Compare minimum step cost with a largest-first shortcut

Minimum step cost can differ from the result of choosing the largest available move first. In this original exact-total model, the allowed moves add one, three or four units, and each move costs one fictional cost unit. To reach six, a largest-first choice uses four, one and one, costing three. Two moves of three reach the same total at a cost of two.

The saving is one cost unit within those declared rules. It is not a product-price comparison, measured time saving or claim about physical work. The exercise asks how few allowed moves reach an exact integer target. A short table, built from smaller targets, supplies a complete way to check that question instead of relying on the largest-first shortcut.

State the target and the cost separately

Begin at total zero. A permitted move adds exactly one, three or four to the accumulated total. Each move adds one to the cost, regardless of its size. You may repeat any move, but you may not pass the target and remove an excess later. The chosen target is a nonnegative whole number.

The total and cost count different things. A move of four adds four to the total while adding one to the cost. A move of one adds one to each. Keep two columns in a written record so the accumulated amount cannot be mistaken for the number of moves used to reach it.

Move sizes have no supply limit in this model, and no move becomes unavailable because an earlier move was used. These conditions allow two threes in the six-unit example. An actual task with limited supplies, changing action costs or a different rule about excess would require its own model and its own calculation.

Follow the largest-first shortcut

Define the shortcut precisely: at each stage, choose the largest allowed move that does not exceed the remaining amount. With target six, four is available and fits, so choose it. The remaining amount becomes two. Neither three nor four fits into two, so choose one, leaving one, then choose one again.

The resulting move list is 4, 1, 1. Its sum is six, so it is a valid way to reach the target. It contains three moves and therefore costs three under the stated unit-cost rule. A correct final total establishes validity; it does not establish that the sequence has the smallest possible cost.

Now compare 3, 3. Its sum is also six, and its two moves cost two. A one-move solution would require a permitted move of six, which is absent from the allowed set. Thus at least two moves are necessary, and the displayed two-threes sequence attains that bound. Two is the minimum for this target.

Use the last move to organize every possibility

Let C(t) mean the minimum number of moves needed to reach exact total t. Define C(0) = 0 because the empty move list already reaches zero. For a positive target, every valid move list has a last move. That move must have size one, three or four, and it cannot be larger than the target.

If the last move has size d, the preceding moves must total t minus d. Their smallest possible cost is C(t minus d). Appending the final move adds one more cost unit. The best result among all eligible last moves therefore gives C(t) = 1 + the minimum of C(t minus d).

Only include d from the allowed set with d at most t. This restriction keeps every predecessor total nonnegative. Compute the table in increasing target order, starting at zero. Every predecessor is smaller than its current target, so its completed minimum is available before the next entry needs it.

Fill the first entries without negative targets

For target one, only the last move of one is eligible. It follows C(0), giving C(1) = 1. For target two, again only a last move of one is eligible. It follows C(1), giving C(2) = 2. The witnesses are the move lists 1 and 1, 1.

For target three, a last move of one would follow target two and give cost three. A last move of three follows target zero and gives cost one. The smaller candidate wins, so C(3) = 1. A move of four is ineligible because it exceeds the target; it does not supply a third candidate.

For target four, the eligible predecessors are three, one and zero, corresponding to last moves one, three and four. Their known costs are one, one and zero. Adding the final move gives candidate costs two, two and one. Therefore C(4) = 1, witnessed by the single move four.

For target five, the predecessor totals are four, two and one. Their costs are one, two and one, so the candidate complete costs are two, three and two. The minimum is two. Both 4, 1 and 1, 4 reach five in two moves; the minimum need not determine a unique order.

Compute six from its three eligible predecessors

For target six, consider a last move of one first. Its predecessor total is five, whose minimum cost is two, so that choice gives complete cost three. A last move of three has predecessor total three with cost one, giving complete cost two. A last move of four has predecessor total two with cost two, giving complete cost three.

The three candidates are therefore three, two and three, associated with last moves one, three and four. Choose the middle candidate. It yields C(6) = 2 and identifies a last move of three after an optimal way to reach three. The earlier table entry supplies one move of three, reconstructing the list 3, 3.

The finished costs for targets zero through six are 0, 1, 2, 1, 1, 2 and 2. Keep the target labels beside those entries. A bare list of seven numbers can otherwise lose its starting zero or shift the correspondence by one position. Each entry describes its own exact target.

Recover a move list from the completed costs

Alongside each positive target, save a last move that attained its minimum. For six, save three, then move back to predecessor three. That entry also saves three and leads back to zero. Reading the saved moves from the starting end gives 3, 3. The predecessor chain supplies a concrete witness rather than leaving the minimum as an unsupported table label.

A tie allows more than one saved choice. For five, a final one after four and a final four after one both give minimum cost two. Saving either choice is sufficient to recover one optimal list. It does not imply that the other choice is invalid, and the number two in C(5) counts moves, not the number of optimal lists. Counting all possible optimal orders would be an additional question with a different output.

Explain why the calculation gives a minimum

Take any minimum-cost list for a positive target and inspect its last move d. Suppose its preceding list were more expensive than the minimum for t minus d. Replacing that prefix with a cheaper list reaching the same predecessor total would leave the same final move available and would lower the complete cost. That would contradict the assumed minimum.

The prefix of an optimal list must consequently be optimal for its predecessor total under these rules. Conversely, take an optimal list for any eligible predecessor and append its associated last move. This creates a valid list for the current target at exactly the candidate cost. The candidates are achievable costs, not merely lower estimates.

Every valid last move is included in the comparison. Beginning with the correct zero entry and proceeding upward therefore establishes each table entry from already established smaller entries. Move one ensures that every nonnegative target can be reached. This reasoning supplies the general rule; checking a few convenient totals alone would not do so.

Keep unavailable states out of the comparison

At target two, subtracting a last move of three would give predecessor negative one. That is not a zero-cost way to complete the calculation. Negative totals are outside the model. Treating them as if they had cost zero would incorrectly invent a one-move solution for two using a move that is too large.

The zero entry has a different status: zero is an allowed exact total reached by taking no moves. Retain that base case while excluding negative predecessors. If a future exercise removes move one or changes the permitted sizes, some nonnegative totals could also become unreachable. They would need an explicit unavailable status rather than an invented zero cost.

Check a second saving and a successful shortcut

At target ten, largest-first chooses four, four, one and one, costing four. The list four, three, three reaches ten in three moves. Two moves can total at most eight because each move is at most four, so ten needs at least three. The three-move witness meets that bound and establishes the minimum.

The shortcut does work for some targets. For seven it chooses four and three, giving two moves. One move cannot reach seven, so two is optimal. For eight it chooses four and four, again giving an optimal two moves. The examples six and ten show that largest-first is not a guaranteed minimum method; they do not show that it always fails.

Record the target and allowed moves beside each comparison. The fact that two threes improve six does not permit a move of three in a different exercise that omits it. Similarly, a result under equal per-move costs does not establish a minimum when one move size carries a larger declared cost than another.

Distinguish other minimum-move rules

The one-position change guide counts mismatched fixed letter positions and corrects each directly. This model instead accumulates an integer total using different move sizes. Its last-move comparison has no fixed set of wrong positions that can each be repaired once.

The adjacent card-swap guide uses inversion changes and parity to test minimum and exact move counts. Here the state is an accumulated total, with no card ordering or swap. Borrowing that parity condition would introduce a movement rule absent from this exercise.

Two native grinding-set accounts

The Eisco Labs twenty-pack listing describes white mortar-and-pestle sets. Its four images include a five-by-four array of sets, a single pair and a dimension diagram. Material, capacity, measurements, surface properties and grinding performance remain unverified listing statements rather than independent test results.

The Vestiray grinding-set listing describes a bowl and pestle. Its nine images include a white grooved interior, dark outer bowl, wood-colored pestle and SYOKAMI packaging. That photographed label does not authenticate the catalog brand. Comparative marketing supplies no verified health, safety, weight, efficiency or cleaning result.

A conditional purchase enquiry stays separate

A reader considering a hypothetical order through republeq.com could ask whether USDT is offered as a cryptocurrency payment method for the intended transaction. An enquiry of that kind establishes no actual merchant acceptance, checkout option or agreed transfer. It could identify the precise offer and quantity so a reply addresses the proposed order. If the method is available, the responsible merchant or checkout would need to state the asset, required network, receiving destination, payable amount and order reference. A familiar token name alone does not settle those details. The reader could also ask about current delivery, cancellation and return terms before deciding whether to proceed, including how a refund due would be handled. Keep each answer attached to the transaction it identifies rather than assuming it governs another purchase. This article supplies no wallet address, transfer procedure, exchange rate, fee estimate or assurance that a transfer can be recovered. If a payment or order status remains uncertain, clarify that status through the responsible merchant before considering another payment. The fictional unit costs in the mathematical model count permitted moves; they are not prices, token amounts or evidence that money has arrived. A complete minimum-cost proof also supplies no evidence that a checkout accepted a transaction. The exercise can be finished without making any purchase. A merchant may decline the requested method, an enquiry may remain unanswered, or the reader may choose not to order. None of those outcomes changes the allowed moves or the exact-target calculation. Any later purchasing decision requires its own current instructions, terms and evidence, independently of the paper model.

Save the minimum with its witness

For target six, retain the allowed moves one, three and four, the unit cost per move, the predecessor costs and the witness 3, 3. Keep the valid but more expensive 4, 1, 1 list as the shortcut comparison. The model's conclusion is minimum cost two under its stated rules, with an explicit way to achieve it. Actual equipment or purchasing decisions require evidence beyond that arithmetic.