First true search repair: keep the known true endpoint

An original integer-search worksheet with explicit endpoint rules.

First true search repair starts with the meaning of the two endpoints. In this original integer worksheet, the left endpoint is known to fail a stated test and the right endpoint is known to pass it. The repair keeps those facts true after every update. When the midpoint passes, it becomes the new right endpoint. Subtracting one from that passing midpoint can discard the answer we are trying to find.

The example concerns a deliberately defined mathematical predicate. It supplies no diagnosis of a particular application, device or service, and it makes no measured claim about search speed. You can check every test directly from the given rule. Keeping the rule beside the endpoint record makes the disputed update easy to inspect.

State the domain and the endpoint promise

Use the integer indices zero through ten, inclusive. Define P(k) to be true exactly when k is at least seven. Thus P(0) through P(6) are false, while P(7) through P(10) are true. Seven is the first true index in this stated domain. We already know that answer from the definition; the worksheet uses it to check the procedure.

Begin with L equal to zero and R equal to ten. The maintained promise is L less than R, P(L) false and P(R) true. The first true index must lie strictly above L and at or below R. Write the candidate interval as (L, R]. The round left bracket excludes the known false endpoint; the square right bracket includes the known true endpoint.

For the starting state, that interval contains the integers one through ten. Zero is excluded because it is known false. Ten remains a candidate because a known true point could itself be the first true point when no earlier true point has yet been established. A passing endpoint does not mean the answer must occur strictly before it.

These endpoint facts are premises of the method. If an initial left endpoint already passes, or an initial right endpoint fails, the stated bracket is invalid. Preserve that problem rather than inventing a result for the procedure. The rule here gives no permission to test an index below zero or above ten.

Write the midpoint and both updates

Continue while R minus L is greater than one. Compute m as the floor of (L plus R) divided by two. Floor means the greatest integer no larger than the quotient. Test P(m) once for this state. A false result replaces L with m. A true result replaces R with m. Leave the other endpoint unchanged in either branch.

The loop condition matters. With an integer width greater than one, this midpoint lies strictly between L and R. For a width of two, the midpoint is exactly one step above L. For a larger width, flooring the average still leaves at least one step above L and at least one step below R. Either update therefore changes an endpoint.

Save each old pair, midpoint, test result and new pair on its own line. That record lets you ask whether the update preserved the required endpoint facts. Merely watching the interval become narrower would miss the faulty branch examined below: an invalid interval can also shrink.

Follow the three tests in the worked example

From L equal to zero and R equal to ten, the midpoint is five. Since five is below seven, P(5) is false. Replace L with five. The new state is L equal to five and R equal to ten. The left endpoint still fails, the right endpoint still passes, and the candidate integers are six through ten.

The width is now five, so another test is required. The average of five and ten is seven and a half; flooring it gives m equal to seven. P(7) is true under the stated rule. Replace R with seven. The new state is L equal to five and R equal to seven. Candidate indices are six and seven.

The width is two. The midpoint is six, and P(6) is false. Replace L with six while retaining R equal to seven. Now the width is one, so stop. There is exactly one integer strictly above six and at or below seven. Return R, which is seven, and retain the final endpoint facts beside that answer.

The sequence of tests is five false, seven true, six false. The endpoint pairs are initially zero and ten, then five and ten, then five and seven, then six and seven. Recording those pairs prevents a midpoint from one state being accidentally combined with an endpoint from another state.

Locate the fault in R equals m minus one

Repeat the first step on a separate record, reaching L equal to five and R equal to ten. At midpoint seven, the test passes. Suppose the true branch instead assigns R equal to m minus one. It sets R to six. But P(6) is false, so the required known true right endpoint has been lost immediately.

The resulting pair is five and six. Its width is one, which would make the stated stopping condition end the loop. Returning R would return six, even though six fails the test. The smaller interval therefore supplies a concrete failure rather than evidence that the procedure is nearly correct.

The passing midpoint seven was still a valid answer candidate. Removing it required a fact that the test did not supply. P(7) true tells us that the first true index is at most seven. It does not establish that the first true index is at most six. Keeping R equal to seven preserves exactly the information obtained.

Changing the stopping condition alone would not restore the missing endpoint fact in this record. The repair belongs at the update that discarded seven. Likewise, starting a new record after the faulty update would hide the cause. Keep the failed line and show the corrected branch separately so another reader can reproduce both outcomes.

Prove that the corrected bracket survives

Consider a valid state with the same monotone threshold form. All indices below its threshold are false and all indices at or above it are true. Assume the first true index lies in (L, R], with L false and R true. The midpoint is strictly inside the endpoints whenever the loop continues.

If P(m) is false, the threshold must be above m. Replacing L with m excludes only indices that cannot be first true under this predicate. R retains its known true value. The threshold consequently remains strictly above the new L and at or below the unchanged R.

If P(m) is true, the threshold is at most m. Replacing R with m keeps that possibility, including the case where m itself is the threshold. The unchanged L remains false. Thus both branches preserve the endpoint promise and the candidate interval's inclusion rule.

The width is a positive integer. Each continuing step moves one endpoint strictly inward, so the width decreases. It cannot decrease through positive integers forever. When it reaches one, there is only one integer in (L, R], namely R. Since the first true index remains inside that interval, R must be the answer.

This argument uses the threshold premise. A test that turns true and later false would require a different justification. Nor does the proof establish an actual program's arithmetic representation, error behavior or implementation details. The worksheet's exact integer operations and fixed predicate are sufficient for the conclusion it makes.

Check boundary cases without widening the domain

An initial pair whose width is already one needs no midpoint test. With L false and R true, its only candidate is R. The endpoint promise therefore explains the immediate answer. There is no need to force an extra iteration or evaluate a midpoint equal to an endpoint.

For an additional explicit check, keep the same domain but set the threshold to one. Starting with zero and ten, the correct method eventually returns one. A threshold of ten is also valid with those endpoints and returns ten. These cases keep the first true point close to each end without asking for a value outside the domain.

A finite control set can examine every threshold from one through ten and every initial pair satisfying zero at most L less than the threshold at most R at most ten. There are 220 such brackets. Applying the corrected updates to each preserves the endpoint facts and returns its own threshold. Those checks support the implementation of this worksheet; the general preservation and termination argument explains why the result holds.

The frozen-row replacement exercise concerns which version of a label supplies a table lookup. Here each step deliberately creates a new endpoint state after testing the current midpoint. There is no simultaneous replacement of every original position. The minimum step cost exercise compares eligible last moves to establish an optimum. This search has no move-cost comparison: it preserves a known false and known true boundary.

For threshold ten, the midpoint sequence from zero and ten is five, seven, eight and nine. Every one of those tests is false. Each becomes the new L, while the known true R stays ten. The final pair is nine and ten. This shows why a right endpoint may remain unchanged through an entire valid run: each new false test narrows the candidates from the left.

For threshold one, the same starting pair tests five true, then two true, then one true. The right endpoint moves from ten to five to two to one, while L stays zero. The final pair is zero and one. Keeping each passing midpoint is necessary even in this edge example; a passing one can be the answer itself.

These two traces also distinguish an unchanged endpoint from an unchanged state. One endpoint can stay fixed while the other moves inward. What termination needs is a strictly decreasing width on every continuing step. Recording the difference after each update checks that property without requiring both endpoints to move.

Keep the native towel accounts within their evidence

The Organic Textiles bathroom-set listing describes three bath towels. Its sole image shows three pale folded or rolled towel textiles, with different displayed sizes. The generic Bathroom Sets title and category remain part of its native account. The image authenticates no GOTS certification, material, GSM, dimensions or performance statement.

The Living Quarters LI4327110 towel listing calls its title color Tan while recording catalog color Multicolor. Its sole image shows one folded tan or brown waffle-textured towel. That catalog conflict does not establish several supplied colors. No maker, size or absorbency result is independently established here.

Keep a possible purchase separate from the worksheet

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 endpoint values in this exercise are invented mathematical inputs; 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 search result cannot establish that funds were sent, accepted or reconciled, and a payment record cannot validate a faulty endpoint update. Each question needs the evidence appropriate to the decision being made.

For the stated threshold seven, retain the corrected sequence ending at six and seven. Its final right endpoint passes, and the only remaining candidate is seven. The failed R equals m minus one line remains useful evidence of the exact mistake: it removed a midpoint that could still be the first true index. Keeping that midpoint as R repairs the endpoint promise that makes the final answer justified.