)
- 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
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
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)