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Suppose that $Z = X \vee Y$, where $X$, $Y$ and $Z$ are 96-bit binary numbers. If I'm given the values of $Z$ and $Y$, is it possible to work out what $X$ is?

I know this is possible with XOR but can it be done with OR?

David Richerby
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SHdotCom
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  • For each 0 bit in Z you know that corresponding bit in X and Y have to be 0. For each 1 bit in Z, X and Y can be any of (1,1) (0,1) (1,0). Now if Y bit is 0 then you can infer X bit as 1 BUT if the Y bit 1 then X bit can be 1 or 0 and there is no way to figure out which one it is. – Ankur Oct 19 '15 at 12:52

1 Answers1

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It is easy to find out that there can be more than one value, used as $X$, to satisfy $Z = X \vee Y$. When a specific bit of $Y$ is $1$, there are two possibilities for such bit in $X$, i.e., $0$ or $1$.

Let's make a simple example with a 2-bit number:

$Y$ = $10$ and $Z$ = $11$

The possible values of $X$ are:

  • $11$
  • $01$

because:

  • $11 \vee 10 = 11$
  • $01 \vee 10 = 11$

In short, you don't have the certainty that the end result of the reverse operation of $\vee$ will be a unique result.

Gentian Kasa
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beginner1010
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  • X, Y, Z are 96 bits binary numbers. EX: Z = 001100010001010000100101011110111111010101110100010110000101011000101010000011100001111000000000 – SHdotCom Oct 19 '15 at 04:08
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    It does not matter how many is the number of bits of $Z$. You can save 96-bit numbers in two 64-bit integers: long long in C++, and Long in Java, or you can preserve it in a string data type. Then, let us assume that $Z = 1111$ and $Y = 1111$. $X$ could be from $0000$ to $1111$; therefore there is no a single value for $X$ to obtain. – beginner1010 Oct 19 '15 at 04:49
  • @beginner1010's point is that the problem is the same whether you are talking about 1 bit or 96 bits, as the bits in two 96 bit number are treated independently by $\lor$. – Dave Clarke Oct 19 '15 at 08:37
  • @Dave Clarke if I have Z = 01111100110110101001011111111101111111111111101101101101011110111011110100000011‌​1110011010101110

    Y= 01101100010100101001010111101001110101011101000101100001010110001010100000000011‌​1100000000001010

    Is it possible to get X value which is equal to : 01011000100010100001001010111101111110101011101000101100001010110001010100000000‌​0110011010100100

    From Z and Y ??

     – SHdotCom Oct 19 '15 at 08:45
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    @SHdotCom: First, think about whether you can do this for a 1-bit number. It's the same problem for a 96-bit number, just 96 times. – Dave Clarke Oct 19 '15 at 08:56
  • So, sometimes you can be 100% sure, but only in the following scenario: Z = 1 and Y = 0 -> X = 1. Generally on such large numbers, you only narrow down the possible options. But it's still better than 2^96. – ryuu9187 Oct 19 '15 at 13:09
  • Dave Clarke, with 1-bit you have only two possibilities to get X 0 or 1 , but with 96bits How many Combinations of (1 and 0) you can make to get one value equal to X?? – SHdotCom Oct 20 '15 at 04:44
  • @SHdotCom: The number of possibilities is equal to $2^{Y_1}$, where $Y_1$ is the number of ones in binary number $Y$. – beginner1010 Oct 20 '15 at 07:07