Talk:Fairshare between two and more: Difference between revisions

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(→‎Fairness example and cycles: For a different set of numbers.)
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:: Great :-) <br>--[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 18:26, 2 February 2020 (UTC)
:: Great :-) <br>--[[User:Paddy3118|Paddy3118]] ([[User talk:Paddy3118|talk]]) 18:26, 2 February 2020 (UTC)
::: I tried to clearify things to me, like [[User:Paddy3118|Paddy3118]] described in his links.Without different values, it makes no sense.<BR>The first will get the highest value of a bucket, the second the maximum of left over and so on.I use a bucket of size Peoplecnt and the values are PeopleCnt downto 1.The choosen people grabs one value from Top.After all people are finished the game starts again ( MOD peoplecnt). <lang pascal>program Fair;
::: I tried to clearify things to me, like [[User:Paddy3118|Paddy3118]] described in his links.Without different values, it makes no sense.<BR>Rest removed, explanation by [[User:Paddy3118|Paddy3118]] ([[User:Horst.h|Horst.h]]) [[User:Horst.h|Horst.h]] ([[User talk:Horst.h|talk]]) 11:42, 26 June 2020 (UTC)
{$IFDEF FPC}
{$MODE DELPHI}
{$OPTIMIZATION ON,ALL}
{$CodeAlign proc=8}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}

const
cntbasedigits = 21;
type
tSumDigit = record
sdSumDig : NativeUint;
sdBase : NativeUint;
sdNumber : NativeUint;
sdDigits : array[0..cntbasedigits-1] of NativeUint;
end;
var
SumWealth : array of NativeUint;
Values : array of NativeUint;

function InitSumDigit(n,base : NativeUint):tSumDigit;
var
sd : tSumDigit;
qt : NativeUint;
i : integer;
begin
with sd do
begin
sdNumber:= n;
sdBase := base;
fillchar(sdDigits,SizeOf(sdDigits),#0);
sdSumDig :=0;
i := 0;
// calculate Digits und sum them up
while n > 0 do
begin
qt := n div sdbase;
{n mod base}
sdDigits[i] := n-qt*sdbase;
inc(sdSumDig,sdDigits[i]);
n:= qt;
inc(i);
end;
end;
InitSumDigit:=sd;
end;

procedure IncSumDigit(var sd:tSumDigit);
var
i,d: integer;
begin
i := 0;
with sd do
begin
inc(sdNumber);
repeat
d := sdDigits[i];
inc(d);
inc(sdSumDig);
//base-1 times the repeat is left here
if d < sdbase then
begin
sdDigits[i] := d;
BREAK;
end
else
begin
sdDigits[i] := 0;
dec(sdSumDig,sdbase);
inc(i);
end;
until i > high( sdDigits);
end;
end;

procedure First25(base:NativeUint);
var
MySumDig : tSumDigit;
cnt: NativeUint;
begin
write(' [',base:5,'] -> ');
MySumDig:=InitSumDigit(0,base);
cnt := 0;
repeat
with MySumDig do
write(sdSumDig MOD sdbase,'-');
inc(cnt);
IncSumDigit(MySumDig);
until cnt >= 25;
writeln('....');
end;

procedure CheckRoundsOfPeople(turns,peopleCnt:NativeUint);
var
MySumDig : tSumDigit;
i,
wholeWealth,
minWealth,
maxWealth : NativeUint;
Begin
setlength(SumWealth,peopleCnt);
setlength(Values,peopleCnt);
//Values[0] = peopleCnt ...Values[peopleCnt-1] = 1
For i := 0 to peopleCnt-1 do
Values[i] := peopleCnt-i;

MySumDig:=InitSumDigit(0,peopleCnt);
i := 0;
while i<turns do
begin
inc(SumWealth[MySumDig.sdSumDig MOD peopleCnt],Values[i MOD peopleCnt]);
IncSumDigit(MySumDig);
inc(i);
end;
setlength(Values,0);
MinWealth := High(MinWealth);
MaxWealth := Low(MaxWealth);
For i := 0 to peopleCnt-1 do
Begin
wholeWealth := SumWealth[i];
IF MaxWealth<wholeWealth then
MaxWealth:=wholeWealth;
IF MinWealth>wholeWealth then
MinWealth := wholeWealth;
end;
writeln(peopleCnt:6,turns:11,MinWealth:10,MaxWealth:10, MinWealth/MaxWealth:10:7);
setlength(SumWealth,0);
end;

procedure CheckRoundsOfPeopleOneByOne(turns,peopleCnt:NativeUint);
var
i,
wholeWealth,
minWealth,
maxWealth : NativeUint;
Begin
setlength(SumWealth,peopleCnt);
setlength(Values,peopleCnt);
//Values[0] = peopleCnt ...Values[peopleCnt-1] = 1
For i := 0 to peopleCnt-1 do
Values[i] := peopleCnt-i;

i := 0;
while i<turns do
begin
//first gets always the max value 0,1,2,3,4..,n
inc(SumWealth[i MOD peopleCnt],Values[i MOD peopleCnt]);
inc(i);
end;
setlength(Values,0);
MinWealth := High(MinWealth);
MaxWealth := Low(MaxWealth);
For i := 0 to peopleCnt-1 do
Begin
wholeWealth := SumWealth[i];
IF MaxWealth<wholeWealth then
MaxWealth:=wholeWealth;
IF MinWealth>wholeWealth then
MinWealth := wholeWealth;
end;
writeln(peopleCnt:6,turns:11,MinWealth:10,MaxWealth:10, MinWealth/MaxWealth:10:7);
setlength(SumWealth,0);
end;

const
cTURNS = 500;
begin
First25(2);First25(3);First25(5); First25(11);
writeln;
writeln('Fair share');
writeln(' people turns MinWealth MaxWealth ratio MinWealth/MaxWealth');
CheckRoundsOfPeople(11*11,11);
CheckRoundsOfPeople(1377 *1377,1377);
CheckRoundsOfPeople(cTURNS*11,11);
CheckRoundsOfPeople(cTURNS*1377,1377);

writeln;
writeln('First gets max value , last gets 1');
writeln(' people turns MinWealth MaxWealth ratio MinWealth/MaxWealth');
CheckRoundsOfPeopleOneByOne(11*11,11);
CheckRoundsOfPeopleOneByOne(1377 *1377,1377);
CheckRoundsOfPeopleOneByOne(cTURNS*11,11);
CheckRoundsOfPeopleOneByOne(cTURNS*1377,1377);
end.</lang>
{{out}}
<pre>
[ 2] -> 0-1-1-0-1-0-0-1-1-0-0-1-0-1-1-0-1-0-0-1-0-1-1-0-0-....
[ 3] -> 0-1-2-1-2-0-2-0-1-1-2-0-2-0-1-0-1-2-2-0-1-0-1-2-1-....
[ 5] -> 0-1-2-3-4-1-2-3-4-0-2-3-4-0-1-3-4-0-1-2-4-0-1-2-3-....
[ 11] -> 0-1-2-3-4-5-6-7-8-9-10-1-2-3-4-5-6-7-8-9-10-0-2-3-4-....

Fair share
people turns MinWealth MaxWealth ratio MinWealth/MaxWealth
11 121 66 66 1.0000000
1377 1896129 948753 948753 1.0000000
11 5500 2985 3015 0.9900498
1377 688500 125250 563750 0.2221729

First gets max value , last gets 1
//[ 11] -> 0-1-2-3-4-5-6-7-8-9-10-0-1-2-3-4-5-6-7-8-9-10-0-1-2-3-....
people turns MinWealth MaxWealth ratio MinWealth/MaxWealth
11 121 11 121 0.0909091
1377 1896129 1377 1896129 0.0007262
11 5500 500 5500 0.0909091
1377 688500 500 688500 0.0007262</pre>
:::This shows the fairness.But a cyclic value of people A to C is sufficient.<BR>The easier sequence ABC_BCA_CAB will get the same results. A,B,C are in every possible position, so square( peopleCnt) is fair too. [[User:Horst.h|Horst.h]]14:16, 25 June 2020 (UTC)

===Fairness example and cycles===
===Fairness example and cycles===
I saw Horsts' Perl program above, and recognized that the idea of fairness is hard to bring across so I thought I might do an example by hand.
I saw Horsts' Perl program above, and recognized that the idea of fairness is hard to bring across so I thought I might do an example by hand.

Revision as of 11:42, 26 June 2020

Perl 6 count of how many turns each person gets

Whilst important to some degree, the sequence minimises any advantage that going first/going earlier might give. I've blogged twice, here, and here about it and the sequence appears many times in science and maths. (Try this paper (PDF), for example. --Paddy3118 (talk) 23:54, 1 February 2020 (UTC)

I have to say, I kind of missed the point of the task initially so was not really sure what it was demonstrating. The actual algorithm was simple, the reason for it escaped me. After reading your links, the lightbulb lit. I removed the "number of times each person goes" which was kind-of pointless, and added a "fairness correlation" calculation showing the relative fairness to the Perl 6 entry. --Thundergnat (talk) 13:43, 2 February 2020 (UTC)
Great :-)
--Paddy3118 (talk) 18:26, 2 February 2020 (UTC)
I tried to clearify things to me, like Paddy3118 described in his links.Without different values, it makes no sense.
Rest removed, explanation by Paddy3118 (Horst.h) Horst.h (talk) 11:42, 26 June 2020 (UTC)

Fairness example and cycles

I saw Horsts' Perl program above, and recognized that the idea of fairness is hard to bring across so I thought I might do an example by hand.

The set of numbers in this case are not a linear progression, so we (maybe), see the emergence of Thue-Morse as being the most "fair" calculated as the spread in final amounts per person.

For all cases we will have have three people A B and C to choose the best at their turn, from the same, ever decreasing pots of money.

18 then 27 Fibonacci numbers

Numbers: [2584, 1597, 987, 610, 377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1]

  Order: ABC_ABC_ABC_ABC_ABC_ABC : Simple Repetition
    A gets: 2584 + 610 + 144 + 34 +  8 +  2 = 3382
    B gets: 1597 + 377 + 89 + 21 +  5 +  1 = 2090
    C gets: 987 + 233 + 55 + 13 +  3 +  1 = 1292
  Maximum difference in amounts = 2090

  Order: ABC-BCA-CAB_ABC-BCA-CAB : Simple Rotation
    A gets: 2584 + 233 + 89 + 34 +  3 +  1 = 2944
    B gets: 1597 + 610 + 55 + 21 +  8 +  1 = 2292
    C gets: 987 + 377 + 144 + 13 +  5 +  2 = 1528
  Maximum difference in amounts = 1416

  Order: ABC-BCA-CAB-BCA-CAB-ABC : Thue-Morse Fairshare
    A gets: 2584 + 233 + 89 + 13 +  5 +  2 = 2926
    B gets: 1597 + 610 + 55 + 34 +  3 +  1 = 2300
    C gets: 987 + 377 + 144 + 21 +  8 +  1 = 1538
  Maximum difference in amounts = 1388

Numbers: [196418, 121393, 75025, 46368, 28657, 17711, 10946, 6765, 4181, 2584, 1597, 987, 610, 377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1]

  Order: ABC_ABC_ABC_ABC_ABC_ABC_ABC_ABC_ABC : Simple Repetition
    A gets: 196418 + 46368 + 10946 + 2584 + 610 + 144 + 34 +  8 +  2 = 257114
    B gets: 121393 + 28657 + 6765 + 1597 + 377 + 89 + 21 +  5 +  1 = 158905
    C gets: 75025 + 17711 + 4181 + 987 + 233 + 55 + 13 +  3 +  1 = 98209
  Maximum difference in amounts = 158905

  Order: ABC-BCA-CAB_ABC-BCA-CAB_ABC-BCA-CAB : Simple Rotation
    A gets: 196418 + 17711 + 6765 + 2584 + 233 + 89 + 34 +  3 +  1 = 223838
    B gets: 121393 + 46368 + 4181 + 1597 + 610 + 55 + 21 +  8 +  1 = 174234
    C gets: 75025 + 28657 + 10946 + 987 + 377 + 144 + 13 +  5 +  2 = 116156
  Maximum difference in amounts = 107682

  Order: ABC-BCA-CAB-BCA-CAB-ABC-CAB-ABC-BCA : Thue-Morse Fairshare
    A gets: 196418 + 17711 + 6765 + 987 + 377 + 144 + 21 +  8 +  1 = 222432
    B gets: 121393 + 46368 + 4181 + 2584 + 233 + 89 + 13 +  5 +  2 = 174868
    C gets: 75025 + 28657 + 10946 + 1597 + 610 + 55 + 34 +  3 +  1 = 116928
  Maximum difference in amounts = 105504

Thue-Morse is best in this case.

Hmmm, I feel a blog coming on...

--Paddy3118 (talk) 03:50, 26 June 2020 (UTC)

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