clrs算法导论题解
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chp1
1.2-1
1.2-2
1.2-3
n=15
**************
chp2
2.1-1 to 2.1-4
2.2-* //*: all has been done.
2.1-1 to 2.1-4
2.2-* //*: all has been done.
2.2-2
O(n^2),O(n^2)
?2-1
?2-1
2-2
***************
chp3
3.1-6
***************
chp3
3.1-6
***************
chp4
? 4-6
4-7
ab
c:
random select row r1<r2, if f(r1) > f(r2), we will have A[r1, f(r1)]+A[r2,f(r2)]<A[r1, f(r2)]+A[r2,f(r1)], which cause conflict with the condition.
d:
nlogm (similar to quicksort)
+++++++++
chp5
5.4-1
assume 355days/year.
1)178+1 people(1 is yourself).
2)n*n-n>=355*177
++++++++++++++++++++++++
chp6
6.1-*
6.5-*
**************
chp7
7.4-2
7.4-4 ? probability analysis
7.6
if [a1,b1] and [a2,b2] overlap, we set these 2 element as equal.
while quicksorting, put those internals equaling to pivot together in 2 place(head and tail).
aftera round of partition is finished, put those equal ones to the middle.
7.8
7.8
*************
chp8
8-6
a,c,d
b ?
*************'
chp10
10.1-*
10.2-1 to 10.2-4
10.2-7
10.2-8
***************
chp12
12.2-2
12.2-3 ？
12.2-6
12.3-3
worst: O(n^2)
best: ?
*************
chp13
13.1-1 ?
13.2-1
//RIGHT-ROTATE(T,y)
x = y.left
y.left = x.right
if(x.right != NIL(T))
p(x.right) =y
p(x) = p(y)
if( p(y) == NIL(T) )
root = x
else
if ( y == p(y).left )
p(y).left = x
else
p(y).right = x
x.right = y
p(y) = x
13.2-2
13.2-3
a: the depth ofa is added by 1.
b: stay the same.
c: substracted by 1.
13.2-4
one right rotation can add one node to the right-going chain, so at most n - 1 right rotations suffice to transform the tree into a right-going chain.
and, left rotation and right rotation are symmetric.
so, for any 2 n-node binary search tree A and B.
A->right-going chain->B can be done in O(n).
13.2-5 ?
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chp15
15.4-2
compare xi and yi.
then compare c[i-1,j] and c[i,j-1].
15.4-3
as figure 15.6, we onlyneed Y[7]to store eachrow of the matrix. so that we stores c[i-1,j-1] and c[i-1,j].
and, we only need a integer to store the value c[i,j-1].
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chp18
In a typical B-tree application, the amount of data handled is so large that all the data do not
fit into main memory at once.
fit into main memory at once.
18.1-1
18.1-2
18.1-3
3: rooted with 2,3,4
18.1-4
(2t)^h-1
****************
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Givena n*nmatrix, whoes elements are all integers.
now, we need to select n elements in this matrix, so that there are neither 2 elements in a single row nor in a single column, and the sum of
these n elements must be minimum.
skill: substract or add the same integer in a row or column would not change the positions of the elements of the result. so we can modify the
matrix until it contains no negative integers.
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C.1-1
C.1-4
O: for odd;
E: for even
there are 4 cases:
a: O O O
b: O O E => sum is even
c: O E E
d: E E E => sum is even
the chances for a, d to happen are the same.
the chances for b, c to happen are the same.
so the answer is C(100,3)/2
C.1-5