Pseudo-random numbers/Middle-square method: Difference between revisions
Pseudo-random numbers/Middle-square method (view source)
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Line 210:
{{output}}
<syntaxhighlight lang="applescript">{959861, 333139, 981593, 524817, 432883}</syntaxhighlight>
=={{header|Amazing Hopper}}==
Flavour BASICO.
<p>Version 1:</p>
<syntaxhighlight lang="c">
#include <basico.h>
algoritmo
s=675248, i=4
iterar
#( s = int(s ^2 / 1000 % 1000000) )
s, NL, imprimir
mientras ' i-- '
terminar
</syntaxhighlight>
{{out}}
<pre>
959861
333139
981593
524817
432883
</pre>
<p>Version 2:</p>
<syntaxhighlight lang="c">
#include <basico.h>
algoritmo
s=675248, i=5
iterar grupo( --i, i, #( s = int(s ^2 / 1000 % 1000000) ) \
s, NL, imprimir )
terminar
</syntaxhighlight>
{{out}}
<pre>
idem.
</pre>
<p>Version 3:</p>
<syntaxhighlight lang="c">
#include <basico.h>
algoritmo
s=675248, i=5
iterar grupo( --i, i, #( int(s ^2 / 1000 % 1000000) ) \
---copiar en 's'---, NL, imprimir )
terminar
</syntaxhighlight>
{{out}}
<pre>
idem.
</pre>
<p>Version 4:</p>
<syntaxhighlight lang="c">
#include <basico.h>
algoritmo
s=675248, i=4
rareti( i, #( int(s ^2 / 1000 % 1000000) ) \
---copiar en 's'---, NL ); entonces imprime
terminar
</syntaxhighlight>
{{out}}
<pre>
idem.
</pre>
=={{header|ARM Assembly}}==
Line 412 ⟶ 477:
4: 524817
5: 432883</pre>
==={{header|Chipmunk Basic}}===
{{works with|Chipmunk Basic|3.6.4}}
<syntaxhighlight lang="qbasic">10 semilla = 675248
20 for i = 1 to 5
30 print i;": ";semilla
31 semilla = ((semilla*semilla)/1000) mod 1000000
40 next i</syntaxhighlight>
==={{header|FreeBASIC}}===
Line 530 ⟶ 603:
524817
432883</pre>
=={{header|BQN}}==
<syntaxhighlight lang="bqn">1e6 |⟜(⌊∘÷⟜1e3 ט)⍟(1+↕5) 675248</syntaxhighlight>
Or:
<syntaxhighlight lang="bqn">675248 (1e6 | ט⊸(⌊÷))` 5 ⥊ 1e3</syntaxhighlight>
{{out}}
<pre>⟨ 959861 333139 981593 524817 432883 ⟩</pre>
=={{header|C}}==
Line 651 ⟶ 731:
432883
</pre>
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
<syntaxhighlight lang="Delphi">
var Seed: int64 = 675248;
function MiddleSquareRandom: int64;
var S: string;
begin
S:=IntToStr(Seed * Seed);
while Length(S)<12 do S:='0'+S;
Seed:=StrToInt(MidStr(S, 4, 6));
Result:=Seed;
end;
procedure ShowMiddleSqrRandom(Memo: TMemo);
var I: integer;
begin
for I:=1 to 5 do
Memo.Lines.Add(IntToStr(MiddleSquareRandom));
end;
</syntaxhighlight>
{{out}}
<pre>
959861
333139
981593
524817
432883
Elapsed Time: 5.610 ms.
</pre>
=={{header|dc}}==
Line 667 ⟶ 785:
global seed .
seed = 675248
func rand
strSeed$ = seed
s$ = seed * seed
Line 675 ⟶ 793:
seed = number substr s$ (len strSeed$ / 2 + 1) len strSeed$
randNum = seed
return randNum
.
for i = 1 to 5
.
</syntaxhighlight>
Line 688 ⟶ 806:
524817
432883
</pre>
=={{header|EDSAC order code}}==
EDSAC (including its name) was inspired by John von Neumann's "First Draft of a Report on the EDVAC", so a task in his honour ought to have an EDSAC solution. As noted in the Discussion, the sequence in the task description falls into a repetition of 625000.
<syntaxhighlight lang="edsac">
[Von Neumann's middle-square pseudo-random number generator, for Rosetta Code.
EDSAC program, Initial Orders 2.]
[Arrange the storage]
T46K P56F [N parameter: library subroutine P7 to print integer]
T47K P134F [M parameter: main routine]
T51K P92F [G parameter: generator for pseudo-random numbers]
[This version of von Neumann's PRNG uses values in the range 0..999999.
Initialize: Call 0G with seed in 0D.
Next term: Call 1G; term is returned in 0D.
41 storage locations, load at even address. Workspace 4D.]
E25K TG GK
G10@ [jump to initialize the generator]
G15@ [jump to return the next term in 0D]
[Instructions to the loader - not executed at runtime]
T2#Z PF [ensure sandwich bit between 2@ and 3@ is zero]
T4#Z PF T6#Z PF [same for 4@ and 5@, 6@ and 7@]
T2Z [resume normal loading at 2@]
[Constants]
[2] M1667D I1208F [2^29/10^9, near enough (see note at end)]
[4] G1327D I393F [2^9/10^3, near enough (see note at end)]
[10^9/2^34 can't be stored using pseudo-orders, so store its negative instead.]
[6] D768F V140D [-10^9/2^34]
[Variable]
[8] PF PF [state of PRNG]
[Initialize the PRNG. Caller passes seed in 0D.]
[10] A3F T14@ [plant return link as usual]
AD T8#@ [copy seed to state]
[14] ZF [(planted) jump back to caller]
[Return the next value in 0D.
Outline: Let X = state, 0 <= X <= 999999, fits into 20 bits.
Calculate Y = X^2 div 10^9.
Deduce Z = X^2 mod 10^9 = X^2 - (10^9)*Y
The next state is Z div 10^3.
See note at end of program for details of integer division.]
[15] A3F T40@ [plant return link as usual]
A8#@ [acc := X/2^34]
L32F L32F [shift 7 + 7 left, acc := X/2^20]
T4D [store X/2^20 in 4D]
H4D V4D [square, acc := (X^2)/2^40]
[Here acc holds sign bit plus 40 binary places of (X^2)/2^40.]
[On storing acc in a 35-bit location, the low 6 bits are lost.]
[This doesn't matter, because 2^6 divides 10^9, so we can evaluate]
[X^2 div 10^9 = (X^2 div 2^6) div (10^9/2^6).]
TD [0D now represents X^2 div 2^6]
H2#@ [mult reg := 2^29/10^9 nearly]
VD
[Commented out: code for 35-bit operations, following note at end.
|R 1024 F| |R 512 F| shift 23 right, as in note
|T D| 0D := Y/2^34 where Y = X^2 div 10^9
|H neg_10_9 #@| |V D| times -10^9/2^34, acc := -10^9*Y/2^68
We need to shift 34 left to restore the scaling after multiplication
|L 1024 F| |L 1024 F| |L 4 F| first shift 28 left, acc := 10^9*Y/2^40]
[More efficient code, possible because Y (= X^2 div 10^9) fits into a 17-bit location.
Shifting is 18 less than the 35-bit version.]
R8F [shift 5 right]
TF [0F := Y/2^16]
H6#@ VF [times -10^9/2^34, acc := -10^9*Y/2^50]
[We need to shift 16 left to restore the scaling after multiplication]
L256F [first shift 10 left, acc := 10^9*Y/2^40]
H4D V4D [4D = X/2^20 from above, so acc := (X^2 - 10^9*Y)/2^40]
L16F TD [shift 6 more left; 0D := Z/2^34 where Z = X^2 - 10^9*Y]
H4#@ [mult reg := 2^9/10^3 nearly]
VD
R128F [shift 9 right]
U8#@ [save next state u]
TD [also return next state to caller in 0D]
[40] ZF [(planted) jump back to caller]
[------------------------------------------------------------------------]
E25K TM GK [M parameter, main routine. Load at even address.]
T#Z PF [clear 35-bit value at relative locations
0 & 1, including the middle ("sandwich") bit]
TZ [resume normal loading at relative location 0]
[We could read the seed and number of terms from a separate tape,
but that would require another subroutine.]
[0] G296F V2046D [-675248 (negative of seed, cf -10^9 above)]
[2] P211F [number of terms, in the address field]
[3] PF [index of term]
[4] #F [teleprinter, set figures mode]
[5] !F [space]
[6] &F [line feed]
[7] @F [carriage return]
[8] K4096F [null]
[9] PF
[Enter with acc = 0]
[10] O4@ [set teleprinter to figures]
S#@ TD [pass seed in 0D]
[13] A13@ GG [call subroutine to initilaize PRNG]
T3@ [index := 0]
[Head of loop; here with acc = 0]
[16] TD [clear parameter for print subroutine]
A3@ S2@ [printed enough terms?]
E34@ [if so, jump to exit]
A2@ [restore acc after test]
A2F U3@ [update index]
RD TF [right-justify for printing; pass in 0D]
[25] A25@ GN [call subroutine to print index]
[27] A27@ G1G [call PRNG; returns next term in 0D]
[29] A29@ GN [print term; clears acc.]
O7@ O6@ [print CR, LF]
E16@ [loop back for next term]
[34] O8@ [print null to flush printer buffer]
ZF [stop]
[----------------------------------------------------------------------]
E25K TN [N parameter, print subroutine]
[Library subroutine P7, prints long strictly positive integer in 0D.]
[10 characters, right justified, padded left with spaces.]
[Even address; 35 storage locations; working position 4D.]
GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSF
L4FT4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@
[----------------------------------------------------------------------]
E25K TM GK [M parameter again]
E10Z [start execution at relative address]
PF
[
Note: Integer division by a constant on EDSAC.
Some programs require the integer quotient N div D, where N is variable and
D is constant throughout the program. Since EDSAC had hardware multiplication
but not hardware division, it makes sense to store 1/D and multiply N by that.
This note describes one way of doing this, and finds a sufficient condition on N
for rounding errors not to affect the result. It's assumed that 35-bit integers are used,
so that an integer N is represented by N*(2^-34); also that N and D are positive.
Let k be the greatest integer such that 2^k < D. We store a multiplier 2^k/D,
rounded up to an integer multiple of 2^-34. Let the stored multiplier be
2^k/D + f*(2^-34), where 0 <= f < 1.
Let N = q*D + r, where q is an integer and 0 <= r < D. To find the integer quotient q,
multiply N*(2^-34) by the stored multiplier and shift right by k bits. This gives
(2^-34)*(q + r/D + N*f*(2^(-k -34)))
in the accumulator. If the top 35 bits of the accumulator are stored in memory,
the value stored will be q*(2^-34), provided
r/D + N*f*(2^(-k-34)) < 1.
Since r <= D - 1, a sufficient condition for this is
N*f*(2^(-k-34)) < 1/D
which is equivalent to
N < (2^k/D)*(2^34/f). (*)
Examples from the middle-square program:
(1) D = 10^9/2^6. Then k = 23 and 2^k/D = 0.536870912. Also
2^(k+34)/D = 2^63/10^9 = 9223372036.854775808
so that the stored multiplier is 9223372037*(2^-34) and f = 0.145224192.
Since f < 2^k/D, the condition (*) is satisfied for all N < 2^34.
(2) D = 10^3. Then k = 9 and 2^k/D = 0.512. Also
2^(k+34)/D = 2^43/10^3 = 8796093022.208
so that the stored multiplier is 8796093023*(2^-34) and f = 0.792.
The condition (*) becomes N < 1.11*(10^10) (approx), which is satisfied
in this program because N < 10^9. ]
</syntaxhighlight>
{{out}}
<pre>
1 959861
2 333139
3 981593
4 524817
5 432883
[...]
207 437510
208 415000
209 225000
210 625000
211 625000
</pre>
Line 773 ⟶ 1,061:
=={{header|J}}==
<syntaxhighlight lang="j">(_6 {. _3 }. ])&.
Or, shorter & faster:
<syntaxhighlight lang="j">}. (1e6 1e3 {.@#: *:)^:(< 7) 675248</syntaxhighlight>
{{out}}
<pre>959861 333139 981593 524817 432883 387691</pre>
=={{header|Java}}==
<syntaxhighlight lang="java">
public final class MiddleSquareTask {
public static void main(String[] aArgs) {
MiddleSquare random = new MiddleSquare(675248);
for ( int i = 0; i < 5; i++ ) {
System.out.println(random.nextInt());
}
}
}
final class MiddleSquare {
public MiddleSquare(int aSeed) {
final int length = String.valueOf(aSeed).length();
if ( length % 2 == 1 ) {
throw new IllegalArgumentException("Seed must have an even number of digits");
}
state = aSeed;
divisor = (int) Math.pow(10, length / 2);
modulus = (int) Math.pow(10, length);
}
public int nextInt() {
state = ( ( state * state ) / divisor ) % modulus;
return (int) state;
}
private long state;
private final int divisor, modulus;
}
</syntaxhighlight>
{{ out }}
<pre>
959861
333139
981593
524817
432883
</pre>
=={{header|jq}}==
Line 825 ⟶ 1,163:
{{trans|C}}
<syntaxhighlight lang="Scheme">
{def W.fill
{lambda {:v :n}
{if {<= :n 0}
then
else :v{W.fill :v {- :n 1}}}}}
-> W.fill
{def W.pad
{lambda {:n :size}
{if {<= {W.length :n} :size}
then :n{W.fill :size {- :size {W.length :n}}}
else {W.slice 0 {- :size {W.length :n}} :n}}}}
-> W.pad
{def randoms
{lambda {:s :n}
Line 834 ⟶ 1,186:
-> randoms
{randoms
-> 959861 333139 981593 524817 432883
</syntaxhighlight>
Line 1,134 ⟶ 1,486:
=={{header|Wren}}==
<syntaxhighlight lang="
var seed = 675248
| |||