0
I am trying to understand the algorithm of bitcoin.
Let A be a node, and let (...,T_0,T_1,T_2) represents a block chain that A has at some time.
Here, T_i designates a block which was created at the time T_i, meaning that the block T_n was created before the block T_{n+1} was created.
Now, say miners C and D created two different blocks T_3 and T_3’ within 10 minutes. Then, A appends these two blocks to the current block chain and store them separately. Following the above notation, these two chains are (...,T_0,T_1,T_2,T_3) and (...,T_0,T_1,T_2,T_3’).
Now, say E sent F some bitcoins and made a transaction which is included in the block T_3.
After some time, if the blockchain (...,T_0,T_1,T_2,T_3’) is chosen to be the correct one, the chain (...,T_0,T_1,T_2,T_3) is removed from server. (Actually, at this point, these two separate blockchains do not appear in this way. That is, they would appear like (...,T_0,T_1,T_2,T_3’,T_4’,...,T_n’) and (...,T_0,T_1,T_2,T_3,T_4,...,T_n) because as time passes new blocks are added.
Assume that the transaction that E sent F some bitcoins was not included in any other blocks except the block T_3. (This is possible if E sent F very small amount of money, so if this transaction costs really tiny small fee, miners would not choose this transaction to be included in their block).
In this case, if E wants to send some bitcoins to F, does he have to make a transaction all over again?
So theoretically, is it possible that some transactions will never be included in a “winning” blockchain? And how many transancations each node can save in mempoola? Is there no limit? – Rubertos – 2018-01-12T22:30:45.847
It is theoretically possible for a transaction to remain unconfirmed forever, regardless of blockchain reorgs. The mempool configurations of each node may be different. The defaults for Bitcoin Core nodes is 300 MB of transactions and transactions will be evicted if they remain in the mempool for 2 weeks. – Andrew Chow – 2018-01-12T22:34:59.003
Oh.. so this is a disadvantage of bitcoin algorithm.. thank you very much! – Rubertos – 2018-01-12T22:40:04.927