In a pipe-cutting model where each cut costs the length of the piece it splits, why does taking the cheapest cut first fail?

## The scenario, stated plainly A pipe runs from position 0 to position L. A customer has marked positions where it must be severed. Each cut costs whatever the *current piece being cut* is long — the saw charges by the material it passes through — and after a cut the two resulting pieces are independent. Every mark must eventually be cut; only the order is yours to choose. Total bill depends entirely on that order. ## Why the greedy instinct is wrong The intuitive move is to sort by something — cut the mark nearest an end, or the one splitting the piece most evenly, or the "cheapest" one — and proceed. All of these fail, and the reason is structural rather than a matter of finding a better greedy key. A greedy algorithm needs the property that a locally best choice is contained in some globally optimal solution, usually shown with an exchange argument: take an optimal solution that disagrees with the greedy choice, swap the greedy choice in, and argue the cost does not increase. Here the swap changes the *prices of the remaining decisions*, because a cut's price is the length of the piece it falls in, and that piece is determined by the cuts already made. The exchange argument has nothing to hold fixed. Every choice reprices the rest of the problem. That also explains why counterexamples are easy to produce but hard to predict: with marks at 1, 3 and 8 on a pipe of length 10, one order costs 10 + 8 + 5 and another 10 + 9 + 2, and which is which depends on the arithmetic, not on any orderable feature of the marks. Note as well that the first cut always costs L no matter which mark you choose, so the very first decision has no local signal at all — a clean demonstration that local cost cannot guide the search. ## The decoupling decision Interval DP works when you can name **one decision that splits the range into two subranges which never interact again**. Here that decision is: *which mark is cut first inside this piece?* Whichever mark `k` you cut first, that cut charges the full length of the piece from `i` to `j`, and afterwards the marks left of `k` and the marks right of `k` are two entirely separate cutting problems on two separate pieces. No later choice on the left can change the price of anything on the right. So with `x[]` the mark positions (including the two ends) and `dp[i][j]` the minimum cost to make every cut strictly between marks `i` and `j`: ``` dp[i][j] = (x[j] - x[i]) + min over i < k < j of ( dp[i][k] + dp[k][j] ) dp[i][i+1] = 0 ``` Filled shortest interval first, this is O(n^3) time, O(n^2) space, and the answer is `dp[0][last]`. ## First versus last — the part that decides the recurrence The subtle craft here is choosing *which* decision to branch on, and it is not always the first one. The rule is: **branch on the decision after which the two sides are independent, and whose own cost you can evaluate at the moment you make it.** - In the cutting family, that is the **first** cut in a piece. Its price is the piece's full length — known immediately — and it separates the sides. - In chain-combination problems, where a sequence of items is combined pairwise and the cost of a combination depends on the two operands' dimensions, the decoupling decision is the **last** combination: the one top-level operation whose two operands are the fully-combined left part and the fully-combined right part. Branching on the *first* combination there does not decouple, because two adjacent items merged early change the operand shape seen by everything around them. - In removal-order problems, where removing an element charges a price that depends on the neighbours still present, the decoupling decision is again the **last** removal within the range: at that instant the range's two outside neighbours are exactly the elements bordering it, which is what makes the two sides independent. Getting this backwards is the single most common way a candidate writes an interval recurrence that looks right and computes the wrong thing. Say the sentence out loud before you write the formula: *after this decision, are the two sides truly independent, and do I know this decision's price now?* ## How to present it under interview pressure First, kill greedy with a reason, not just a counterexample — "the cost of a cut depends on the piece it lands in, so each decision reprices the others; there is no invariant for an exchange argument." Second, name the state as a subrange. Third, name the decoupling decision and justify it in one sentence. Fourth, give the recurrence, base case, fill order and cost profile. An interviewer asking this question is testing formulation judgment, not arithmetic. ## The honest caveat Greedy is not always wrong on cost-of-combination problems — there is a well-known family where items are repeatedly combined and the cheapest pair is provably the right choice at every step. The distinguishing feature is whether you are free to reorder the items. When the sequence order is *fixed* and only the grouping varies, you are in interval DP; when items may be combined in any order, a different technique may apply. Naming that boundary is what separates a senior answer from a rehearsed one.