In the decrease-key operation of a Fibonacci Heap, if it is allowed to lose s > 1 children before cutting a node and melding it to the root list (promote the node), does this alter the overall runtime complexity? I think there are no changes in the complexity since the change in potential will be the same. But I am not sure if I am right.
And how can this be proved by the amortized analysis?
Changing the number of children that a node in the Fibonacci heap can lose does affect the runtime, but my suspicion is that if you're careful with how you do it you'll still get the same asymptotic runtime.
You're correct that the potential function will be unchanged if you allow each node to lose multiple children before being promoted back up to the root. However, the potential function isn't the source of the Fibonacci heap's efficiency. The reason that we perform cascading cuts (promoting multiple nodes back up to the root level during a decrease-key) is to ensure that a tree that has order n has a number of nodes in it that is exponential in n. That way, when doing a dequeue-min operation and coalescing trees together such that there is at most one tree of each order, the total number of trees required to store all the nodes is logarithmic in the number of nodes. The standard marking scheme ensures that each tree of order n has at least Θ(φn) nodes, where φ is the Golden Ratio (around 1.618...)
If you allow more nodes to be removed out of each tree before promoting them back to the root, my suspicion is that if you cap the number of missing children at some constant that you should still get the same asymptotic time bounds, but probably with a higher constant factor (because each tree holds fewer nodes and therefore more trees will be required). It might be worth writing out the math to see what recurrence relation you get for the number of nodes in each tree in case you want an exact value.
Hope this helps!