《算法导论(第二版)》(中文版)课后答案
算法导论课程作业答案
算法导论课程作业答案Introduction to AlgorithmsMassachusetts Institute of Technology 6.046J/18.410J Singapore-MIT Alliance SMA5503 Professors Erik Demaine,Lee Wee Sun,and Charles E.Leiserson Handout10Diagnostic Test SolutionsProblem1Consider the following pseudocode:R OUTINE(n)1if n=12then return13else return n+R OUTINE(n?1)(a)Give a one-sentence description of what R OUTINE(n)does.(Remember,don’t guess.) Solution:The routine gives the sum from1to n.(b)Give a precondition for the routine to work correctly.Solution:The value n must be greater than0;otherwise,the routine loops forever.(c)Give a one-sentence description of a faster implementation of the same routine. Solution:Return the value n(n+1)/2.Problem2Give a short(1–2-sentence)description of each of the following data structures:(a)FIFO queueSolution:A dynamic set where the element removed is always the one that has been in the set for the longest time.(b)Priority queueSolution:A dynamic set where each element has anassociated priority value.The element removed is the element with the highest(or lowest)priority.(c)Hash tableSolution:A dynamic set where the location of an element is computed using a function of the ele ment’s key.Problem3UsingΘ-notation,describe the worst-case running time of the best algorithm that you know for each of the following:(a)Finding an element in a sorted array.Solution:Θ(log n)(b)Finding an element in a sorted linked-list.Solution:Θ(n)(c)Inserting an element in a sorted array,once the position is found.Solution:Θ(n)(d)Inserting an element in a sorted linked-list,once the position is found.Solution:Θ(1)Problem4Describe an algorithm that locates the?rst occurrence of the largest element in a?nite list of integers,where the integers are not necessarily distinct.What is the worst-case running time of your algorithm?Solution:Idea is as follows:go through list,keeping track of the largest element found so far and its index.Update whenever necessary.Running time isΘ(n).Problem5How does the height h of a balanced binary search tree relate to the number of nodes n in the tree? Solution:h=O(lg n) Problem 6Does an undirected graph with 5vertices,each of degree 3,exist?If so,draw such a graph.If not,explain why no such graph exists.Solution:No such graph exists by the Handshaking Lemma.Every edge adds 2to the sum of the degrees.Consequently,the sum of the degrees must be even.Problem 7It is known that if a solution to Problem A exists,then a solution to Problem B exists also.(a)Professor Goldbach has just produced a 1,000-page proof that Problem A is unsolvable.If his proof turns out to be valid,can we conclude that Problem B is also unsolvable?Answer yes or no (or don’t know).Solution:No(b)Professor Wiles has just produced a 10,000-page proof that Problem B is unsolvable.If the proof turns out to be valid,can we conclude that problem A is unsolvable as well?Answer yes or no (or don’t know).Solution:YesProblem 8Consider the following statement:If 5points are placed anywhere on or inside a unit square,then there must exist two that are no more than √2/2units apart.Here are two attempts to prove this statement.Proof (a):Place 4of the points on the vertices of the square;that way they are maximally sepa-rated from one another.The 5th point must then lie within √2/2units of one of the other points,since the furthest from the corners it can be is the center,which is exactly √2/2units fromeach of the four corners.Proof (b):Partition the square into 4squares,each with a side of 1/2unit.If any two points areon or inside one of these smaller squares,the distance between these two points will be at most √2/2units.Since there are 5points and only 4squares,at least two points must fall on or inside one of the smaller squares,giving a set of points that are no more than √2/2apart.Which of the proofs are correct:(a),(b),both,or neither (or don’t know)?Solution:(b)onlyProblem9Give an inductive proof of the following statement:For every natural number n>3,we have n!>2n.Solution:Base case:True for n=4.Inductive step:Assume n!>2n.Then,multiplying both sides by(n+1),we get(n+1)n!> (n+1)2n>2?2n=2n+1.Problem10We want to line up6out of10children.Which of the following expresses the number of possible line-ups?(Circle the right answer.)(a)10!/6!(b)10!/4!(c) 106(d) 104 ·6!(e)None of the above(f)Don’t knowSolution:(b),(d)are both correctProblem11A deck of52cards is shuf?ed thoroughly.What is the probability that the4aces are all next to each other?(Circle theright answer.)(a)4!49!/52!(b)1/52!(c)4!/52!(d)4!48!/52!(e)None of the above(f)Don’t knowSolution:(a)Problem12The weather forecaster says that the probability of rain on Saturday is25%and that the probability of rain on Sunday is25%.Consider the following statement:The probability of rain during the weekend is50%.Which of the following best describes the validity of this statement?(a)If the two events(rain on Sat/rain on Sun)are independent,then we can add up the twoprobabilities,and the statement is true.Without independence,we can’t tell.(b)True,whether the two events are independent or not.(c)If the events are independent,the statement is false,because the the probability of no rainduring the weekend is9/16.If they are not independent,we can’t tell.(d)False,no matter what.(e)None of the above.(f)Don’t know.Solution:(c)Problem13A player throws darts at a target.On each trial,independentlyof the other trials,he hits the bull’s-eye with probability1/4.How many times should he throw so that his probability is75%of hitting the bull’s-eye at least once?(a)3(b)4(c)5(d)75%can’t be achieved.(e)Don’t know.Solution:(c),assuming that we want the probability to be≥0.75,not necessarily exactly0.75.Problem14Let X be an indicator random variable.Which of the following statements are true?(Circle all that apply.)(a)Pr{X=0}=Pr{X=1}=1/2(b)Pr{X=1}=E[X](c)E[X]=E[X2](d)E[X]=(E[X])2Solution:(b)and(c)only。
《算法导论(第二版)》(中文版)课后答案
26.1-2
26.2-1
26.2-4
14
《算法导论（第二版） 》参考答案 26.3-1 s–1–6–t s–2–8–t s–3–7–t 26.3-3
26.4-2 根据 26.22 对每个顶点执行 relabel 操作的次数最多为 2|V|-1 所以对每条边检测次数最多为 2*（2|V|-1）次 O（V） 在剩余网络中最多 2|E|条边 O（E） 所以为 O（VE）
f1[ j ] f 2 [ j 1] t2, j 1 a1, j f 2 [ j ] f1[ j 1] t1, j 1 a2, j
所以有：
f1[ j ] f 2 [ j ] f 2 [ j 1] t2, j 1 a1, j f1[ j 1] t1, j 1 a2, j
6
《算法导论（第二版） 》参考答案
最终答案：((A1A2)((A3A4)(A5A6))) 15.2-2
15.3-2 没有重叠的子问题存在 15.3-4 这题比较简单,由 P328 的(15.6)(15.7)展开两三步就可以看出来. 15.4-3 先初始化 c 数组元素为无穷大
7
《算法导论（第二版） 法， 最好是像书上画一颗递归树然后进行运算。
1
《算法导论（第二版） 》参考答案
注意题目中的要求使用递归树的方法，最好是像书上画一颗递归树然后进行运 算。 4.2.2 证略 4.2.3 由 2 i n 得 i=lgn
T (n) 2 i cn cn
i 0
由课本 P328（15.6） (15.7)式代入，可得：
min( f1[ j 1] a1, j , f 2 [ j 1] t2, j 1 a1, j ) min( f 2 [ j 1] a2, j , f1[ j 1] t1, j 1 a2, j ) min( f1[ j 1], f 2 [ j 1] t2, j 1 ) a1, j min( f 2 [ j 1], f1[ j 1] t1, j 1 ) a2, j f 2 [ j 1] t2, j 1 a1, j f1[ j 1] t1, j 1 a2, j
计算机科学导论原书第二版答案第十二章汇编
}
return true
}
32. Algorithm S12.32 shows the pseudocode. Algorithm S12.32 Exercise 32
Algorithm: CompareStack(S1, S2)
Purpose: Check if two stacks are the same Pre: Given: S1 and S2 Post: Return: true (S1 = S2) or false (S1 ≠ S2) {
}
if (NOT empty (S1) or NOT empty (S2))
flag ← false
while (NOT empty (Temp1) and NOT empty (Temp2))
{ pop (Temp1, x) push (S1, x) pop (Temp2, y) push (S2, y)
算法语言书后习题参考答案(第二版)
算法语言书后习题参考答案(第二版)第一章1(1)A (2)D (3)D (4)AC第二章1、(1)AC (2)BD (3)CD (4)C (5)B (6)D2、(2)Hi FrankWelcome(3)3 Apr Wed第三章1(1) A (2) AB (3) AD2(5)方法MyMethod1的重载形式不合法，因为访问限制修饰符和返回类型都不是方法的标识。
方法MyMethod2的重载形式合法。
(6) 运行结果为x=0, y=5, z=0x=0, y=5, z=5(8)应将public static readonly InnerClasses inner;改成public static readonly InnerClasses inner=new InnerClasses();运行结果为7913第四章1（1）C（2）A（3）BC（4）A（5）A第(5)涉及到对负数的移位，不要求掌握对负数的移位，但要求掌握对正数的移位。
2（10）true true false true false第五章1（1）没有一个是正确的（2）C（3）D（4）A（5）B（6）D（7）BC第(6)题应改为：下面语句所计算的x数学表达式为_______。
for (int x = 0, y = 1, z = 1; z <= 10; x += y, y *= ++z)第(7)的选项A应改为：A. int x = 1, y = 0; while (true) if ((x += (y++)) > 100) break;第六章1（1）D（2）A（3）D（4）不作要求（5）AC2（7）没有错误。
但它在属性的get访问函数中修改了所封装的字段值，这违反了访问函数的设计原则，容易引起误解。
应避免使用这类代码。
第七章1（1）C（2）没有一个正确（3）B（4）AC (5) ABD(不作要求)2 (9) 代码中存在一处错误，派生类B的属性Y在重载基类A的属性Y时，必须同时重载其get和set访问函数。
算法设计与分析第二版课后习题及解答（可编辑）
算法设计与分析第二版课后习题及解答算法设计与分析基础课后练习答案习题1.14.设计一个计算的算法,n是任意正整数。
除了赋值和比较运算,该算法只能用到基本的四则运算操作。
算法求 //输入:一个正整数n2//输出:。
step1:a1; step2:若a*an 转step 3,否则输出a; step3:aa+1转step 2;5. a.用欧几里德算法求gcd(31415,14142)。
b. 用欧几里德算法求gcd(31415,14142),比检查min{m,n}和gcd(m,n)间连续整数的算法快多少倍?请估算一下。
a. gcd31415, 14142 gcd14142, 3131 gcd3131, 1618 gcd1618, 1513 gcd1513, 105 gcd1513, 105 gcd105, 43 gcd43, 19 gcd19, 5 gcd5, 4 gcd4, 1 gcd1, 0 1.b.有a可知计算gcd(31415,14142)欧几里德算法做了11次除法。
连续整数检测算法在14142每次迭代过程中或者做了一次除法,或者两次除法,因此这个算法做除法的次数鉴于1?14142 和 2?14142之间,所以欧几里德算法比此算法快1?14142/11 ≈1300 与2?14142/11 ≈ 2600 倍之间。
6.证明等式gcdm,ngcdn,m mod n对每一对正整数m,n都成立.Hint:根据除法的定义不难证明:如果d整除u和v, 那么d一定能整除u±v;如果d整除u,那么d也能够整除u的任何整数倍ku.对于任意一对正整数m,n,若d能整除m和n,那么d一定能整除n和rm mod nm-qn;显然,若d能整除n和r,也一定能整除mr+qn和n。
数对m,n和n,r具有相同的公约数的有限非空集,其中也包括了最大公约数。
故gcdm,ngcdn,r7.对于第一个数小于第二个数的一对数字,欧几里得算法将会如何处理?该算法在处理这种输入的过程中,上述情况最多会发生几次?Hint:对于任何形如0mn的一对数字,Euclid算法在第一次叠代时交换m和n, 即gcdm,ngcdn,m并且这种交换处理只发生一次.8.a.对于所有1≤m,n≤10的输入, Euclid算法最少要做几次除法?1次b. 对于所有1≤m,n≤10的输入, Euclid算法最多要做几次除法?5次gcd5,8习题1.21.农夫过河P?农夫W?狼 G?山羊 C?白菜2.过桥问题1,2,5,10---分别代表4个人, f?手电筒4. 对于任意实系数a,b,c, 某个算法能求方程ax^2+bx+c0的实根,写出上述算法的伪代码可以假设sqrtx是求平方根的函数算法Quadratica,b,c//求方程ax^2+bx+c0的实根的算法//输入:实系数a,b,c//输出:实根或者无解信息If a≠0D←b*b-4*a*cIf D0temp←2*ax1←-b+sqrtD/tempx2←-b-sqrtD/tempreturn x1,x2else if D0 return ?b/2*ael se return “no real roots”else //a0if b≠0 return ?c/belse //ab0if c0 return “no real numbers”else return “no real roots”5. 描述将十进制整数表达为二进制整数的标准算法a.用文字描述b.用伪代码描述解答:a.将十进制整数转换为二进制整数的算法输入:一个正整数n输出:正整数n相应的二进制数第一步:用n除以2,余数赋给Kii0,1,2,商赋给n第二步:如果n0,则到第三步,否则重复第一步第三步:将Ki按照i从高到低的顺序输出b.伪代码算法 DectoBinn//将十进制整数n转换为二进制整数的算法//输入:正整数n//输出:该正整数相应的二进制数,该数存放于数组Bin[1n]中i1while n!0 doBin[i]n%2;nintn/2;i++;while i!0 doprint Bin[i];i--;9.考虑下面这个算法,它求的是数组中大小相差最小的两个元素的差.算法略对这个算法做尽可能多的改进.算法 MinDistanceA[0..n-1]//输入:数组A[0..n-1]//输出:the smallest distance d between two of its elements 习题1.3考虑这样一个排序算法,该算法对于待排序的数组中的每一个元素,计算比它小的元素个数,然后利用这个信息,将各个元素放到有序数组的相应位置上去.a.应用该算法对列表”60,35,81,98,14,47”排序b.该算法稳定吗?c.该算法在位吗?解:a. 该算法对列表”60,35,81,98,14,47”排序的过程如下所示:b.该算法不稳定.比如对列表”2,2*”排序c.该算法不在位.额外空间for S and Count[]4.古老的七桥问题第2章习题2.17.对下列断言进行证明:如果是错误的,请举例a. 如果tn∈Ogn,则gn∈Ωtnb.α0时,Θαgn Θgn解:a这个断言是正确的。
计算方法教程(第2版)习题答案
《计算方法教程(第二版)》习题答案第一章 习题答案1、浮点数系),,,(U L t F β共有 1)1()1(21++---L U t ββ 个数。
3、a ．4097b ．62211101110.0,211101000.0⨯⨯c ．6211111101.0⨯ 4、设实数R x ∈，则按β进制可表达为：,1,,,3,2,011)11221(+=<≤<≤⨯++++++±=t t j jd d l t t d t t d dd x βββββββ按四舍五入的原则，当它进入浮点数系),,,(U L t F β时，若β211<+t d ，则 l tt d dd x fl ββββ⨯++±=)221()(若 β211≥+t d ，则 l tt d d d x fl ββββ⨯+++±=)1221()(对第一种情况：t l lt l t t d x fl x -++=⨯≤⨯+=-βββββ21)21(1)()(11对第二种情况：t l lt l t t d x fl x -++=⨯≤⨯--=-ββββββ21)21(1)(11就是说总有： tl x fl x -≤-β21)( 另一方面，浮点数要求 β<≤11d ， 故有l x ββ1≥，将此两者相除，便得t x x fl x -≤-121)(β 5、a ． 5960.1 b ． 5962.1 后一种准确6、最后一个计算式：00025509.0原因：避免相近数相减，避免大数相乘，减少运算次数7、a ．]!3)2(!2)2(2[2132 +++=x x x yb ．)21)(1(22x x x y ++=c ．)11(222-++=x x x yd ． +-+-=!2)2(!6)2(!4)2(!2)2(2642x x x x y e ．222qp p q y ++=8、01786.098.5521==x x9、 m )10(m f - 1 233406.0- 3 20757.0- 5 8.07 710计算宜采用：])!42151()!32141()!22131[()(2432+⨯-+⨯-+⨯--=x x x f第二章 习题答案1、a ．Tx )2,1,3(= b ．Tx )1,2,1,2(--= c ．无法解 2、a ．与 b ．同上， c ．T T x )2188.1,3125.0,2188.1,5312.0()39,10,39,17(321---≈---=7、a ．⎪⎪⎪⎪⎭⎫ ⎝⎛--⎪⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎪⎭⎫ ⎝⎛--⎪⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛---14112111473123247212122123211231321213122 b ． ⎪⎪⎪⎪⎪⎭⎫⎝⎛--⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎪⎪⎪⎭⎫⎝⎛----333211212110211221213231532223522121⎪⎪⎪⎪⎪⎭⎫ ⎝⎛--⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-=111211212130213221219、T x )3415.46,3659.85,1220.95,1220.95,3659.85,3415.46(1= T x )8293.26,3171.7,4390.2,4390.2,3171.7,8293.26(2= 10、T LDL 分解：)015.0,579.3,9.1,10(diag D =⎪⎪⎪⎪⎪⎭⎫⎝⎛=16030.07895.05.018947.07.019.01L Cholesky 分解⎪⎪⎪⎪⎪⎭⎫⎝⎛=1225.01408.10833.15811.18918.12333.12136.23784.18460.21623.3G 解：)1,1,2,2(--=x 12、16,12,1612111===∞A A A611,4083.1,61122212===∞A A A2)(940)()(12111===∞A Cond A Cond A Cond524)(748)()(22221===∞A C o n d A C o n d A C o n d⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛=--180.0000180.0000- 30.0000 180.0000- 192.0000 36.0000- 30.0000 36.0000- 9.0000,0.0139 0.1111- 0.0694- 0.1111- 0.0556 0.1111- 0.0694- 0.1111- 0.0139 1211A A1151.372,1666.0212211==--A A15、 1A ：对应 Seidel Gauss - 迭代收敛，Jacobi 迭代不收敛； 2A ：对应 Seidel Gauss - 迭代收敛，Jacobi 迭代不收敛； 3A ：对应 Seidel Gauss - 迭代收敛，Jacobi 迭代收敛；第三章 习题答案1、Lagrange 插值多项式：)80.466.5)(20.366.5)(70.266.5)(00.166.5()80.4)(20.3)(70.2)(00.1(7.51)66.580.4)(20.380.4)(70.280.4)(00.180.4()66.5)(20.3)(70.2)(00.1(3.38)66.520.3)(80.420.3)(70.220.3)(00.120.3()66.5)(80.4)(70.2)(00.1(0.22)66.570.2)(80.470.2)(20.370.2)(00.170.2()66.5)(80.4)(20.3)(00.1(8.17)66.500.1)(80.400.1)(20.300.1)(70.200.1()66.5)(80.4)(20.3)(70.2(2.14)(4--------⨯+--------⨯+--------⨯+--------⨯+--------⨯=x x x x x x x x x x x x x x x x x x x x x L Newton 插值多项式：)80.4)(20.3)(70.2)(00.1(21444779.0)20.3)(70.2)(00.1(527480131.0)70.2)(00.1(855614973.2)00.1(117647059.22.14)(4----+------+-+=x x x x x x x x x x x N2、设)(x y y =，其反函数是以y 为自变量的函数)(y x x =，对)(y x 作插值多项式：)1744.0)(1081.0)(4016.0)(7001.0(01253.0)1081.0)(4016.0)(7001.0(01531.0)4016.0)(7001.0(009640.0)7001.0(3350.01000.0)(----+---+--+--=y y y y y y y y y y y N 3376.0)0(=N 是0)(=x y 在]4.0,3.0[中的近似根。
《算法导论》习题答案
n/2
n! nn , n! o(nn )
3.2.4 是否多项式有界 lg n !与 lg lg n ！
设lgn=m，则 m! 2 m ( )m e2m ( )m em(ln m1) mln m1 nln ln n
∴lgn！不是多项式有界的。
T (n) O(lg n)
4.1.2 证明 T (n) 2T (n) n 的解为 O(n lg n)
设 T (n) c n lg n
T (n) 2c n lg n n c lg n n n c(n 1) lg(n / 2) n cn lg n c lg n cn n cn(lg n 1) n c(lg n 2n)
虽然用二分查找法可以将查找正确位置的时间复杂度降下来，但 是移位操作的复杂度并没有减少， 所以最坏情况下该算法的时间复杂 度依然是 (n2 )
2.3-7 给出一个算法， 使得其能在 (n lg n) 的时间内找出在一个 n 元
素的整数数组内，是否存在两个元素之和为 x
首先利用快速排序将数组排序，时间 (n lg n) ，然后再进行查找：
sin(n / 2) 2 1，所以 af (n / b) cf (n) 不满足。 2(sin n 2)
4.1.6 计算 T (n) 2T (
令 m lg n, T (2 ) 2T (2
m m/ 2
n ) 1 的解
) 1
令 T(n)=S(m),则 S (m) 2S (m / 2) 1 其解为 S (m) (m),T (n) S (m) (lg n)
4.2 The recursion-tree method 4.2.1 4.2.2 4.2.3 4.2.5 略
算法导论32章答案
算法导论32章答案32 String Matching32.1-2Suppose that all characters in the pattern P are different. Show how to accelerate NAIVE-STRING-MATCHER to run in timeO.n/ on an n-character text T.Naive-Search(T,P)for s = 1 to n – m + 1j = 0while T[s+j] == P[j] doj = j + 1if j = m return ss = j + s;该算法实际只是会扫描整个字符串的每个字符⼀次，所以其时间复杂度为O(n).31.1-3Suppose that pattern P and text T are randomly chosen strings of length m and n, respectively, from the d-ary alphabet ∑d ={0,1,2,..,d-1},where d ≧ 2.Show that the expected number of character-to-character comparisons made by the implicit loop inline 4 of the naive algorithm isover all executions of this loop. (Assume that the naive algorithm stops comparing characters for a given shift once it finds amismatch or matches the entire pattern.) Thus, for randomly chosen strings, the naive algorithm is quite efficient.当第4⾏隐含的循环执⾏i次时，其概率P为：P = 1/K i-1 * (1-1/k), if i < mP = 1/K m-1 * (1-1/k) + 1/K m , if i = m可以计算每次for循环迭代时，第4⾏的循环的平均迭代次数为：[1*(1-1/k)+2*(1/K)*(1-1/k)+3*(1/k2)(1-1/k)+…+(m-1)*(1-k m-2)(1-1/k) +m*(1/k m-1)(1-1/k) + m*(1/k m)]= 1 - 1/k + 2/k - 2/k2 + 3/k2 - 3/k3 +...+ m/k m-1 - m/k m + m/k m= 1 + 1/k + 1/k2 +...+ 1/k m-1= (1 - 1/K m) / (1 - 1/k)≤ 2所以，可知，第4⾏循环的总迭代次数为：(n-m+1) * [(1-1/K m) / (1-1/k)] ≤ 2 (n-m+1)31.1-4Suppose we allow the pattern P to contain occurrences of a gap character } that can match an arbitrary string of characters(even one of zero length). For example, the pattern ab}ba}c occurs in the text cabccbacbacab asand asNote that the gap character may occur an arbitrary number of times in the pattern but not at all in the text. Give a polynomial-time algorithm to determine whether such a pattern P occurs in a given text T, and analyze the running time of your algorithm.该算法只是要求判断是否模式P出现在该字符串中，那么问题被简化了许多。