Self adapting - try to break it yourself with a different passcode! (i have left the alt-tab etc. keys in for now)

Self adapting to methods of password breaking + randomized effects. try it with a different password and see how far you get!

                                    ` This code was downloaded from The Game Creators
                                    ` It is reproduced here with full permission
                                    ` http://www.thegamecreators.com
                                    
                                    rem ****************************
rem By Mark Boyall 22/1/05
rem Version 5 
rem ****************************

a = 1024

do
print "enter password"
input b
if b = a
print "enter password"
else 
b = c
a = a^9999999^b
endif

z = rnd(4)

if z = 1
a = a^b^c^d^x^y^9
endif
if z = 2
a = a*99999999
endif
if z = 3
a = c
endif
if z = 4
a = ab^99999 
endif

print "enter password"
input b
if b = a
print "enter password"
else 
b = d
a = a^9999999^b
endif

z = rnd(4)

if z = 1
a = a^b^c^d^x^y^9
endif
if z = 2
a = a*99999999
endif
if z = 3
a = c
endif
if z = 4
a = ab^99999 
endif

print "enter password"
input b
if b = a
print "enter password"
else 
b = e
a = a^9999999^b
endif

z = rnd(4)

if z = 1
a = a^b^c^d^x^y^9
endif
if z = 2
a = a*99999999
endif
if z = 3
a = c
endif
if z = 4
a = ab^99999 
endif

print "enter password"
input b
if b = a
goto label
else 
a = a^9999999^b
endif

x = d - c 

if b = c + x + x + x 
if b = d + x + x
if b = e + x 
a = a^b^c^d
endif
endif
endif

y = d/c

if b = c*y*y*y
if b = d*y*y
if b = e*y
a = a^b^c^d
endif
endif
endif

z = rnd(4)

if z = 1
a = a^b^c^d^x^y^9
endif
if z = 2
a = a*99999999
endif
if z = 3
a = c
endif
if z = 4
a = ab^99999 
endif

loop

label:
end