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Alan
Guest
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 Posted: Thu Feb 15, 2007 4:30 am Post subject: Shuffling zero-based arrays |
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I'm having a problem using the following shuffling algorithm I
found searching the Internet and newsgroups:
for J := Max downto 2 do Swap(A[J], A[Succ(Random(J))]) ;
When using the above algorithm with a zero-based array (dynamic
array), the first element of the array always seems to be
consistent. I have called Randomize once in my program, and the
array is sorted using QuickSort.
Here's a brief sample of the shuffle code I am using:
type
TStrArray = array of string;
procedure ShuffleStrs(var a: TStrArray);
var
j, x : integer;
temp : string;
begin
for j := High(a) downto 2 do
begin
x := Succ(Random(j));
temp := a[j];
a[j] := a[x];
a[x] := temp;
end;
end; |
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Les Pawelczyk
Guest
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| Posted: Thu Feb 15, 2007 5:02 am Post subject: Re: Shuffling zero-based arrays |
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| Quote: | I'm having a problem using the following shuffling algorithm I
found searching the Internet and newsgroups:
for J := Max downto 2 do Swap(A[J], A[Succ(Random(J))]) ;
When using the above algorithm with a zero-based array (dynamic
array), the first element of the array always seems to be
consistent. I have called Randomize once in my program, and the
array is sorted using QuickSort.
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Succ(Random(J)) is never zero.
Les. |
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Marc Rohloff [TeamB]
Guest
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| Posted: Thu Feb 15, 2007 7:49 am Post subject: Re: Shuffling zero-based arrays |
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On 14 Feb 2007 14:30:08 -0800, Alan wrote:
| Quote: | I'm having a problem using the following shuffling algorithm I
found searching the Internet and newsgroups:
for J := Max downto 2 do Swap(A[J], A[Succ(Random(J))]) ;
|
How about:
for J := Max-1 downto 1 do Swap(A[J], A[Random(J)]) ;
--
Marc Rohloff [TeamB]
marc -at- marc rohloff -dot- com |
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John Herbster
Guest
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| Posted: Thu Feb 15, 2007 8:55 am Post subject: Re: Shuffling zero-based arrays |
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"Marc Rohloff [TeamB]" wrote
| Quote: | for J := Max-1 downto 1 do Swap(A[J], A[Random(J)]) ;
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What does the last step "Swap(A[0], A[Random(0)])" do?
What does the step "Swap(A[1], A[Random(1)])" do?
<g>
Maybe we should start thinking backwards::
Swap(A[1],A[random(2)]);
looks all right, so let's proceed that with
Swap(A[2],A[random(3)]);
and so on to
Swap(A[NbrElem-1],A[random(NbrElem)]);
Now then, we can put this all in a loop like this
For j := NbrElem downto 2 do Swap(A[j-1],A[random(j)]);
HTH, JohnH |
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