Fixed interval cover cost: start at the leftmost point

Fixed interval cover cost can be minimized by placing each new cover at the leftmost point still uncovered. In this original fictional model, every cover is a closed interval of length three and costs one unit. The target coordinates are zero, two, three, four, six, seven, eight and eleven. Three covers suffice, and a short lower-bound argument proves fewer cannot work.

The model describes mathematical points on a line. Its units are invented, with no product prices, material coverage, measured equipment performance or physical installation implied. An interval may start at any real coordinate, and overlapping covers are allowed. The task is to cover every listed point at least once while minimizing the declared total cost.

Fix the interval rule and cost rule

A cover beginning at a consists of all real points from a through a plus three, including both endpoints. Write it as [a, a+3]. Its length is three regardless of where it begins. Every chosen interval adds exactly one cost unit, so minimizing cost is equivalent to minimizing the number of intervals.

Only the listed target points require coverage. The exercise does not require covering every real point between zero and eleven. A gap between two covers is permitted if it contains no target point that remains uncovered. Keep that requirement beside the drawing so a continuous-strip problem is not substituted for the stated finite-point problem.

The targets are already in increasing order. Each coordinate identifies a required location, rather than an interval count or a cost. Mark the eight positions on a working line and keep a separate list of the intervals selected. This gives another reader enough information to check both feasibility and the final cost.

Place the first cover at zero

Zero is the leftmost uncovered target. Begin the first interval there, giving [0, 3]. It covers zero, two and three. The right endpoint three is included because the interval is closed. The next target four is outside this interval and remains uncovered.

There is no need to cover an unlisted point at coordinate one separately. It already lies inside the interval, but its coverage has no independent effect on the count. Nor do the three target points covered by this first interval make it three purchases in the model. One interval costs one unit regardless of how many listed targets it contains.

Remove the covered targets from the remaining list. The required positions still uncovered are four, six, seven, eight and eleven. Keep the first interval in the saved solution; removing its points from the working list does not erase the coverage it provides.

Repeat from the next uncovered target

The leftmost point now remaining is four. Start the second interval at four, obtaining [4, 7]. It covers four, six and seven. Again the right endpoint belongs to the cover, so seven is removed from the remaining targets. Eight and eleven remain.

Start the third interval at eight, the next leftmost uncovered point. The interval [8, 11] contains both eight and eleven. There are no targets left afterward. The chosen cover list is [0, 3], [4, 7] and [8, 11], with total cost three.

Every listed target appears in at least one selected interval. That proves the three-interval solution is feasible. Establishing feasibility alone leaves open whether a clever placement of two intervals could be cheaper. The separated-point check below supplies the missing lower bound.

Use three separated targets as a minimum certificate

Focus on targets zero, four and eight. The distance from zero to four is four, and from four to eight is also four. Each separation exceeds the allowed interval length three. Zero and eight are even farther apart. Thus no single permitted interval can cover two of these three selected targets.

Any feasible solution must cover all three selected targets. Since each interval can account for at most one of them, every feasible solution needs at least three intervals. The displayed solution uses exactly three. Its cost therefore meets the lower bound and is minimum under the declared equal-unit-cost rule.

The lower-bound targets are also the successive left endpoints chosen by the procedure. This provides a compact record: keep zero, four and eight beside the three covering intervals. The interval list proves attainment, while the separated target list proves that no smaller count is possible.

The certificate concerns separation between required points. It does not depend on how large the blank areas of a drawing look or on a scale guessed from an illustration. The stated coordinates and fixed length are enough to establish each comparison.

Explain why starting at the leftmost point is safe

Consider any nonempty remaining target set, and call its smallest coordinate x. Every feasible cover of that set contains some interval [a, a+3] covering x. Therefore a is at most x, and x is at most a plus three. Replace that one interval by [x, x+3].

Take any remaining target y that the old interval covered. Because x is the smallest remaining target, y is at least x. Because the old interval covered y, y is at most a plus three. Since a is at most x, the value a plus three is at most x plus three. Hence y still belongs to the replacement interval.

Moving the interval's start to x has therefore lost no required point from the remaining set and has not increased the number of intervals. The other intervals can stay where they were. In particular, a minimum cover can be arranged to begin with the procedure's choice [x, x+3].

After making that choice, remove the targets it covers and solve the smaller remaining problem. Each step removes at least x itself, so the finite list eventually becomes empty. Repeating the same exchange reasoning at each step justifies the complete procedure rather than only its first placement.

At later steps, a point removed earlier is already covered by an interval retained in the solution prefix. The remaining problem need not cover it again. This explains why the exchange checks only the current remaining targets, even if an old candidate interval happened to extend into previously covered territory.

Read the general lower bound directly from the procedure

There is also a direct proof for any finite target set with this fixed-length rule. Record each leftmost uncovered point when the procedure chooses its interval. After choosing [x, x+3], every remaining target is strictly greater than x plus three. The next recorded leftmost point is therefore more than three units away.

Every pair of consecutive recorded points has this separation, and points farther apart in the list are separated by still more. No permitted interval can contain two recorded points. Any cover needs at least as many intervals as there are recorded points, while the procedure uses exactly one interval for each.

This reasoning works with fractional or negative coordinates as well as integers. It uses ordering and separation, with no assumption that targets sit on a unit grid. A different positive fixed length can replace three throughout the same argument, provided each interval has that length and each has the same unit cost.

Compare a more expensive placement rule

Consider a separate rule that centers a length-three interval on each next uncovered target. From zero it chooses [-1.5, 1.5], covering only zero from our list. The next uncovered target is two, so the second interval is [0.5, 3.5], covering two and three.

The next four intervals center on four, six, eight and eleven. They are [2.5, 5.5], [4.5, 7.5], [6.5, 9.5] and [9.5, 12.5]. Together the six centered intervals cover the targets, but their cost is six instead of three. Under this fictional unit-cost model, the leftmost-start solution saves three cost units relative to that particular rule.

Centering the next target spends half of the interval on positions to its left. In the remaining problem there is no earlier required target still needing coverage. Starting at the target instead retains it and extends the right endpoint farther, which is the advantage formalized by the exchange argument.

The six-interval comparison is a worked counterexample to that centering rule being always minimum. It does not claim every centered placement on every target set is more expensive. The minimum proof for our procedure comes from preservation and separation, rather than from assuming another rule must always fail.

Handle boundaries and changes of model explicitly

If the target set is empty, select no intervals and record cost zero. If it contains only one target, one interval beginning there covers it and costs one. Repeated copies of the same coordinate do not require separate covers, because the condition is that each location be covered at least once.

Keep the closed-endpoint convention. Coordinate three is covered by [0, 3], and eleven is covered by [8, 11]. A right-open interval would have different membership at those endpoints. Do not silently borrow the displayed coverage list after changing that convention.

The free-placement assumption also matters. If a separate exercise restricts starts to a prescribed list, [x, x+3] might not be permitted. If intervals have different costs, minimizing their number need not minimize their total cost. Variable lengths or additional setup charges would likewise require a new argument.

A finite arithmetic control checks every subset of integer targets zero through eleven, including the empty set, for each fixed length one through four. Across 16,384 cases, the procedure matches an independent minimum-cover calculation that considers all candidate integer-start intervals. The exchange and separated-point proofs establish the broader real-coordinate conclusion; the finite checks support the worked implementation.

Distinguish the neighboring optimization questions

The minimum step cost exercise reaches an exact integer total with allowed move sizes and compares last-move predecessors. This cover problem may span gaps and overlap, so its feasibility rule is different. It minimizes the number of fixed-length intervals needed to contain target locations.

The holiday meeting-point exercise chooses one real location to minimize a sum of absolute distances. Here no distance sum is minimized. The task chooses several closed intervals, and the lower bound comes from target points that cannot share one interval.

Keep the native bowl accounts within their evidence

The Adcraft SBL-30 bowl listing has one original photograph showing a complete plain metallic-looking bowl. Its Tools & Gadgets category and mixing-bowl description are native account details. Stainless steel, capacity, dimensions, finish and maker authenticity are not independently established by the photograph.

The five-piece nesting-bowl listing has three images showing five silver-looking bowls, nested capacity markings and bowls staged with food. Markings do not authenticate volume or measurement accuracy. Material, nonslip bases, care, weight and storage-saving claims remain unverified. Food, utensil and other context props are not established supplied stock.

Keep a possible transaction separate from the model

A reader considering an unrelated purchase on republeq.com could ask whether USDT is offered as a cryptocurrency payment method for the intended order. Such an enquiry would need to identify the exact offer and quantity so any answer concerns that transaction. The costs in this exercise are invented mathematical values; they are neither prices nor token amounts. Before deciding whether to proceed, the reader could review the actual choices and terms presented for the proposed purchase and ask the responsible merchant to clarify essential missing information. If a cryptocurrency option is available, the transaction information would need to specify the asset, network, receiving destination, amount and reference linking a payment to the order. A token name alone would not settle those details. Any delivery, cancellation, return or refund question would also need an answer for that same offer, including the applicable refund denomination and conditions. An answer about an earlier basket would not establish the terms of a changed order. This article gives no wallet address, transfer instruction, conversion rate, fee estimate or assurance of recovery. It promises no current USDT acceptance or transaction outcome. The reader can finish the mathematical worksheet without making a purchase. If an order or payment status is uncertain, clarify that status through the responsible merchant before considering another payment. Keep any eventual acknowledgement and transaction record with its own order information. A correct cover minimum cannot establish that funds were sent, accepted or reconciled, and a payment record cannot validate a proposed interval list. Each question needs the evidence appropriate to the decision being made.

For the stated targets, retain [0, 3], [4, 7] and [8, 11] with cost three. Keep zero, four and eight as the minimum certificate. The solution covers every required point, and their separation proves any solution needs at least three intervals. That complete pair of checks justifies the fixed interval cover cost within its declared rules.