notrand()

You are given a black-box function, notrand(), which returns 0 with 60% probability and 1 with 40% probability. Create a new function, rand(), which returns 0 and 1 with equal probability (i.e. 50% each) using only notrand() as your source of randomness.

I've heard many interesting approaches to the problem. I'll post my own solution later after hearing some from the community!

#Percentage #ComputerScience

Note by Brian Rodriguez
6 years, 10 months ago

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Comments

Here is a possible function for rand():

  1. Run notrand() twice.
  2. If the outputs are 0 and 1, return 0.
  3. If the outputs are 1 and 0, return 1.
  4. Otherwise, go back to step 1.

Jon Haussmann - 6 years, 10 months ago

let a0, a1 use notrand() to assign a value. so a0 = {0: 60%, 1: 40%}, a1 = {0: 60%, 1: 40%}.

Consider a0 + a1:

0(60%) + 0(60%) = 0b00

0(60%) + 1(40%) = 0b01

1(40%) + 0(60%) = 0b01

1(40%) + 1(40%) = 0b10

Note the least significant bit (which can be found with the xor of a0 and a1) has a different weight than the notrand() function, namely:

a0 ⊕ a1 = {0: 52%, 1: 48%}

Now if we combine two of these results (4 calls to notrand() in total) the percentages change again to:

0(52%) ⊕ 0(52%) = 0

0(52%) ⊕ 1(48%) = 1

1(48%) ⊕ 0(52%) = 1

1(48%) ⊕ 1(48%) = 0

or {0: 50.08%, 1: 49.92%}

This pattern will approach 50/50 as the sub iterations get larger, however, the associative property of xor gives one more detail:

((a ⊕ b) ⊕ (c ⊕ d)) ⊕ ((i ⊕ j) ⊕ (k ⊕ l)) = (a ⊕ b ⊕ c ⊕ d ⊕ i ⊕ j ⊕ k ⊕ l)

and so xor'ing multiple calls of notrand() actually brings the function closer to 50/50 regardless of the number of calls used and the order they are combined in.

Brian Rodriguez - 6 years, 9 months ago
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