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From Wikipedia
A number of logic puzzles exist that are based on the balancing of similarlooking items, often coins, to determine which one is of a different value within a limited number of uses of the balance scales. These differ from puzzles where items are assigned weights, in that only the relative mass of these items is relevant.
Premise
A wellknown example has nine (or fewer) items, say coins (or balls), that are identical in weight save for one, which in this example we will say is lighter than the others—a counterfeit (an oddball). The difference is only perceptible by using a pair of scales balance, but only the coins themselves can be weighed, and it can only be used twice in total.
Is it possible to isolate the counterfeit coin with only two weighings?
Solution
To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one we can compare any two coins, leaving the third unweighed. If the two coins tested weigh the same, then the lighter coin must be the one not on the balance  otherwise it is the one indicated as lighter by the balance.
Now, assume we have three coins wrapped in a bigger coinshaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of 3*3 = 9.
Note that we could reason along the same line, further, to see that in three weighings one can find the oddlighter coin among 27 coins and in 4 weighings, from 81 coins.
The twelvecoin problem
A more complex version exists where there are twelve coins, eleven of which are identical and one of which is different, but it is not known whether it is heavier or lighter than the others. This time the balance may be used three times to isolate the unique coin and determine its weight relative to the others.
Solution
The procedure is less straightforward for this problem, and the second and third weighings depend on what has happened previously, although that need not be the case (see below).
 Four coins are put on each side. There are two possibilities:
 1. One side is heavier than the other. If this is the case, remove three coins from the heavier side, move three coins from the lighter side to the heavier side, and place three coins that were not weighed the first time on the lighter side. (Remember which coins are which.) There are three possibilities:
 1.a) The same side that was heavier the first time is still heavier. This means that either the coin that stayed there is heavier or that the coin that stayed on the lighter side is lighter. Balancing one of these against one of the other ten coins will reveal which of these is true, thus solving the puzzle.
 1.b) The side that was heavier the first time is lighter the second time. This means that one of the three coins that went from the lighter side to the heavier side is the light coin. For the third attempt, weigh two of these coins against each other: if one is lighter, it is the unique coin; if they balance, the third coin is the light one.
 1.c) Both sides are even. This means the one of the three coins that was removed from the heavier side is the heavy coin. For the third attempt, weigh two of these coins against each other: if one is heavier, it is the unique coin; if they balance, the third coin is the heavy one.
 2. Both sides are even. If this is the case, all eight coins are identical and can be set aside. Take the four remaining coins and place three on one side of the balance. Place 3 of the 8 identical coins on the other side. There are three possibilities:
 2.a) The three remaining coins are lighter. In this case you now know that one of those three coins is the odd one out and that it is lighter. Take two of those three coins and weigh them against each other. If the balance tips then the lighter coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is lighter.
 2.b) The three remaining coins are heavier. In this case you now know that one of those three coins is the odd one out and that it is heavier. Take two of those three coins and weigh them against each other. If the balance tips then the heavier coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is heavier.
 2.c) The three remaining coins balance. In this case you know that the unweighed coin is the odd one out. Weigh the remaining coin against one of the other 11 coins and this will tell you whether it is heavier or lighter.
With some outside the box thinking, such as assuming that there are authentic (genuine) coins at hand, a solution may be found quicker. In fact if there is one authentic coin for reference then the suspect coins can be thirteen. Number the coins from 1 to 13 and the authentic coin number 0 and perform these weightings in any order:
 0,1,4,5,6 against 7,10,11,12,13
 0,2,4,10,11 against 5,8,9,12,13
 0,3,8,10,12 against 6,7,9,11,13
If only one weighting is off balance then it must be one of the coins 1,2,3 which only appear in one weighting. If all weightings are off balance then it is one of the coins 1013 that appear in all weightings. Picking out the one counterfeit coin corresponding to each of the 27 outcomes is always possible (13 coins one either too heavy or too light is 26 possibilities) except when all weightings are ballanced, in which case there is no counterfeit coin (or its weight is correct). If coins 0 and 13 are deleted from these weightings they give one generic solution to the 12coin problem.
In literature
Niobe, the protagonist of Piers Anthony's novelWith a Tangled Skein, must solve the twelvecoin variation of this puzzle to find her son inHell: Satan has disguised the son to look identical to eleven other demons, and he is heavier or lighter depending on whether he is cursed to lie or able to speak truthfully. The solution in the book follows the given example 1.c.
In mathematics, a recurrence relation is an equation that recursively defines a sequence: each term of the sequence is defined as a function of the preceding terms.
The term difference equationsometimes (and for the purposes of this article) refers to a specific type of recurrence relation. Note however that "difference equation" is frequently used to refer to any recurrence relation.
An example of a recurrence relation is the logistic map:
 x_{n+1} = r x_n (1  x_n) \,
Some simply defined recurrence relations can have very complex (chaotic) behaviours, and they are a part of the field of mathematics known as nonlinear analysis.
Solving a recurrence relation means obtaining a closedform solution: a nonrecursive function of n.
Example: Fibonacci numbers
The Fibonacci numbers are defined using the linear recurrence relation
 F_n = F_{n1}+F_{n2} \,
with seed values:
 F_0 = 0 \,
 F_1 = 1 \,
Explicitly, recurrence yields the equations:
 F_2 = F_1 + F_0 \,
 F_3 = F_2 + F_1 \,
 F_4 = F_3 + F_2 \,
etc.
We obtain the sequence of Fibonacci numbers which begins:
 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
It can be solved by methods described below yielding the closed form expression which involve powers of the two roots of the characteristic polynomial t^{2} = t + 1; the generating function of the sequence is the rational function
 \frac{t}{1tt^2}.
Structure
Linear homogeneous recurrence relations with constant coefficients
An order d linear homogeneous recurrence relation with constant coefficients is an equation of the form:
 a_n = c_1a_{n1} + c_2a_{n2}+\cdots+c_da_{nd} \,
where the d coefficients c_{i}(for all i) are constants.
More precisely, this is an infinite list of simultaneous linear equations, one for each n>d−1. A sequence which satisfies a relation of this form is called a linear recursive sequence or LRS. There are d degrees of freedom for LRS, the initial values a_0,\dots,a_{d1} can be taken to be any values but then the linear recurrence determines the sequence uniquely.
The same coefficients yield the characteristic polynomial (also "auxiliary polynomial")
 p(t)= t^d  c_1t^{d1}  c_2t^{d2}\cdotsc_{d}\,
whose d roots play a crucial role in finding and understanding the sequences satisfying the recurrence. If the roots r_{1}, r_{2}, ... are all distinct, then the solution to the recurrence takes the form
 a_n = k_1 r_1^n + k_2 r_2^n + \cdots + k_d r_d^n,
where the coefficients k_{i}are determined in order to fit the initial conditions of the recurrence. When the same roots occur multiple times, the terms in this formula corresponding to the second and later occurrences of the same root are multiplied by increasing powers of n. For instance, if the characteristic polynomial can be factored as (x − r)^{3}, with the same root r occurring three times, then the solution would take the form
 a_n = k_1 r^n + k_2 n r^n + k_3 n^2 r^n.\,
Rational generating function
Linear recursive sequences are precisely the sequences whose generating function is a rational function: the denominator is the auxiliary polynomial (up to a transform), and the numerator is obtained from the seed values.
The simplest cases are periodic sequences, a_n = a_{nd}, n\geq d, which have sequence a_0,a_1,\dots,a_{d1},a_0,\dots and generating function a sum of geometric series:
\begin{align} & \frac{a_0 + a_1 x^1 + \cdots + a_{d1}x^{d1}}{1x^d} \\[6pt] & = \left(a_0 + a_1 x^1 + \cdots + a_{d1}x^{d1}\right) \\[3pt] & {} \quad + \left(a_0 + a_1 x^1 + \cdots + a_{d1}x^{d1}\right)x^d \\[3pt] & {} \quad + \left(a_0 + a_1 x^1 + \cdots + a_{d1}x^{d1}\right)x^{2d} + \cdots. \end{align}
More generally, given the recurrence relation:
 a_n = c_1a_{n1} + c_2a_{n2}+\cdots+c_da_{nd} \,
with generating function
 a_0 + a_1x^1 + a_2 x^2 + \cdots,
the series is annihilated at a_d and above by the polynomial:
 1 c_1x^1  c_2 x^2  \cdots  c_dx^d. \,
That is, multiplying the generating function by the polynomial yields
 b_n = a_n  c_1 a_{n1}  c_2 a_{n2}  \cdots  c_d a_{nd} \,
as the coefficient on x^n, which vanishes (by the recurrence relation) for n \geq d. Thus
 (a_0 + a_1x^1 + a_2 x^2 + \cdots {} ) (1 c_1x^1  c_2 x^2  \cdots  c_dx^d) = (b_0 + b_1x^1 + b_2 x^2 + \cdots + b_{d1} x^{d1})
so dividing yields
 a_0 + a_1x^1 + a_2 x^2 + \cdots =
\frac{b_0 + b_1x^1 + b_2 x^2 + \cdots + b_{d1} x^{d1}}{1 c_1x^1  c_2 x^2  \cdots  c_dx^d},
expressing the generating function as a rational function.
The denominator is x^d p\left(x^{1}\right), a transform of the auxiliary polynomial (equivalently, reversing the order of coefficients); one could also use any multiple of this, but this normalization is chosen both because of the simple relation to the auxiliary polynomial, and so that b_0 = a_0.
Relationship to difference equations narrowly defined
Given an ordered sequence \left\{a_n\right\}_{n=1}^\infty of real numbers: the first difference \Delta(a_n)\, is defined as
 \Delta(a_n) = a_{n+1}  a_n\,.
The second difference \Delta^2(a_n)\, is defined as
 \Delta^2(a_n) = \Delta(a_{n+1})  \Delta(a_n)\,,
which can be simplified to
 \Delta^2(a_n) = a_{n+2}  2a_{n+1} + a_n\,.
More generally: the k^{th} difference of the sequence a_n\, is written as \Delta^k(a_n)\, is defined recursively as
 \Delta^k(a_n) = \Delta^{k1}(a_{n+1})  \Delta^{k1}(a_n)\,.
The more restrictive definition of difference equation is an equation composed of a_{n}and its k^{th} differences. (A widely used broader definition treats "difference equation" as synonymous with "
This article 'Algorithm examples supplementsAlgorithm and Algorithm characterizations.
An example: Algorithm specification of addition m+n
Choice of machine model:
There is no â€œbestâ€�, or â€œpreferredâ€� model. The Turing machine, while considered the standard, is notoriously awkward to use. And different problems seem to require different models to study them. Many researchers have observed these problems, for example:
 â€œThe principal purpose of this paper is to offer a theory which is closely related to Turing's but is more economical in the basic operationsâ€� (Wang (1954) p. 63)
 â€œCertain features of Turing machines have induced later workers to propose alternative devices as embodiments of what is to be meant by effective computability.... a Turing machine has a certain opacity, its workings are known rather than seen. Further a Turing machine is inflexible ... a Turing machine is slow in (hypothetical) operation and, usually complicated. This makes it rather hard to design it, and even harder to investigate such matters as time or storage optimization or a comparison between efficiency of two algorithms.â€� (Melzak (1961) p. 281)
 ShepherdsonSturgis (1963) proposed their registermachine model because â€œthese proofs [using Turing machines] are complicated and tedious to follow for two reasons: (1) A Turing machine has only one head... (2) It has only one tape....â€� They were in search of â€œa form of idealized computer which is sufficiently flexible for one to be able to convert an intuitive computational procedure into a program for such a machineâ€� (p. 218).
 â€œI would prefer something along the lines of the random access computers of Angluin and Valiant [as opposed to the pointer machine of SchÃ¶nhage]â€� (Gurivich 1988 p. 6)
 â€œShowing that a function is Turing computable directly...is rather laborious ... we introduce an ostensibly more flexible kind of idealized machine, an abacus machine...â€� (BoolosBurgessJeffrey 2002 p.45).
About all that one can insist upon is that the algorithmwriter specify in exacting detail (i) the machine model to be used and (ii) its instruction set.
Atomization of the instruction set:
The Turing machine model is primitive, but not as primitive as it can be. As noted in the above quotes this is a source of concern when studying complexity and equivalence of algorithms. Although the observations quoted below concern the Random access machine model â€“ a Turingmachine equivalent â€“ the problem remains for any Turingequivalent model:
 â€œ...there hardly exists such a thing as an â€˜innocentâ€™ extension of the standard RAM model in the uniform time measure; either one only has additive arithmetic, or one might as well include all multiplicative and/or bitwise Boolean instructions on small operands....â€� (van Emde Boas (1992) p. 26)
 â€œSince, however, the computational power of a RAM model seems to depend rather sensitively on the scope of its instruction set, we nevertheless will have to go into detail...
 â€œOne important principle will be to admit only such instructions which can be said to be of an atomistic nature. We will describe two versions of the socalled successor RAM, with the successor function as the only arithmetic operation....the RAM0 version deserves special attention for its extreme simplicity; its instruction set consists of only a few one letter codes, without any (explicit) addressing.â€� (SchÃ¶nhage (1980) p.494)
Example #1: The most general (and original) Turing machine â€“ singletape with leftend, multisymbols, 5tuple instruction format â€“ can be atomized into the Turing machine of BoolosBurgessJeffrey (2002) â€“ singletape with no ends, two "symbols" { B,  } (where B symbolizes "blank square" and  symbolizes "marked square"), and a 4tuple instruction format. This model in turn can be further atomized into a PostTuring machineâ€“ singletape with no ends, two symbols { B,  }, and a 0 and 1parameter instruction set ( e.g. { Left, Right, Mark, Erase, Jumpifmarked to instruction xxx, Jumpifblank to instruction xxx, Halt } ).
Example #2: The RASP can be reduced to a RAM by moving its instructions off the tape and (perhaps with translation) into its finitestate machine â€œtableâ€� of instructions, the RAM stripped of its indirect instruction and reduced to a 2 and 3operand â€œabacusâ€� register machine; the abacus in turn can be reduced to the 1 and 2operand Minsky (1967)/ShepherdsonSturgis (1963) counter machine, which can be further atomized into the 0 and 1operand instructions of SchÃ¶nhage (and even a 0operand SchÃ¶nhagelike instruction set is possible).
Cost of atomization:
Atomization comes at a (usually severe) cost: while the resulting instructions may be â€œsimplerâ€�, atomization (usually) creates more instructions and the need for more computational steps. As shown in the following example the increase in computation steps may be significant (i.e. orders of magnitude â€“ the following example is â€œtameâ€�), and atomization may (but not always, as in the case of the PostTuring model) reduce the usability and readability of â€œthe machine codeâ€�. For more see Turing tarpit.
Example: The single register machine instruction "INC 3" â€“ increment the contents of register #3, i.e. increase its count by 1 â€“ can be atomized into the 0parameter instruction set of SchÃ¶nhage, but with the equivalent number of steps to accomplish the task increasing to 7; this number is directly related to the register number â€œnâ€� i.e. 4+n):
More examples can be found at the pages Register machine and Random access machine where the addition of "convenience instructions" CLR h and COPY h_{1},h_{1} are shown to reduce the number of steps dramatically. Indirect addressing is the other significant example.
Precise specification of Turingmachine algorithm m+n
As described in Algorithm characterizations per the specifications of BoolosBurgessJeffrey (2002) and Sipser (2006), and with a nod to the other characterizations we proceed to specify:
 (i) Number format: unary strings of marked squares (a "marked square" signfied by the symbol 1) separated by single blanks (signified by the symbol B) e.g. â€œ2,3â€� = B11B111B
 (ii) Machine type: Turing machine: singletape leftended or noended, 2symbol { B, 1 }, 4tuple instruction format.
 (iii) Head location: See more at â€œImplementation Descriptionâ€� below. A symbolic representation of the head's location in the tape's symbol string will put the current state to the right of the scanned symbol. Blank squares may be included in this protocol. The state's number will appear with brackets around it, or subscripted. The head is shown as
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Answers:John is 2 years older than Jack. The sum of their age is 22. Find their ages. answer 10 and 12 years old . . . . solve this on your own A is 100 meters ahead to B from the start. The rate of A is 10 meters /sec and B rate is 12 meters/sec. At what distance will B overtake A and at what time. answer t = 50 sec , distance of B is 600 A number of two digit has a unit twice the tens. if the arrangement of the numbers is interchanged the new number is 36 higher than the original number. answer 48 A quarter dollar is 1 more in number than the dimes. The total money is 270 cents. Find the number of dimes and quarter. answer 7 and 8 A can finished the work alone in 6 days. B can finished the same work alone in 4 days. If they work together from start, In how many days can they finished the job? answer 2.4 days
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