Math related question

[milan1612]

I came across a job interview question some time ago that got me thinking.

Given a function that returns a random integer within the range between 1 and 5,
write a function that uses this random() function and returns a random integer
between 1 and 7.

So something like this:

Code: Select all

Procedure Random5()
  ProcedureReturn Random(4) + 1
EndProcedure

Procedure Random7()
  ; magic happens here, using only the above function
EndProcedure

I have absolutely no idea how this could be done, do you?

[Matt]

There are tons of ways:

http://stackoverflow.com/questions/1377 ... rite-a-fun

[srod]

Here's one I just knocked up.

Code: Select all

Procedure Random5()
  ProcedureReturn Random(4) + 1
EndProcedure

Procedure Random7()
  Repeat
    row = Random5()
    col = Random5()
    total = (row-1)*5 +col - 1
  Until total <= 20
  ProcedureReturn total / 3 + 1
EndProcedure

For i = 1 To 100
  Debug Random7()
Next

All I do is use two random numbers between 1 and 5 to produce a random number between 0 and 24 (the value of total in the above code). Repeat until you actually get something between 0 and 20 (which gives 21 numbers; a multiple of 7). Then, if the resulting value (total) is between 0 and 2 we return a result of 1. If the value is between 3 and 5 then we return a result of 2 and so on. This gives us a result between 1 and 7; each of which is equally likely.

[citystate]

Code: Select all

Procedure.l Random7()
  result.l = 0
  For i = 1 To 7
    result + Random5()
  Next i
  ProcedureReturn result/5
EndProcedure

[STARGÅTE]

sry citystate thats no right random:

Code: Select all

Procedure Random5()
  ProcedureReturn Random(4) + 1
EndProcedure

Procedure.l Random7()
  result.l = 0
  For i = 1 To 7
    result + Random5()
  Next i
  ProcedureReturn result/5
EndProcedure 

Dim Test(7)

For n = 1 To 70000
 i = Random7()
 Test(i) + 1
Next

For n = 1 To 7
 Debug Test(n)
Next

[idle]

What gets me is that they still ask these illogical questions at interviews!

[edit] 1 to 7 [/edit]

Code: Select all

Procedure Random5()
  ProcedureReturn Random(4) + 1
EndProcedure

Procedure Random7()
  ;common multiple 5*7 
  For a = 1 To 7
    b + Random5()
  Next 
  b % 7 + 1 
  ProcedureReturn b 
EndProcedure  

Dim ct(7)

For a = 1 To 10000 
  x = random7()
  ct(x)+1
Next   

For a = 1 To 7 
 Debug StrF((ct(a)/10000)*100.0,3) 
Next 
;Even odds 14.28% 

[citystate]

in that case STARGÅTE,

Code: Select all

Procedure random7()
  result.l = 0
  ;generate a number between 0 and 20 ( 5 x [0 to 4] )
  For i = 1 To 5
    result + (random5() - 1)
  Next i
  ProcedureReturn result/3 + 1
EndProcedure

[idle]

no that don't work either the odds are all skewed
I fixed your solution with a mod

[Michael Vogel]

Why you're not doing a sum using seven calls of Random(5) each time?

Code: Select all

Procedure Random7()
	;common multiple 5*7
	For a = 0 To 6
		b + Random5()
	Next
	b%7+1
	ProcedureReturn b
EndProcedure

And, if the random numbers don't have to be more than (very ) pseudo numbers, what's about something like this:

Code: Select all

Procedure Random7()
	ProcedureReturn (Random5()+3)%7+1
EndProcedure

PS edited, bold tags () in code do not work...

[citystate]

the question didn't ask for an even probability spread, just a function that would produce a random number between 1 and 7

[idle]

Why you're not doing a sum using seven calls of Random(5) each time?

Code: Select all

Procedure Random7()
	;common multiple 5*7
	For a = 0 To 6
		b + Random5()
	Next
	b%7+1
	ProcedureReturn b
EndProcedure

And, if the random numbers don't have to be more than (very ) pseudo numbers, what's about something like this:

Code: Select all

Procedure Random7()
	ProcedureReturn (Random5()+3)%7+1
EndProcedure

PS edited, bold tags () in code do not work...

That doesn't work!

The common multiple is 35 so you need to do random5() seven times
and then mod it

A random distribution is supposed to be a normal distribution
all city states function needed was a mod to make it work properly

[Michael Vogel]

That doesn't work!
The common multiple is 35 so you need to do random5() seven times
and then mod it

Hi, I've really thought this are seven times...

Code: Select all

Procedure Random5()
	ProcedureReturn Random(4)+1
EndProcedure

Procedure Random7()
	;common multiple 5*7
	For a = 0 To 6
		Debug Str(a+1)+" time"+Mid("s",1,a)
		b + Random5()
	Next
	b%7+1
	ProcedureReturn b
EndProcedure
Debug Random7()

A random distribution is supposed to be a normal distribution

Yes, of course (should have added some more above) - otherwise all these things would be correct:

Code: Select all

Procedure Random7()
	CompilerSelect #RandomType
	CompilerCase #AbsolutePseudo
		ProcedureReturn 2
	CompilerCase #Simple
		ProcedureReturn Random5()
	CompilerCase #SimpleAndCentered
		ProcedureReturn Random5()+1
	CompilerCase #Crazy
		ProcedureReturn (ElapsedMilliseconds()*Random5())%7+1
	CompilerEndSelect
EndProcedure

[idle]

sorry I see what you mean now!
I had 0 to 7 while I was meaning 1 to 7, though 0 to 6 would suffice
well I did at least kind of explain the reason.

[Michael Vogel]

sorry I see what you mean now!
I had 0 to 7 while I was meaning 1 to 7, though 0 to 6 would suffice
well I did at least kind of explain the reason.

You're right, I just stole your explanation (a good strategy for normal) to get the score

Michael

[idle]

it wasn't my strategy I would have opted for crazy