There is an implementation difficulty in the SPOW picture. You say that large mining pools shouldn't reuse nonces of other large mining pools. But how are you going to strictly define "large" to something the protocol understands? How will you differentiate between mining pools, and miners (at the protocol level!) who are possibly using the same mining software?
Remember that this algorithm has to make it just as fair for tiny miners to find a block as it is for a miner with say 90% of the global hashrate. Since there are two hashes to be found and a miner has to pick one of them to find, it only takes two different miners like two S19's to find both hashes, since there is no protocol-level distance between mining pools and solo miners.
First of all thank you for your comment
At the protocol level, large pools will follow the process of finding a hash to determine which nonce range will run, for example if the nonce is 500 and there is a pool among 400 600, that nonce will be found at normal speed. But if there were more miner pools operating in the 400 600 range, that means more power and that nonce would be found faster. In other words, pools can find the nonce range they will work in this way. Another way is for miner pools to clearly state this range, if the miner pools explicitly say this information will not lose anything, even more will get more because no more competitors will come in this nonce range, and new pools that know this information will choose another nonce range that suits them. This actually looks like this; There is a mine field in the middle and everyone is looking for the mine by buying some area from this field. The benefit of doing it this way is that honest miners work as if the total hash power is the same. If we were not using the normal mining system like this, the total hash power of honest miners would be divided.For example, if the total hash power is 100 and the number of large miner pools is 10, our hash power is actually 10 because all of the miners start from the same nonce. This means repeating the same job, and as a result, the union of forces is broken. If we distribute the nonce intervals in a certain order, as I said, we combine all our power.
The purpose of the two hash systems is to detect a potential attack or ofspring blockchain excess. If there is a baby chain enough to divide the miners into two, the merkle root in the hash found by the miners defending the a blockchain and the merkle root in the hash that the miners defending the b chain will find will not be the same. In this situation, a rumor process is started in order to stabilize the network. This can be called a kind of voting. However, no miner tries to do this without giving up the transactions they have. Another possibility is the 51 percent attack, if the attacker uses the hash power and targets the first or the second hash, or both are powerful for a certain amount of time, here the majority of miners choose the first hash as honest miners randomly choose which hash to try to find, and if the attacker has that. Honest miners will surely find a hash if the majority cannot guess correctly in percentage terms. In this case, as the miners find a hash, the attacker's block chain is on a rival chain, and in this case, the network enters the process of finding the right chain and rumor again, honest miners will not accept the attacker's incomplete chain, as the attacker wants to delete the existing transactions. , and all efforts of the attacker will be wasted So is there no probability the attacker won? There is, but very unlikely. The attacker needs to be able to find all the hashes to win, so that it is not a rival blockchain. He can only do this if he correctly predicts the dispersion rate of honest miners, but if the attacker is very strong (eg 80 percent) he can win, even if he knows the dispersion rate of honest miners with a bit of error. However, remember that 100 percent of the attackers wins when the cost of 51 attacks today is covered, there is always a risk for the attacker in this algorithm, and a high amount.