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Hermite's problem is an open problem in mathematics posed by Charles Hermite in 1848. He asked for a way of expressing real numbers as sequences of natural numbers , such that the sequence is eventually periodic precisely when the original number is a cubic irrational .
Closer to the Collatz problem is the following universally quantified problem: Given g, does the sequence of iterates g k (n) reach 1, for all n > 0? Modifying the condition in this way can make a problem either harder or easier to solve (intuitively, it is harder to justify a positive answer but might be easier to justify a negative one).
One strategy for solving this version of the hat problem employs Hamming codes, which are commonly used to detect and correct errors in data transmission. The probability for winning will be much higher than 50%, depending on the number of players in the puzzle configuration: for example, a winning probability of 87.5% for 7 players.
The "nine dots" puzzle. The puzzle asks to link all nine dots using four straight lines or fewer, without lifting the pen. The nine dots puzzle is a mathematical puzzle whose task is to connect nine squarely arranged points with a pen by four (or fewer) straight lines without lifting the pen.
Alternatively, one might solve the problem by using another reference to zeroth-order logic. In classical propositional logic, the material conditional is false if and only if its antecedent is true and its consequent is false. As an implication of this, two cases need to be inspected in the selection task to check whether we are dealing with a ...
The Fibonacci sequence is constant-recursive: each element of the sequence is the sum of the previous two. Hasse diagram of some subclasses of constant-recursive sequences, ordered by inclusion In mathematics , an infinite sequence of numbers s 0 , s 1 , s 2 , s 3 , … {\displaystyle s_{0},s_{1},s_{2},s_{3},\ldots } is called constant ...
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Separate the counting sequences according to the first vote. Any sequence that begins with a vote for B must reach a tie at some point, because A eventually wins. For any sequence that begins with A and reaches a tie, reflect the votes up to the point of the first tie (so any A becomes a B, and vice versa) to obtain a sequence that begins with B.
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