Let $S_{Ms},S_{Md},S_{Bs},S_{Bd}$ be salt mass fractions, with reservoir water masses denoted by the corresponding $M$ symbols. Let $S_A$ be Atlantic inflow salinity and $S_D$ the Danube inflow salinity, approximately zero for fresh water. With the diagram endpoint convention from part i, instantaneous mixing givesEvery outgoing salt transport uses the salinity of its donor box. The water masses are constant because the water balances are at steady state, even while the salinities evolve. If reservoir water masses varied, the left side would instead be $d(M_j{S}_j)/dt$.

Summing yields $d({\sum }_j{M}_j{S}_j)/dt=F_1{S}_A+F_9{S}_D-F_2{S}_{Ms}$, a direct whole-system salt check. Fresh water carries essentially no salt, but its dilution effect is still present through the water balance and the resulting transport terms.

There are four well-mixed reservoirs: Mediterranean surface/deep $(M_s,M_d)$ and Black Sea surface/deep $(B_s,B_d)$. Use the actual drawing’s arrow endpoints: $F_2$ leaves $M_s$, $F_5$ enters $M_d$ from $B_s$, and $F_6$ starts in $M_s$ and enters $B_d$. With $F_i$ as positive water-mass transport rates, the four steady conservation of mass equations areAdding them cancels every internal transport and gives the whole-system requirement $F_1+F_9=F_2$. There is no stated evaporation, precipitation or other external transport to add.

The schematic endpoints are unusual for the conventional two-layer exchange picture: its Gibraltar outflow is drawn in the surface box, and its Black Sea surface inflow descends into the Mediterranean deep box. Moving $F_2$ to $M_d$, $F_5$ to $M_s$, and the source of $F_6$ to $M_d$ would define a different, more conventional idealization. The displayed equations consistently follow the supplied drawing; the same endpoint convention is used in the following two parts.

Permitted; returns "C". Both receiver and argument have static type C, so the most-specific applicable overload is show(C). That method is defined in C and returns its own one-letter result without calling super. The six results, in source order, are therefore AB, AB, AB, BC, BC, C.

Permitted; returns "BC". The receiver's static type is C, but the argument's static type is only B. Thus show(C) is inapplicable; among show(A) and show(B), the latter is more specific. Its implementation is C.show(B), whose superclass call returns "B" before "C" is appended. The argument happens to reference a C object, but method overloading uses its static type.

Permitted; returns "BC". Through static type B, both show(A) and show(B) accept an argument of type C; method overloading selects the more specific show(B). The receiver's runtime type is C, so dynamic dispatch invokes C.show(B). Its super.show(o) calls B.show(B), returning "B", and the override appends "C". The show(C) overload is not visible through receiver type B.

Permitted; returns "AB". A receiver of static type B exposes show(A) and show(B), but the argument's static type A is not a subtype of B. Only show(A) is applicable. Its effective implementation on the C receiver is B.show(A), which concatenates the superclass result "A" and "B".

Permitted; returns "AB". Through a receiver of static type A, only show(A) is visible. The argument's static type C is a subtype of A, so it can be passed to that method. Dynamic dispatch selects B.show(A) on the receiver object, exactly as in part (i). The argument's more specific type cannot expose overloads absent from the receiver's static type.

Permitted; returns "AB". The receiver's static type is A, which offers show(A). The argument also has static type A, so that signature is selected. All three variables refer to the same object of runtime type C; the most-derived implementation of show(A) is inherited from B. Its super.show(o) calls A.show(A), producing "A", and it appends "B". C overrides show(B), not this selected signature.

The strongly connected components areThe directed cycle $1\to 3\to 5\to 1$ makes its three vertices mutually reachable. Vertex $6$ has no outgoing edge. Vertex $2$ reaches that cycle but has no return path from it, and vertex $4$ reaches the other components but has no incoming path from them. These facts exclude every further merger.

The only possibilities are $f[2]\in \{10,11\}$. Vertex $2$ reaches all of $1,3,5,6$. In depth-first search, all vertices reachable from an active vertex must have been discovered and finished before its own finishing event, except for active ancestors. None of these four can be an ancestor of $2$, because none can reach $2$. Thus all four are finished before $2$ finishes.

Vertex $4$ cannot already be finished at that point: its outgoing edge to $2$ would have caused $2$ to be visited before it could finish. It also cannot be a descendant of $2$, since $2$ cannot reach $4$. Therefore $4$ is either still undiscovered or an active ancestor. In the first case the only remaining events are discovery and finishing of $4$, giving $f[2]=10$. In the second case the only remaining event is finishing of $4$, giving $f[2]=11$.

Both occur. Starting the outer traversal with $2$ leaves $4$ undiscovered and gives time $10$. Starting with $4$ and taking its edge to $2$ first makes $4$ an ancestor and gives time $11$. The argument does not assume any particular order on the other edges.

Increment a single clock on each discovery and finish, starting with clock zero. Lexicographic depth-first search givesThe first search tree has discovery edges $1\to 3$, $3\to 5$, $3\to 6$. The edge $5\to 1$ returns to an active ancestor. Once that tree finishes, vertex $2$ is a new root whose neighbour $1$ is already finished, and vertex $4$ is the final root. This accounts for all twelve events in the DFS discovery and finishing times.

Use source $1$, as in the preceding part. The Dijkstra algorithm extracts vertices in the order $1,3,4,2$. Distances after relaxing the outgoing edges of each extracted vertex areThe shortest path to $2$ is $1\to 3\to 4\to 2$, with weight $2+1+1=4$; the direct route to $2$ and the route via $3\to 2$ are both heavier.

Start with $(d[1],d[2],d[3],d[4])=(0,\infty ,\infty ,\infty )$. With in-place edge relaxation, the Bellman-Ford algorithm makes the following changes during its first lexicographic pass:The next pass makes no change; any remaining fixed-count passes also make none. Thus the final vector is $(0,4,2,3)$. These are sequential updates within a pass, rather than the simultaneous values of the Synchronous Bellman-Ford recurrence.

Part (a) gives $\mathbb{P}(X\in \mathbb{Z})=0$. Therefore the ceiling function and floor function satisfy $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely. Both are integrable, since $0\le \lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$. Thus $\mathbb{E}[\lceil X\rceil ]=\zeta (t-1)+1$.

Now take expectations in $\lfloor X\rfloor \le X\le \lceil X\rceil$:Writing $s=t-1$, which ranges over every $s>1$, and rearranging yieldsThese are the zeta bounds from rounded Pareto variables: rounding below and above converts one elementary expected value into bounds on an infinite series.

The nonnegative integer-valued tail-sum formula for expectation givesThis is finite because $t-1>1$. Here $\zeta$ is the Riemann zeta function on its convergent real domain. The floor function is nonnegative on the support, so the tail-sum identity applies directly.

For $k\ge 1$, the event $\{\lfloor X\rfloor \ge k\}$ equals $\{X\ge k\}$. Integrating the Pareto distribution's density givesThe mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in \mathbb{N}$, include the separate endpoint values $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely. Substituting $k=0$ into the power expression would be invalid.

The expected value is finite because $t>2$:The condition on $t$ is precisely the first moment threshold for this parameterization of the Pareto distribution.

Normalization of the probability density function requires $1=c{\int }_1^{\infty }{x}^{-t}dx=c/(t-1)$. Hence $c=t-1$. It is positive for the given $t>2$, and the density is nonnegative, so this does define a probability distribution. It is a Pareto distribution with scale $1$ and shape $t-1$.

The events $\{X=k\}$, for $k\in \mathbb{Z}$, are pairwise disjoint. Their union is $\{X\in \mathbb{Z}\}$, and the integers form a countable set. By countable additivity,Countability is the necessary justification: one cannot apply this argument to an arbitrary uncountable union of zero-probability singletons. This is a special case of countable sets have zero probability under a density.

An absolutely continuous probability distribution assigns zero probability to a singleton, because its density has zero integral over that set. Thus $\mathbb{P}(X=x)=0$ for every real $x$, including points where a chosen version of $f$ is nonzero.