Pernicious numbers: Difference between revisions
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3 5 6 7 9 10 11 12 13 14 17 18 19 20 21 22 24 25 26 28 31 33 34 35 36
888888877 888888878 888888880 888888883 888888885 888888886
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=={{header|Action!}}==
Action! integers are limited to 16 bits, so this implements 32 bit addition and multiplication by 8-bit values to handle the larger numbers.
<syntaxhighlight lang="action!">
;;; find some pernicious numbers - numbers where the population count is prime
;;; As the task requires 32 bit integers, this implements 32-bit unsigend
;;; integer addition and multiplication by an 8-bit integer.
;;; The 32-bit values are stored in 4 separate bytes
;;; returns the population (number of bits on) of the non-negative integer n
BYTE FUNC population( CARD n )
CARD number
BYTE result
number = n
result = 0;
WHILE number > 0 DO
IF number AND 1 THEN result ==+ 1 FI
number ==/ 2
OD
RETURN( result )
;;; returns TRUE if n is a prime; n must be <= 32
BYTE FUNC isSmallPrime( BYTE n )
BYTE result
IF n = 2 THEN result = 1
ELSEIF ( n AND 1 ) = 0 THEN result = 0
ELSEIF n = 1 OR n = 9 OR n = 15
OR n = 21 OR n = 25 OR n = 27
THEN result = 0
ELSE result = 1
FI
RETURN( result )
;;; returns TRUE if n is pernicious, FALSE otherwise
BYTE FUNC isPernicious( CARD n ) RETURN( isSmallPrime( population( n ) ) )
;;; returns TRUE if the 32 bit integer in i1, i2, i3, i4 is pernicious,
;;; FALSE otherwise
BYTE FUNC isPernicious32( BYTE i1, i2, i3, i4 )
BYTE p
p = population( i1 ) + population( i2 )
+ population( i3 ) + population( i4 )
RETURN( isSmallPrime( p ) )
;;; adds b to the 32 bit unsigned integer in i1, i2, i3 and i4
PROC i32add8( BYTE POINTER i1, i2, i3, i4, BYTE b )
CARD c1, c2, c3, c4
c1 = i1^ c2 = i2^ c3 = i3^ c4 = i4^
c4 ==+ b
i4 ^= c4 MOD 256
c3 ==+ c4 / 256
i3 ^= c3 MOD 256
c2 ==+ c3 / 256
i2 ^= c2 MOD 256
c1 ==+ c2 / 256
i1 ^= c1 MOD 256
RETURN
;;; multiplies the 32 bit unsigned integer in i1, i2, i3 and i4 by b
PROC i32mul8( BYTE POINTER i1, i2, i3, i4, BYTE b )
CARD c1, c2, c3, c4, r
c1 = i1^ c2 = i2^ c3 = i3^ c4 = i4^
r = c4 * b
i4 ^= r MOD 256
r = ( c3 * b ) + ( r / 256 )
i3 ^= r MOD 256
r = ( c2 * b ) + ( r / 256 )
i2 ^= r MOD 256
r = ( c1 * b ) + ( r / 256 )
i1 ^= r MOD 256
RETURN
;;; find the first 25 pernicious numbers
PROC Main()
BYTE perniciousCount, i
BYTE i81, i82, i83, i84
BYTE p81, p82, p83, p84
perniciousCount = 0
i = 0
WHILE perniciousCount < 25 DO
IF isPernicious( i ) THEN
; found a pernicious number
PrintB( i )Put(' )
perniciousCount ==+ 1
FI
i ==+ 1
OD
PutE()
; find the pernicious numbers between 888 888 877 and 888 888 888
; form 888 888 800 in i81, i82, i83 and i84
i81 = 0 i82 = 0 i83 = 0 i84 = 88 ; 88
i32mul8( @i81, @i82, @i83, @i84, 100 ) ; 8 800
i32add8( @i81, @i82, @i83, @i84, 88 ) ; 8 888
i32mul8( @i81, @i82, @i83, @i84, 100 ) ; 888 800
i32add8( @i81, @i82, @i83, @i84, 88 ) ; 888 888
i32mul8( @i81, @i82, @i83, @i84, 10 ) ; 8 888 880
i32add8( @i81, @i82, @i83, @i84, 8 ) ; 8 888 888
i32mul8( @i81, @i82, @i83, @i84, 100 ) ; 888 888 800
FOR i = 77 TO 88 DO
p81 = i81 p82 = i82 p83 = i83 p84 = i84
i32add8( @p81, @p82, @p83, @p84, i )
IF isPernicious32( p81, p82, p83, p84 )
THEN
print( "8888888" )PrintB( i )Put(' )
FI
OD
PutE()
RETURN
</syntaxhighlight>
{{out}}
<pre>
3 5 6 7 9 10 11 12 13 14 17 18 19 20 21 22 24 25 26 28 31 33 34 35 36
888888877 888888878 888888880 888888883 888888885 888888886
</pre>
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