Imagine a scenario when we all unit weights $$$w = \begin{Bmatrix} 1,1,1,\cdots,1 \end{Bmatrix} $$$ and arbitrary $$$v$$$ value array.
Similarly if we have all unit values $$$v = \begin{Bmatrix} 1,1,\cdots,1 \end{Bmatrix} $$$ and arbitrary weight array.
How should we fill the array ? Can we convert original problem to this?

Since we are allowed to have fractional values and weights. Following the hints, lets divide every $$$v_i$$$ with its corresponding weights $$$w_i$$$, then the problem becomes problem like we have new $$$v$$$ with all unit weighs and contribution of each can be upto $$$w_i$$$.

Let $$$G$$$ = pick the highest value with highest count as possible, say its $$$v_n, v_{n-1}, \cdots , v_m $$$, then with this strategy $$$v_m$$$ the last one will be the only one whose contribution is $$$\leq w_m$$$ .

Let $$$O$$$ be the optimal set = $$$ \begin{Bmatrix} (v_i,u_i) \mid u_i \leq w_i \end{Bmatrix}$$$. We need to prove that $$$O$$$ is same as $$$G$$$.

Claim: Any value $$$v_i \notin O$$$ must be less than every value in $$$O$$$.
Proof: By Contradiction ( Exchange argument ).
If there was such value such that $$$v_j \in O$$$ with contribution $$$u_j$$$, such that $$$v_j \lt v_i $$$ then replacing $$$v_j$$$ by any amount of contribution from $$$v_i$$$ would increase the final total value.

Hence all the values with greater value must be present in $$$O$$$ will full contribution as possible. And only possible partially filled will be smallest value in $$$O$$$. Which means $$$O$$$ will look exactly like $$$G$$$.

If Greedy $$$G$$$ = $$$g_1, g_2, \cdots, g_k$$$, such that if im at $$$g_i$$$ then next stop $$$g_{i+1}$$$ is the furthest value such that $$$g_{i+1} \leq g_i + M$$$ . And $$$O$$$ = $$$o_1, o_2, \cdots o_m$$$,

Then what inequality between $$$g_1$$$ and $$$o_1$$$ can we guarantee ?
Can we generalize it ?

Optimal solution takes $$$m$$$ steps, which means $$$m \leq k$$$. By definition, $$$ o_1 \leq g_1 $$$.

Claim: $$$o_i \leq g_i $$$ for all $$$i \leq k$$$ .
Proof: By Induction.
Our Base case $$$o_1 \leq g_1 $$$ is trivially true.
Let assume Hypothesis $$$o_i \leq g_i $$$, is true all $$$i \lt k$$$ .
Lets assume $$$o_{i+1} \gt g_{i+1}$$$ , then by definition $$$g_{i+1}$$$ should be the furthest possible refreshment stop, which is a contradiction. Hence $$$o_{i+1} \leq g_{i+1}$$$ .

Using this proof we can also say that if $$$o_m$$$ is the last refreshment stop, then we our $$$G$$$'s last stop must at $$$m$$$-th or before. Hence $$$k \leq m$$$. But since $$$m$$$ is the optimal value then $$$k=m$$$. Which means $$$G$$$ performs not worse than $$$O$$$.

What if problem was $$$ \sum_{i=1}^{n} a_i \cdot b_i $$$ .
What if problem was $$$ \sum_{i=1}^{n} lg(a_i) \cdot b_i $$$ .
What if problem was $$$ \sum_{i=1}^{n} f(a_i) \cdot g(b_i) $$$ . $$$f$$$ is monotonically increasing function and $$$g$$$ is monotonically decreasing function.
Will the argument remains the same ?

Lets sort array $$$a$$$ and reorder the respective elements in $$$b$$$. Strategy $$$G$$$ be, sort $$$b$$$ such that high values of $$$a$$$ gets high value of $$$b$$$. Let optimal sequence be $$$O$$$.

Claim: Sequence $$$O$$$ must not contain any inversion.
Proof: By Contradiction ( Exchange argument ).
If there was inversion, say at $$$i \lt j$$$ and $$$b_i \gt b_j$$$ . Currently final product looks like $$$ = \cdots a_{i}^{b_i} \cdots a_{j}^{b_j} \cdots $$$ and after swapping $$$b_i$$$ and $$$b_j$$$ product will be $$$ = \cdots a_{i}^{b_j} \cdots a_{j}^{b_i} \cdots $$$. If we divide the products Before upon After $$$ = a_{i}^{b_i - b_j} / a_{j}^{b_i - b_j} = \left( a_i / a_j \right)^{b_i - b_j}$$$ . This fraction is greater than $$$1$$$ and the integer in power is positive integer. Hence correcting every inversion will increase the value. Hence $$$O$$$ must not contain any inversion.

Not having any inversion is the sufficient condition to make $$$O$$$ sequence look exactly like $$$G$$$. Hence $$$G$$$ is as optimal as $$$O$$$.

Convert this problem to P3.

In every permutation of activities, the first activity's time will be summed $$$n$$$ times, second activity's time will be summed $$$n-1$$$ times, and so on. So Its just like P3. with $$$a$$$ as $$$p$$$ time array, and $$$b = [1,2,3, \cdots n]$$$ array.

Let the Bold be the set $$$O$$$, the set of optimal activities. And dotted are the remaining activities.

Which $$$b_i$$$ and $$$a_j$$$ i can i safely swap?
1. We can replace $$$b_2,b_4$$$ and $$$b_5$$$, in place of $$$a_2,a_3$$$ and $$$a_4$$$ ? maybe the least overlapping, like $$$b_2$$$ ? or the shortest interval because of $$$b_5$$$? Can i do it always ?
2. We can replace $$$b_1$$$ with $$$a_1$$$, or $$$b_6$$$ with $$$a_7$$$ ? Can i do it always ?

Greedy strategy should be something thats reliable always.

• To test the shortest duration hypothesis lets draw samples of only 2 types of length, if it dont work for 2 different types of length then it will surely not work for $$$ \gt 2$$$ types of length, say this

and we can easily see, shortest duration is not good strategy.

• To test, the least overlapping hypothesis lets draw a lot of samples with same length, maybe this

here we can also see, least overlapping don't work.

Lets have an optimal set $$$O= \begin{Bmatrix} o_1,o_2, \cdots , o_m \end{Bmatrix}$$$. In the order of finishing times. $$$O$$$ contains non-overlapping ranges.

Claim: If there exist an activity $$$a_i \notin O$$$ and $$$a_{i}.\text{finish} \leq o_{1}.\text{finish}$$$ . Then there exist some optimal set $$$O'$$$ which contains $$$a_i$$$.

Proof: By Exchange Argument, its actually easy to see, if i replace $$$o_1$$$ with $$$a_i$$$, then $$$a_{i}.\text{finish} \leq o_{2}.\text{start}$$$ it wont be an overlapping, and hence a valid set. And its size did not decrease, hence a new optimal valid set with at least 1 different element.

Lets complete the proof, let the greedy strategy be $$$G = \begin{Bmatrix} g_1,g_2, \cdots g_k \end{Bmatrix} $$$ where we take the earliest non-overlapping finished activity one by one. And let $$$O$$$ be the optimal strategy. Since we $$$O$$$ has the maximal size $$$m$$$, so we have inequality $$$k \leq m$$$.

The rest of the Proof is exactly like P2..

Claim: Every $$$g_{i}.\text{finish}$$$ $$$ \leq $$$ $$$ o_{i}.\text{finish}$$$ for all $$$i \leq k$$$.
Proof: By Induction,
Base case: $$$g_{1}.\text{finish} \leq o_{1}.\text{finish}$$$, by definition.
Hypothesis: Lets say it stays true till $$$i$$$-th, $$$g_{i}.\text{finish}$$$ $$$\leq$$$ $$$o_{i}.\text{finish}$$$ .
Induction: Lets assume $$$g_{i+1}.\text{finish}$$$ $$$ \gt $$$ $$$o_{i+1}.\text{finish}$$$ , then by definition used in $$$G$$$, $$$o_{i+1}$$$ would have been picked, hence a contradiction, which means our initial assumption is false.

Given our claim is true, by the $$$m$$$-th step, since $$$g_m$$$ is along side or behind $$$o_m$$$, which means $$$m \leq k$$$ . Hence our $$$G$$$ is at least as good as any other optimal set $$$O$$$.

Let $$$T$$$ be the minimum spanning tree. $$$S$$$ be the some arbitrary set of nodes and $$$A$$$ be the set of edges denoted in blue belongs to $$$T$$$. $$$S$$$ is made such that edges from $$$ S $$$ to $$$V \setminus S$$$ are not part of $$$A$$$, hence not denoted in blue.

Will any of the $$$u_i$$$ edge be the part of $$$T$$$ ?
If yes, then which $$$u_i$$$ edge should i pick ? we had drawn boundary around $$$A$$$ because in bucket of choices of edges we don't want edges that are already in $$$T$$$.

Will any of the $$$u_i$$$ edge be the part of $$$T$$$ ? Yes, otherwise final tree would be disconnected.

If yes, then which $$$u_i$$$ edge should i pick ? Though there can many crossed edge present in the final MST, can we guarantee that the minimum one will always be present in some MST ? . Lets say minimum edge is $$$u_3$$$ with vertices $$$(x,y)$$$.

Claim: Yes we can guarantee.
Proof: By Construction ( Exchange argument ).
Lets say we have some MST $$$T'$$$ that dont contain edge between $$$(x,y)$$$, instead say it contains edge between $$$(u,v)$$$. If we connect $$$(x,y)$$$ it will form a graph with a cycle, and cycle contains $$$(u,v)$$$ and $$$(x,y)$$$ both, because these 2 edges are among the edges that connects $$$S$$$ with $$$V \setminus S$$$. Then removing edge from this cycle will result in tree, and if i remove $$$(u,v)$$$ then cost of this new tree wont increase, hence resulting tree is a valid MST.

Im not saying $$$T'$$$ is not a MST, im saying there exists a valid MST with $$$(x,y)$$$ being one of the its edge.

Now solution is simple. We have $$$A$$$ a set of edges that are present in MST. We know a way to add new edges to this set $$$A$$$. We should keep adding edges until we form a single tree. If you have noticed carefully what we just did was Proof by Induction. Just like we did in previous problems, instead of array now we have a graph $$$H$$$.

Base Case: $$$A = \phi$$$, we will take no edges as zero weight tree, which is lowest weight possible with non-negative edges. Hence its trivially true.
Hypothesis: We have assumed $$$A$$$ is part of edges present in MST. And by above hint we have proved, we can grow $$$A \cup (x,y)$$$ with minimum choice. Hence cost of the tree made by Greedy wont be worse than MST made by any other algorithm.

Practically you can either grow, by making $$$S = A$$$ every time, since at first $$$A = \phi$$$, we can make any partition and add a single edge into it, and continue like this

Or, since we know we are picking minimum edges all the time, so we can sort them, and try to add into $$$A$$$ one by one. In this case at every iteration, say i have an edge $$$(x,y)$$$, if $$$x$$$ and $$$y$$$ are already connected through some path from edges in $$$A$$$, then i cannot form a partition $$$S$$$. Otherwise, if MST is not completely built, then there are still some disconnected parts not joined by edges from $$$A$$$, so partition is possible you just need to keep looking. At the end through induction we are guaranteed to have minimum spanning tree.

Do these 2 algorithms sounds familiar ?

Second best MST, differ in say how many edges from The best MST ? $$$1$$$ , or $$$2$$$ ?

It should contain 1 different edge from the best MST, because if there were 2 bad edges, then we can always form a partition $$$S$$$ to remove one of them make the cost better than previous but still worse than best.

Just try with brute force all the edges not in $$$T$$$, in each try you will add edge and it will form a loop we we remove the max edge possible from $$$T$$$ to form the minimum cost possible tree in this situation. Answer will be the minimum of such costs.

These are the exact same steps to prove the Dijkstra as well. Can you form the argument right ?

Base Case, Hypothesis, and Inductive Step.

This is exactly what Rethink the Dijkstra algorithm -- Let's go deeper in his blog realized, that the same greedy logic can be used in creative ways.