Editing Secret sharing (section)

=== ''t'' = ''n'' ===
There are several {{nowrap|(''t'', ''n'')}} secret-sharing schemes for {{nowrap|1=''t'' = ''n''}}, when all shares are necessary to recover the secret:

# Encode the secret as a [[Binary numeral system|binary]] number ''s'' of any length. For each player ''i'', where ''i'' is one fewer than the total number of players, give a random binary number ''p<sub>i</sub>'' of the same length as ''s''. To the player without a share, give the share calculated as {{nowrap|1=''p''<sub>''n''</sub> = ''s'' ⊕ ''p''<sub>1</sub> ⊕ ''p''<sub>2</sub> ⊕ ... ⊕ ''p''<sub>''n''−1</sub>}}, where ⊕ denotes [[bitwise operation#XOR|bitwise exclusive or]]. The secret is the bitwise exclusive-or of all the players' numbers (''p''<sub>''i''</sub>, for 1 ≤ ''i'' ≤ ''n'').
# Instead, (1) can be performed using the binary operation in any [[Group (mathematics)|group]].  For example, take the cyclic group of integers with addition modulo 2<sup>32</sup>, which corresponds to 32-bit integers with addition defined with the binary overflow being discarded.  The secret ''s'' can be partitioned into a vector of ''M'' 32-bit integers, which we call ''v''<sub>secret</sub>.  Then {{nowrap|(''n'' − 1)}} of the players are each given a vector of ''M'' 32-bit integers that is drawn independently from a uniform probability distribution, with player ''i'' receiving ''v<sub>i</sub>''.  The remaining player is given ''v<sub>n</sub>'' = ''v''<sub>secret</sub> − ''v''<sub>1</sub> − ''v''<sub>2</sub> − ... − ''v''<sub>''n''−1</sub>.  The secret vector can then be recovered by summing across all the players' vectors.