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Course: CS 504, Fall 2008
School: Uni. Worcester
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: Ex Bucket Sorting (Gonnet, 4.2.3, pg. 176) (1-pass version) Assume you know the distribution of a set of keys to be sorted. That is, assume we have M buckets that partition the key space, M keys, and Pr{key belongs in bucket bm} = 1/M , 1m M . for m :=1 to M do initialize bucket bm (M) for each key do INSERT(key, appropriate bucket) (M) InsertionSort each bucket (length of bucket2) Concatenate buckets 1,2,...M...

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: Ex Bucket Sorting (Gonnet, 4.2.3, pg. 176) (1-pass version) Assume you know the distribution of a set of keys to be sorted. That is, assume we have M buckets that partition the key space, M keys, and Pr{key belongs in bucket bm} = 1/M , 1m M . for m :=1 to M do initialize bucket bm (M) for each key do INSERT(key, appropriate bucket) (M) InsertionSort each bucket (length of bucket2) Concatenate buckets 1,2,...M (M) InsertionSort requires time quadratic in # items in bucket. Let nm, 1m M , be a r.v. denoting the number of keys in bucket bm. Total expected time to InsertionSort the M buckets: 1m M 2 E[nm ] V(nm) = E[nm2] - (E[nm])2 ==> E[nm2] = V (nm) + (E[nm])2 E[nm]? V (nm)? A Bernoulli Trial is an experiment with two possible outcomes: success, and failure, with Pr{success}=p, Pr{failure}=1-p =q. For Bucket Sort, a trial entails picking a bucket, b*, and testing whether a key -> b*. Define rv Xm = if keym ->b* then 1 else 0. nb = Xm 0 for k <0 or k >M . M k M -k Binomial Distribution : Pr{nb =k}= p q k M k M -k Note that it's legal since Pr{nb =k}= 0 for all k and p q k M M k pk q M -k = k pk q M -k = ( p + q) M = 1 k 0k M M E[nb]= k * Pr{nb = k} = k p k q M -k k 0k M 0k M M M M - 1 Get rid of k multiplier: Absorption/extraction = k k k -1 M - 1 k M -k M - 1 k M -k E[nb]= M =M p q p q k -1 k -1 0k M 0k M (make exponent of p match lower term of binomial) M - 1 k -1 M -k = Mp p q k -1 0k M (replace k by k+1) M - 1 k M -1- k M - 1 k M -1- k =Mp = Mp p q p q 0 k +1 M k -1 k M k What is Pr{nb =k}? 1m M 1 M - 1 = 0) -1 M - 1 k M -1- k Mp =Mp p q k 0 k M 0 k M Another approach: nb = Xm , so = (because M - 1 k M p q k -1 -1- k = M p (p + q )M -1 =Mp (Xm) = p- p2 = p(1-p) = pq Theorem : If the M r.v.s are independent, then V Xm = V(Xm ) 1m M 1m M So for Bucket Sort, V Xm = Mpq , and the expected time to Insertion 1m M Sort bucket bm = E[nm2]=V (nm)+(E[nm])2=Mpq + Mp =Mp(2-p) 1 1 Noting that p = , this is 2 - = constant time to Insertion Sort each M M bucket, & (expected) linear time to Bucket Sort the array. Ex : Assume n bits to be transmitted across a channel, Pr{error-free transmission of a bit} = p, & errors are independent. If code capable of detecting/correcting up to e errors, then probability of correct n transmission of the n bits = k q k pn-k 0ke Ex : Choosing a leader in a ring Itai & Rodeh, Symmetry Breaking in nd FOCS, 1981 Distributive Networks , 22 Given a ring of n identical processors, in the same state, but each processor endowed with a (non-psuedo) random number generator, they must all agree on a leader. Each processor can only transmit 1-bit message each time. Initially, all processors are active. loop Each processor draws from U(1..#active processors ) independently of the others. All processors that chose 1 send this message. if 0 processors chose 1 then repeat the loop if 1 processor chose 1 then it's the leader if >1 processors chose 1 then processors that didn't choose 1 become inactive & the loop is repeated n 1 k n - 1 n-k (n, k) = Let Pr{k processors chose 1}= k n n Let (n) - expected number of passes through the loop E[nb ] = E Xm = E[Xm ] = p = Mp 1m M 1m M 1m M 2] - (E[Xm])2 V (nm)? V (Xm) = E[Xm E[X 2] = 12p + 02q = p, (E[X ])2 = p2, V 1m M m m 2 (n) = 1 + (n,0)*(n) + ...Results: For all n 2, (n)<e =2.718... 2kn (n, k) * (k) ==> (n) = n 1+ 1 - (n,0) - (n,n) lim (n) < 2.442 2kn-1 (n, k) * (k) Ex : Counting distinct words Flajolet & Martin, "Probabilistic Counting Algorithms for Database Applications", J. Comp. Syst. Sciences , 1985 Assume there are n (not necessarily distinct) words in a text. Let there {0,1}5+lgn . If s is a string of bits, let be a hashing function h :{words} ( x,b), b {0,1}, denote the index of the leftmost bit of x equal to b signature =s1s2...s5+lgn = 00...0 for each word x do signature[(h(x),1)] := 1 return((signature,0)) 2k If algorithm returns k, estimate number of distinct words= 1.5470 If (signature,0)=0, then leftmost bits of signature =1110. If there are 16 distinct words on the tape, 16 15 =.356 Pr{none of hash codings begins 0001}= 16 Ex :2-statistic Given a sample space with K events (for dice, the K =11 events are Xi {2,...,12}) and a series of n independent experiments, we expect each event i to occur n*Pr{Xi} times. n =10000 throws of dice Event Pr{Xi} n*Pr{Xi} Observed Number Yi 2 1/36 277.777780 269 3 1/18 555.555560 520 4 1/12 833.333358 849 5 1/9 1111.111119 1088 6 5/36 1388.888955 1382 7 1/6 1666.666716 1664 8 5/36 1388.888955 1429 9 1/9 1111.111119 1112 10 1/12 833.333358 836 11 1/18 555.555560 581 12 1/36 277.777780 270 2 Statistic: (Yi - n * Pr{Xi}) For loaded dice, this statistic should be large; 1i K Problem : If Pr{Xi}>>Pr{Xj}, then probably (Yi - n *Pr{Xi})2 >> (Yj - n *Pr{Xj})2 . (Yi - n * Pr{Xi})2 2-statistic: n * Pr{X } i 1i K fair small. 3 For above dice data, 2-statistic = 5.917396 99% 95% 75% 50% 25% 5% 1% n=1 .00016 .00393 .1015 .4549 1.323 3.841 6.635 n=9 2.088 3.325 5.899 8.343 11.39 16.92 21.67 n=10 2.558 3.940 6.737 9.342 12.55 18.31 23.21 For dice data, there are K =11 categories, hence n=10 degrees of freedom (Y11 depends upon Y1,...,Y10). The 2-statistic should be > 5.917396 95% between & 75% of the time, so our deviation is perfectly reasonable (we can not reject the hypothesis that our dice are fair). Geometric Distribution : Let r.v. X denote the number of Bernoulli trials until success . X has values in {1,2,3,...}. Pr{X =k} = qk-1p It's a legal distribution since Pr{X =k} 0 k 0 and p p k -1 k -1 k Pr{X = k} = q p = p q = p q = 1 - q = p = 1 1k 0k 1k 1k p k -1 k -1 E[X] = kq p = p kq = p kq k -1 = kq k q 0k 1k 0k 1k Tangent : kq k = q +2q2+3q3+... 0k 1 = 1+q +q2+q3+... 1- q d d f-f* g g* 1 d 1 d f dx dx 2 3 = = 2 =1+2q +3q +4q +... because 2 dq 1 - q (1 - q) dx g g q = q +2q2+3q3+4q4+... (end of tangent ) (1 - q)2 p q q V (X )= 2 E[X] = 2 = 1/p , q (1 - q) p Geometric waiting times are memoryless (& the geometric is the only memoryless distribution). If equipment's life span is geometric, then it doesn't age: -waiting time in flipping a coin until heads -duration of phone conversations within a city -radioactive decay -life expectancy of an adult fish -aging of glass labware Theorem : If a (discrete) process is memoryless, then it's geometric. Proof : If memoryless, Pr{X =k |X >k -1} = Pr{X =1}, 1k Pr{X = k} By Bayes's Rule, Pr{X =k |X >k -1} = = Pr{X =1} Pr{X > k - 1} Pr{X =k} = Pr{X >k-1} - Pr{X >k} (Pr{X >k-1} - Pr{X >k})/Pr{X >k-1} = Pr{X=1} 1-Pr{X=1} = Pr{X >k}/Pr{X >k-1} = Pr{X >k|X >k -1} = q for all k . 4 Ex : In hashing (to b buckets) with collision resolution by chaining, how many keys must be inserted until every bucket contains a key? A hit is th stage is the sequence of when a bucket gets its first key, and the k st hit until the k th hit. The first stage consists inserts between the (k-1) th stage, Pr{hit}= b - k + 1 . Let nk be a r.v. of the first insert. For the k b th stage. The number of inserts denoting the number of inserts during the k for every bucket to get a hit is n = nk . Because each nk is geometrically b - k +1 b distributed (with Pr{success}= ), E[nk]= . By linearity of b b - k +1 b 1 expectation, E[n] = E nk = E[nk ] = =b = bHb b - k +1 k 1kb 1kb 1kb 1kb 1kb Poisson Distribution Let X be an r.v. denoting the number of events in a given time interval (or region). X can assume values {0,1,2,...}. Let pk(t)=Pr{X =k during time interval t}. Assume: -# events in non-overlapping intervals independent. -For small dt : p1(dt) = dt + o(dt) pk (dt) = o(dt) 2k Given independence, p0(t +dt) = p0(t)*p0(dt) p0(dt) = 1 pk (dt) = 1 - ( dt +o (dt)+o (dt))=1- dt -2o(dt) p0(t +dt) = p0(t)*[1- 1k dt -2o(dt)] p0 (t + dt) - p0 (t) 2o(dt) p0 (t) = - p0(t) - p0(t) dt dt Combined with p0(0)=1 yields p0(t)=e -t p1(t +dt) = p1(t)p0(dt) + p0(t )p1(dt ) = p1(t)[1-p1(dt)] + p0(t)p1(dt) p1(t +dt) - p1(t) = - p1(t)p1(dt) + p0(t)p1(dt) = - p1(t) dt + p0(t) dt p1 (t + dt) - p1 (t) = - p1 + p0(t) dt ( t)n - t e (Poisson Distribution) n! p1(t) = ( t)e -(t) ... pn(t) = Legal distribution since: - pn(t)0 for all n 0 ( t)n - t - t ( t)n - pn (t) = e =e = e -t e t = 1 n! n! 0n 0n 0n 5 Ex :Flying bomb hits in London during WW2 in regions of area 1/4 km2 t = 0.9323 km2 n 0 1 2 3 4 5 Yn 229 211 93 35 7 1 npn 226.74 211.39 98.54 3...

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Uni. Worcester - CS - 504
Read : GKP 8.1,8.2 Quicksort is the sorting technique of choice although its worst-case behavior is inferior to that of many others. Why? What does average case mean? Seeking a phone number (which exists) in a phone book sees &quot;on average&quot; half the el
Uni. Worcester - CS - 504
Given set S, assume each s S has a weight |s|. A GF of S is F ( z) = S an = {s S: s = n} , F ( z) = an z n S NOTE: Since S = a n , S = FS (1) .n ns Szs. LettingEx: In how many ways can we give money using 1 coin? (z + z 5 + z 10 + z 25
Uni. Worcester - CS - 504
CS504: Analysis of Computations and Systems Spring 1999Homework VIDue: March 8 or 10, 1999One more recurrence1. Solve the second-order linear recurrence xn 1 xn 312 xn 32nn2x0a x1bGenerating Functions2. Here is a different w
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems Spring 1999Homework VIDue: March 8 or 10, 1999One more recurrence1. Solve the second-order linear recurrence xn 1 xn 312 xn 32nn2x0a x1bGenerating Functions2. Here is a different w
Uni. Worcester - CS - 9906
CS504: Analysis of Computations and Systems Spring 1999Homework VIDue: March 8 or 10, 1999One more recurrence1. Solve the second-order linear recurrence xn 1 xn 312 xn 32nn2x0a x1bGenerating Functions2. Here is a different w
Uni. Worcester - CS - 504
C.S.504 H.W. #7Due: November 5, 1991 1. (3 points) Use the method of characteristic roots to solve the tn = tn -2 + 4n , n 2 recurrence: t 0 = 1, t 1 = 4 2. (3 points) Use the method of characteristic roots to solve the recurrence: tn = 2*tn -1 + (n
Uni. Worcester - CS - 91
C.S.504 H.W. #7Due: November 5, 1991 1. (3 points) Use the method of characteristic roots to solve the tn = tn -2 + 4n , n 2 recurrence: t 0 = 1, t 1 = 4 2. (3 points) Use the method of characteristic roots to solve the recurrence: tn = 2*tn -1 + (n
Uni. Worcester - CS - 504
C.S.504 H.W. #4Due: October 8, 19921. (1 point) Evaluate1 2 n2 +10n2 5 2. (1 point) Evaluate0m kn 3. (4 points) Assume that you're sequentially seeking a card in a perfectly shuffled deck of 106 distinct cards. The card is in the deck. A)W
Uni. Worcester - CS - 92
C.S.504 H.W. #4Due: October 8, 19921. (1 point) Evaluate1 2 n2 +10n2 5 2. (1 point) Evaluate0m kn 3. (4 points) Assume that you're sequentially seeking a card in a perfectly shuffled deck of 106 distinct cards. The card is in the deck. A)W
Uni. Worcester - CS - 504
C.S.504 H.W. #1Due: September 23, 19931. (2 points) What does the following algorithm compute? function f(n,m : integer) : integer; X m; K 0; while Xn do X X*m; K K+1; return(K); 2. (9 points) We want to find the maximum and the minimum element
Uni. Worcester - CS - 93
C.S.504 H.W. #1Due: September 23, 19931. (2 points) What does the following algorithm compute? function f(n,m : integer) : integer; X m; K 0; while Xn do X X*m; K K+1; return(K); 2. (9 points) We want to find the maximum and the minimum element
Uni. Worcester - CS - 504
C.S.504 H.W. #5Due: October 15, 1991 1.(5 points) Prove the equality on page 72 of our text that for direct chaining hashing,s 2 ( A n' ) =2. (3 points) Given two sorted lists,n ( m -1) m2a 1 ,., a n a b 1,.,b nbshow that any algorithm whi
Uni. Worcester - CS - 91
C.S.504 H.W. #5Due: October 15, 1991 1.(5 points) Prove the equality on page 72 of our text that for direct chaining hashing,s 2 ( A n' ) =2. (3 points) Given two sorted lists,n ( m -1) m2a 1 ,., a n a b 1,.,b nbshow that any algorithm whi
Uni. Worcester - CS - 504
C.S.504H.W. #4Due: March 25/26, 1998Read SECTIONS 5.1, 5.2, 5.3 (Trick 1) Do EXERCISEs 5.1, 5.2, 5.4, 5.15 from GKP. Do not submit your solutions, but check them with the answers from the back of the text. 1. (3 points) Find a closed form for coe
Uni. Worcester - CS - 98
C.S.504H.W. #4Due: March 25/26, 1998Read SECTIONS 5.1, 5.2, 5.3 (Trick 1) Do EXERCISEs 5.1, 5.2, 5.4, 5.15 from GKP. Do not submit your solutions, but check them with the answers from the back of the text. 1. (3 points) Find a closed form for coe
Uni. Worcester - CS - 504
C.S.504 H.W. #1Due: September 17, 1991 Assume that you're sequentially seeking a card in a perfectly shuffled deck of 106 distinct cards. Use Chebyshev's inequality to give a bound on the probability that you'll have to examine at least 900,000 card
Uni. Worcester - CS - 91
C.S.504 H.W. #1Due: September 17, 1991 Assume that you're sequentially seeking a card in a perfectly shuffled deck of 106 distinct cards. Use Chebyshev's inequality to give a bound on the probability that you'll have to examine at least 900,000 card
Uni. Worcester - CS - 504
C.S.504SOLUTION FOR H.W. #1k n n n n( n +1) n( n - 1) n 1 - 1 = ( n - k) = n - k = n 2 = 1. 1 = 2 2 k =1 j =k+ 1 k =1 j =1 j =1 k =1 k =1 k =1 n n n2. In fact, Gn = (-1)n+1Fn.0, if n = 0 Gn = if n = 1 1, G - G , if n &gt;1 n -2 n -13.
Uni. Worcester - CS - 98
C.S.504SOLUTION FOR H.W. #1k n n n n( n +1) n( n - 1) n 1 - 1 = ( n - k) = n - k = n 2 = 1. 1 = 2 2 k =1 j =k+ 1 k =1 j =1 j =1 k =1 k =1 k =1 n n n2. In fact, Gn = (-1)n+1Fn.0, if n = 0 Gn = if n = 1 1, G - G , if n &gt;1 n -2 n -13.
Uni. Worcester - CS - 504
C.S.504 H.W. #2Due: September 24, 1992(6 points) Consider the following procedure void InsertionSort(int A, int n) {int i, j,temp; for (i=1; i&lt;n; i+) { /* A[0.i-1] already sorted */ temp=A[i]; for (j=i-1; j&gt;=0 &amp; temp&lt;A[j]; j-) A[j+1]=A[j]; /*&lt;-*/ A
Uni. Worcester - CS - 92
C.S.504 H.W. #2Due: September 24, 1992(6 points) Consider the following procedure void InsertionSort(int A, int n) {int i, j,temp; for (i=1; i&lt;n; i+) { /* A[0.i-1] already sorted */ temp=A[i]; for (j=i-1; j&gt;=0 &amp; temp&lt;A[j]; j-) A[j+1]=A[j]; /*&lt;-*/ A
Uni. Worcester - CS - 504
C.S.504 H.W. #3Due: October 21, 1993 1. (5 points) We define an Isaac tree In recursively by: -I0 consists of a single node, -the Isaac tree In, n1, consists of two Isaac trees In-1 such that the root of one is the rightmost child of the root of the
Uni. Worcester - CS - 93
C.S.504 H.W. #3Due: October 21, 1993 1. (5 points) We define an Isaac tree In recursively by: -I0 consists of a single node, -the Isaac tree In, n1, consists of two Isaac trees In-1 such that the root of one is the rightmost child of the root of the
Uni. Worcester - CS - 504
C.S.504 H.W. #10Due: Tuesday, December 15, 1992 1. (8 points) One way to estimate the size of a set X ={ x 1 ,., x n } is to sample the elements of X from a uniform distribution with replacement until an element is sampled twice. The number of eleme
Uni. Worcester - CS - 92
C.S.504 H.W. #10Due: Tuesday, December 15, 1992 1. (8 points) One way to estimate the size of a set X ={ x 1 ,., x n } is to sample the elements of X from a uniform distribution with replacement until an element is sampled twice. The number of eleme
Uni. Worcester - CS - 504
C.S.504SOLUTION FOR H.W. #5 1. (A) If we restrict the composition to have one part (k=1), then the GF is z + + z r . For 1 an arbitrary number of parts (no restrictions on k), the GF is . 1- z - - z r 1 (B) fn ,2 = z n F2 ( z) = z n . Recognizing, y
Uni. Worcester - CS - 98
C.S.504SOLUTION FOR H.W. #5 1. (A) If we restrict the composition to have one part (k=1), then the GF is z + + z r . For 1 an arbitrary number of parts (no restrictions on k), the GF is . 1- z - - z r 1 (B) fn ,2 = z n F2 ( z) = z n . Recognizing, y
Uni. Worcester - CS - 504
C.S.504H.W. #2Due: February 11/12, 19981. (6 points) Find a closed form solution for the linear first-order nonhomogeneous recurrence with nonconstant coefficients 0,if n = 0 xn = n + 3 x + n + 3,if n &gt; 0 n +1 n -1 2 The first three terms are
Uni. Worcester - CS - 98
C.S.504H.W. #2Due: February 11/12, 19981. (6 points) Find a closed form solution for the linear first-order nonhomogeneous recurrence with nonconstant coefficients 0,if n = 0 xn = n + 3 x + n + 3,if n &gt; 0 n +1 n -1 2 The first three terms are
Uni. Worcester - CS - 504
C.S.504 H.W. #3Due: October 1,1991 1.(2 points) In Section 3.1.1 of our text, what is E[An ] when Pr{An =i } = if 1i n -1 then (1/2)i else if i =n then (1/2)n -1. 2. (3 points) Prove that0 k &lt;n 0k &lt;n for n 0. 3. (2 points) Suppose a program has
Uni. Worcester - CS - 91
C.S.504 H.W. #3Due: October 1,1991 1.(2 points) In Section 3.1.1 of our text, what is E[An ] when Pr{An =i } = if 1i n -1 then (1/2)i else if i =n then (1/2)n -1. 2. (3 points) Prove that0 k &lt;n 0k &lt;n for n 0. 3. (2 points) Suppose a program has
Uni. Worcester - CS - 504
C.S.504H.W. #5Due: April 8/9, 1998Read SECTIONS 7.1, 7.2, 7.3, 7.5 Do EXERCISEs 7.1, 7.2, 7.3, 7.4, 5.15 from GKP. Do not submit your solutions, but check them with the answers from the back of the text. 1. (4 points) Define an r-composition of n
Uni. Worcester - CS - 98
C.S.504H.W. #5Due: April 8/9, 1998Read SECTIONS 7.1, 7.2, 7.3, 7.5 Do EXERCISEs 7.1, 7.2, 7.3, 7.4, 5.15 from GKP. Do not submit your solutions, but check them with the answers from the back of the text. 1. (4 points) Define an r-composition of n
Uni. Worcester - CS - 504
C.S.504 H.W. #3Due: October 1, 19921. (6 points) You should compare three techniques for evaluating the 1 integral 4 1- x 2 d x . For each of the techniques, you should test the 0 rate of convergence by comparing the influence of n upon the accu
Uni. Worcester - CS - 92
C.S.504 H.W. #3Due: October 1, 19921. (6 points) You should compare three techniques for evaluating the 1 integral 4 1- x 2 d x . For each of the techniques, you should test the 0 rate of convergence by comparing the influence of n upon the accu
Uni. Worcester - CS - 504
Due: December 16, 1993C.S.504 H.W. #71. (5 points) Solve the recurrence kfn k , if &gt; 0 n 0k fn = 1, n if = 0 0, if &lt; 0 n ( Hints : -One solution could parallel the text's solution to its recurrence (7.41). -The summation is a convolution. )
Uni. Worcester - CS - 93
Due: December 16, 1993C.S.504 H.W. #71. (5 points) Solve the recurrence kfn k , if &gt; 0 n 0k fn = 1, n if = 0 0, if &lt; 0 n ( Hints : -One solution could parallel the text's solution to its recurrence (7.41). -The summation is a convolution. )
Uni. Worcester - CS - 504
C.S.504 H.W. #5Due: October 15, 1992 1. (3 points) Find simple forms for 1k n( 2k - 1 )and0k(k- 1 ) . 2k2. (2 points) Show that1k 3. (5 points) Robin hood hashing is used to reduce the variance of the expected successful search time for
Uni. Worcester - CS - 92
C.S.504 H.W. #5Due: October 15, 1992 1. (3 points) Find simple forms for 1k n( 2k - 1 )and0k(k- 1 ) . 2k2. (2 points) Show that1k 3. (5 points) Robin hood hashing is used to reduce the variance of the expected successful search time for
Uni. Worcester - CS - 504
C.S.504 H.W. #1Due: September 17, 1992 1. (1 point) What is (0.125)-2/3? 2. (4 points) Given array A[1.n] of integers such that |A[k+1]-A[k]|1 for 1k&lt;n and given integer x such that A[1]xA[n], we seek a j, 1jn, such that A[j]=x. Describe an algorith
Uni. Worcester - CS - 92
C.S.504 H.W. #1Due: September 17, 1992 1. (1 point) What is (0.125)-2/3? 2. (4 points) Given array A[1.n] of integers such that |A[k+1]-A[k]|1 for 1k&lt;n and given integer x such that A[1]xA[n], we seek a j, 1jn, such that A[j]=x. Describe an algorith
Uni. Worcester - CS - 504
C.S.504 H.W. #9Due: December 3, 1992 1. (3 points) A) Give a generating function, F m ( z ), for the number of ways to put indistinguishable balls into m distinguishable boxes, such that each box contains at least one ball. [z n ] F m ( z ) is the n
Uni. Worcester - CS - 92
C.S.504 H.W. #9Due: December 3, 1992 1. (3 points) A) Give a generating function, F m ( z ), for the number of ways to put indistinguishable balls into m distinguishable boxes, such that each box contains at least one ball. [z n ] F m ( z ) is the n
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems Spring 1999Homework II with Solution Probability Problems1. Two true dice are rolled. One is known to show an even number. What is the probability that the sum of both is 8? That it is 9? Solution: Sinc
Uni. Worcester - CS - 9902
CS504: Analysis of Computations and Systems Spring 1999Homework II with Solution Probability Problems1. Two true dice are rolled. One is known to show an even number. What is the probability that the sum of both is 8? That it is 9? Solution: Sinc
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems Spring 1999Homework IIDue: February 8 or 10, 1999Probability Problems1. Two true dice are rolled. One is known to show an even number. What is the probability that the sum of both is 8? That it is 9?
Uni. Worcester - CS - 9902
CS504: Analysis of Computations and Systems Spring 1999Homework IIDue: February 8 or 10, 1999Probability Problems1. Two true dice are rolled. One is known to show an even number. What is the probability that the sum of both is 8? That it is 9?
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems - Spring 1999Homework VIII - with solution1. Let an be the number of ways to obtain the score n by throwing a die. For example, a3 4. This corresponds to the possible throw sequences 1,1,1; 1,2; 2,1; 3. F
Uni. Worcester - CS - 9908
CS504: Analysis of Computations and Systems - Spring 1999Homework VIII - with solution1. Let an be the number of ways to obtain the score n by throwing a die. For example, a3 4. This corresponds to the possible throw sequences 1,1,1; 1,2; 2,1; 3. F
Uni. Worcester - CS - 504
CS504: Analysis of Computations and Systems - Spring 1999Homework VI - with SolutionThis may be as good a place as any to repeat that when you submit a home assignment or an exam you should show all work. Otherwise, when a result seems to appear &quot;o
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems - Spring 1999Homework VI - with SolutionThis may be as good a place as any to repeat that when you submit a home assignment or an exam you should show all work. Otherwise, when a result seems to appear &quot;o
Uni. Worcester - CS - 504
CS504Name_ Midterm ExamDate : October 22, 1992 All documentation permitted 1(10 points) The product of polynomials A=an xn + an -1xn-1 +.+ a1x +a0 and B=bmxm + bm -1xm-1 +.+ b1x +b0 is polynomial C=cn +mxn+m + cn+m -1xn+m-1 +.+ c1x +c0 where ci =
Uni. Worcester - CS - 504
CS504Name_ Final ExamDate : December 10, 1991 All documentation permitted 1(25 points) Consider the following program. procedure Friendly (n : integer); var k : 1.n ; begin if n &gt; 0 then for k := 1 to n do begin write(&quot;hello&quot;); Friendly (k -1) en
Uni. Worcester - CS - 504
CS504Name_ Final ExamDate : December, 1992 All documentation permitted 1. (25 points) A) Given a set of positive integers A={a1,.,am}, find a generating function FA(z) such that [zn]FA(z) is the number of subsets of A which add up to n. For exampl
Uni. Worcester - CS - 2
CS504: Analysis of Computations and Systems Spring 1999Final test Worcester Campus 5/5/99 All books and notes are allowed but do not pass them around! Name:1. A right-binary tree is a binary tree in which no node may have a left child only a
Uni. Worcester - CS - 504
CS504: Analysis of Computations and Systems Spring 1999Final test Worcester Campus 5/5/99 All books and notes are allowed but do not pass them around! Name:1. A right-binary tree is a binary tree in which no node may have a left child only a
Uni. Worcester - CS - 2
CS504: Analysis of Computations and Systems Spring 1999Solution of Midterm test Worcester Campus 3/10/1999Clearly the test surprised many of you. I am still not sure why, but most (all?) performed under what I believe is their normal capabilty.
Uni. Worcester - CS - 504
CS504: Analysis of Computations and Systems Spring 1999Solution of Midterm test Worcester Campus 3/10/1999Clearly the test surprised many of you. I am still not sure why, but most (all?) performed under what I believe is their normal capabilty.
Uni. Worcester - CS - 504
CS504: Analysis of Computations and Systems Spring 1999Homework X &amp; Additional ProblemsThe rst 3 Problems are Due: May 3 or 5, 1999 1. (a) Find the number of all words of size n over the alphabet a, b which begin with the string ab and end with th
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems Spring 1999Homework X &amp; Additional ProblemsThe rst 3 Problems are Due: May 3 or 5, 1999 1. (a) Find the number of all words of size n over the alphabet a, b which begin with the string ab and end with th
Uni. Worcester - CS - 99
CS504: Analysis of Computations and Systems Spring 1999Homework IVDue: February 22 or 24, 1999RecurrencesSince recurrences are a major tool in the analysis of algorithms, this assignment deals with them exclusively. 1. Consider the rst order li