Fibonacci heaps achieve O(1) amortized decrease-key while binary heaps require O(log n). Why does this matter for Dijkstra's algorithm?
AIt reduces Dijkstra's space usage from O(n^2) to O(n)
BDijkstra performs O(m) decrease-key operations and O(n) delete-min operations; with Fibonacci heaps, the total cost becomes O(m * 1 + n * log n) = O(m + n log n), improving over the O(m log n) of binary heaps
CIt allows Dijkstra to process negative-weight edges
DIt eliminates the need for the relaxation step in Dijkstra's algorithm
Dijkstra's algorithm performs one delete-min per vertex (O(n) total) and one decrease-key per edge relaxation (O(m) total). With a binary heap, both operations cost O(log n), giving O((n + m) log n) = O(m log n) total. With a Fibonacci heap, delete-min remains O(log n) amortized but decrease-key drops to O(1) amortized, giving O(n log n + m). For dense graphs where m = Theta(n^2), this improves from O(n^2 log n) to O(n^2), which is optimal since you must examine all edges. The same improvement applies to Prim's MST algorithm, which has the same operation profile.
Question 2 Short Answer
The potential function for Fibonacci heap analysis is Phi(H) = t(H) + 2*m(H), where t(H) is the number of trees in the root list and m(H) is the number of marked nodes. Explain why decrease-key has O(1) amortized cost despite potentially triggering a cascade of cuts.
Think about your answer, then reveal below.
Model answer: When decrease-key triggers a cascading cut that cuts c nodes from their parents, the actual cost is O(c). But each cut: (1) adds 1 tree to the root list (increasing t by 1), (2) unmarks the cut node (decreasing m by 1, so -2 from 2*m(H)). The net potential change per cascading cut is 1 - 2 = -1. Over c cascading cuts, the actual cost is O(c) and the potential change is at most 4 - c (the initial cut adds at most 4 to potential: +1 tree, +2 for possibly marking the parent, +1 for the new root). The amortized cost is O(c) + (4 - c) = O(1). The cascading cuts are 'paid for' by the potential released from unmarking nodes — the marks are savings deposited by earlier operations.
This is the essence of the potential method: expensive operations (cascading cuts) are amortized against potential that was built up by cheap operations (earlier decrease-keys that marked nodes). The factor of 2 in 2*m(H) is precisely calibrated to pay for cascading cuts: each unmarking releases 2 units, which covers the O(1) actual cost of the cut.
Question 3 Multiple Choice
The aggregate method, accounting method, and potential method are three equivalent frameworks for amortized analysis. Which statement correctly distinguishes them?
AThe aggregate method assigns different costs to different operations; the accounting method sums all costs and divides by n; the potential method tracks a bank balance
BThe aggregate method computes the total cost of n operations and divides by n; the accounting method assigns an amortized cost to each operation type (overcharging cheap operations to prepay for expensive ones); the potential method defines a function of the data structure's state, and amortized cost = actual cost + potential change
CThe three methods can give different amortized bounds for the same data structure
DThe potential method is strictly more powerful than the other two
All three methods are equivalent in power and always give the same amortized bound (when optimally applied). The aggregate method is simplest: sum all actual costs over a worst-case sequence of n operations, divide by n. The accounting method is more flexible: assign amortized costs to each operation type, with the constraint that total amortized costs >= total actual costs (the excess is 'credit' stored in the structure). The potential method is the most general formulation: define Phi mapping states to non-negative reals, set amortized_i = actual_i + Phi(D_i) - Phi(D_{i-1}). The total amortized cost telescopes to total actual cost + Phi(final) - Phi(initial) >= total actual cost when Phi(final) >= Phi(initial).
Question 4 True / False
A Fibonacci heap with n nodes can have at most O(log n) children per node. This bound follows from the Fibonacci sequence — specifically, a node of degree k has at least F_{k+2} >= phi^k descendants, where phi is the golden ratio.
TTrue
FFalse
Answer: True
This is the structural property that gives Fibonacci heaps their name and ensures O(log n) delete-min. When a node x has degree k, its children were added in some order. The i-th child had degree at least i-1 when linked to x (since we link trees of equal degree during consolidation). After being linked, each child may have lost at most one child (since losing two triggers a cascading cut that removes it from x). So the i-th child has degree at least max(0, i-2). By induction, a tree rooted at a degree-k node has at least F_{k+2} nodes, where F_k is the k-th Fibonacci number. Since F_{k+2} >= phi^k (where phi = (1+sqrt(5))/2 ≈ 1.618), a degree-k node roots a subtree of size >= phi^k. Therefore k <= log_phi(n), meaning the maximum degree is O(log n).
Question 5 True / False
Binary heaps are always preferred over Fibonacci heaps in practice because Fibonacci heaps have better theoretical bounds.
TTrue
FFalse
Answer: False
This statement has the right conclusion for the wrong reason. Fibonacci heaps DO have better theoretical amortized bounds (O(1) decrease-key vs O(log n)), but in practice, binary heaps (or their variants like pairing heaps) are usually faster due to: (1) large constant factors in Fibonacci heap operations, (2) poor cache behavior from pointer-heavy tree structure vs array-based binary heap, (3) the improvement only matters when m >> n (dense graphs), which is not always the case. However, the statement is false because it's not ALWAYS the case that binary heaps are preferred — for very dense graphs with millions of vertices, or in theoretical contexts where asymptotic complexity matters, Fibonacci heaps (or pairing heaps, which match the bounds empirically) are the right choice. The practical relevance depends on the specific problem instance.