Algorithm for cutting rods with minimum waste

Given a set of cuts and their lengths we need to find out the minimum number of rods (of constant length) and the cuts required which will lead to minimum wastage.

Here we bundle the rods and cut them all at once. So we can have a bundle with any number of rods.

For example:

Input data – Consider a rod of length 120 inches

( Quantity of Cuts Required, Length (in inches) ) = (5,16") , (5,30") , (24,36") , (4,18") , (4,28") , (6,20")

So here we required cuts such that we get 5 rods of 16 inches, 5 rods of 30 on.


Imagine each row (in the image) is a rod of 120 inches and each table is a bundle with rows as the number of rods in that bundle. So the first table is a bundle with 5 rods with cuts [16,30,36,36] and second table is a bundle of 4 rods with cuts [18,28,36,36] and so on. We can see that we have satisfied the input data we get (5,16") five rods of sixteen inches and so on.

enter image description here

Given input with (just like above) number of cuts and their lengths. how do we find the bundle of rods and their cuts having minimum amount of wastage?

Cutting Words timing when playing online

Lore Bard’s Cutting Words feature states the following:

Also at 3rd level, you learn how to use your wit to distract, confuse, and otherwise sap the confidence and competence of others. When a creature that you can see within 60 feet of you makes an attack roll, an ability check, or a damage roll, you can use your reaction to expend one of your uses of Bardic Inspiration, rolling a Bardic Inspiration die and subtracting the number rolled from the creature’s roll. You can choose to use this feature after the creature makes its roll, but before the DM determines whether the attack roll or ability check succeeds or fails, or before the creature deals its damage. The creature is immune if it can’t hear you or if it’s immune to being charmed.

What exactly does that mean. At what time do I have to say my DM I interrupt them?

When playing at a table, my DM usually goes this way: “The goblin attacks [player character] with their bow and [rolls] hits”. I can hear the dice roll and I say I want to know the result in order to decide if I want to use Cutting Words.

Now with the current sanitary situation, we play online, and the DM usually rolls their own physical dice and tells whether it hits or not because they have all our ACs registered. This basically forbids me to use my Cutting Words.

So what is the exact timing where I can interrupt the DM, online, and use my Cutting Words?

Optimal substructure of rod cutting?

How do you show the optimal substructure of the rod cutting problem(defined as in Optimal substructure and dynamic programming for a variant of the rod cutting problem). I am trying to follow the guideline steps enter image description here

So suppose someone told us one of the cuts of the rod $ R$ in an optimal solution OPT for $ R$ .

This cut of OPT paritions $ R$ into two smaller rods $ R_1$ and $ R_2$ , one of length $ i$ , and the other of length $ n-1$ .

Here is the only argument that I can think of: (Again, trying to follow steps 3 and 4 now but the argument does not seem like one that would suffice)

Suppose that $ R_1$ is not an optimal solution in itself, then OPT would not have given an optimal cut in the first place which contradicts our starting supposition that OPT is an optimal solution.

Also, the argument seems too “easy” to appear like something that is not just informal.

How would you prove the optimal substructure?

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Why this greedy algorithm fails in rod cutting problem?

Recall the rod cutting problem.

Given a rod of length $ n$ inches and a table of prices $ p_{i}$ for $ i=1,2,3,4,\,.\,.\,.$ determine the maximum revenue $ r_n$ obtainable by cutting up the rod and selling the pieces. You may assume no cutting cost.

The following is a question from CLRS 15.1-2.

Show, by means of a counterexample, that the following ‘greedy’ strategy does not always determine an optimal way to cut rods. Define the density of a rod of length i to be p_i/i, that is, its value per inch. The greedy strategy for a rod of length n cuts off a first piece of length i, where 1≤i≤n, having maximum density. It then continues by applying the greedy strategy to the remaining piece of length n-i.

Consider prices up to length $ 4$ are $ p_1=1, p_2=5, p_3= 8, p_4 = 9$ respectively. So, densities are $ d_1=1, d_2 = 2.5, d_3= 2.67, d_4 = 2.25$ respectively.

If we are given a rod of length $ 4,$ then an optimal way of cutting the rod to maximize revenue is $ 2$ pieces of length $ 2$ each, generating $ 5+5 = 10.$

However, greedy algorithm above will suggest cutting the rod into 2 pieces of length $ 3$ and $ 1$ , generating revenue $ 8+1=9.$

I obtain this example by merely following the same prices given in CLRS but do not understand why such greedy algorithm fails to provide optimal way of cutting the rod.