Is there an efficient 2 out of 3 secret sharing scheme based on XOR? I want to know if there is an efficient secret sharing scheme based on XOR such that a share of the message is given to three people and at least two must combine their share in order to construct the message?

What I have tried so far:

Suppose I got a message $M$, and three people A, B and C. Then what I do is


A gets
$a_1 = M \oplus \alpha$ and $a_2 = \beta$
B gets
$b_1 = M \oplus \beta$ and $b_2 = \alpha$
C gets
$c_1 = \alpha$ and $c_2 = \beta$


Where $\alpha$ and $\beta$ are two randomly generated bits the size of the message.

This works fine (with some modification), but the problem is that each person have to store twice the size of the message. Is there a more efficient one?
Here is a simple (2, 3) secret sharing scheme that used only xor's, and each share is the same size as the secret (assuming that the secret is an even number of bits long)

First, we divide the secret $s$ into two equal sized halves, $s_1, s_2$.

And, we pick two random values of the same size, $r_1, r_2$

Then, the Alice shares are $a_1 = s_1 \oplus r_1$, $a_2 = s_2 \oplus r_2$

The Bob shares are $b_1 = s_1 \oplus r_2$, $b_2 = s_2 \oplus r_1 \oplus r_2$

And the Carol shares are $c_1 = s_1 \oplus r_1 \oplus r_2$, $c_2 = s_2 \oplus r_1$.

It is straight-forward to show that any two parties can recover the secrets $s_1, s_2$

With the Alice and Bob shares, we have $s_1 = a_1 \oplus a_2 \oplus b_2$, $s_2 = a_1 \oplus b_1 \oplus b_2$

With the Alice and Carol shares, we have $s_1 = a_2 \oplus c_1 \oplus c_2$, $s_2 = a_1 \oplus a_2 \oplus c_1$

With the Bob and Carl shares, we have $s_1 = b_1 \oplus b_2 \oplus c_2$, $s_2 = b_1 \oplus c_1 \oplus c_2$

And, with a single share, neither Alice nor Bob nor Carol gain any information about the value of the secret (assuming $r_1, r_2$ was picked randomly).

And, if the total length of the secret is n bits (and so each half of the secret is $\lceil n/2 \rceil$ bits long), the total length of each share is $2 \lceil n/2 \rceil$ bits, which is $n$ bits if $n$ is even.