lecture 10

CS 332: Algorithms Linear-Time Sorting Continued Medians and Order Statistics

Review: Comparison Sorts Comparison sorts: O(n lg n) at best Model sort with decision tree Path down tree = execution trace of algorithm Leaves of tree = possible permutations of input Tree must have n! leaves, so O(n lg n) height

Review: Counting Sort Counting sort: Assumption: input is in the range 1..k Basic idea: Count number of elements k  each element i Use that number to place i in position k of sorted array No comparisons! Runs in time O(n + k) Stable sort Does not sort in place: O(n) array to hold sorted output O(k) array for scratch storage

Review: Counting Sort 1 CountingSort(A, B, k) 2 for i=1 to k 3 C[i]= 0; 4 for j=1 to n 5 C[A[j]] += 1; 6 for i=2 to k 7 C[i] = C[i] + C[i-1]; 8 for j=n downto 1 9 B[C[A[j]]] = A[j]; 10 C[A[j]] -= 1;

Review: Radix Sort How did IBM get rich originally? Answer: punched card readers for census tabulation in early 1900’s. In particular, a card sorter that could sort cards into different bins Each column can be punched in 12 places Decimal digits use 10 places Problem: only one column can be sorted on at a time

Review: Radix Sort Intuitively, you might sort on the most significant digit, then the second msd, etc. Problem: lots of intermediate piles of cards (read: scratch arrays) to keep track of Key idea: sort the least significant digit first RadixSort(A, d) for i=1 to d StableSort(A) on digit i Example: Fig 9.3

Radix Sort Can we prove it will work? Sketch of an inductive argument (induction on the number of passes): Assume lower-order digits {j: j<i}are sorted Show that sorting next digit i leaves array correctly sorted If two digits at position i are different, ordering numbers by that digit is correct (lower-order digits irrelevant) If they are the same, numbers are already sorted on the lower-order digits. Since we use a stable sort, the numbers stay in the right order

Radix Sort What sort will we use to sort on digits? Counting sort is obvious choice: Sort n numbers on digits that range from 1.. k Time: O( n + k ) Each pass over n numbers with d digits takes time O( n+k ), so total time O( dn+dk ) When d is constant and k= O( n ), takes O( n ) time How many bits in a computer word?

Radix Sort Problem: sort 1 million 64-bit numbers Treat as four-digit radix 2 16 numbers Can sort in just four passes with radix sort! Compares well with typical O( n lg n ) comparison sort Requires approx lg n = 20 operations per number being sorted So why would we ever use anything but radix sort?

Radix Sort In general, radix sort based on counting sort is Fast Asymptotically fast (i.e., O( n )) Simple to code A good choice To think about: Can radix sort be used on floating-point numbers?

Summary: Radix Sort Radix sort: Assumption: input has d digits ranging from 0 to k Basic idea: Sort elements by digit starting with least significant Use a stable sort (like counting sort) for each stage Each pass over n numbers with d digits takes time O( n+k ), so total time O( dn+dk ) When d is constant and k= O( n ), takes O( n ) time Fast! Stable! Simple! Doesn’t sort in place

Bucket Sort Bucket sort Assumption: input is n reals from [0, 1) Basic idea: Create n linked lists ( buckets ) to divide interval [0,1) into subintervals of size 1/ n Add each input element to appropriate bucket and sort buckets with insertion sort Uniform input distribution  O(1) bucket size Therefore the expected total time is O(n) These ideas will return when we study hash tables

Order Statistics The i th order statistic in a set of n elements is the i th smallest element The minimum is thus the 1st order statistic The maximum is (duh) the n th order statistic The median is the n /2 order statistic If n is even, there are 2 medians How can we calculate order statistics? What is the running time?

Order Statistics How many comparisons are needed to find the minimum element in a set? The maximum? Can we find the minimum and maximum with less than twice the cost? Yes: Walk through elements by pairs Compare each element in pair to the other Compare the largest to maximum, smallest to minimum Total cost: 3 comparisons per 2 elements = O(3n/2)

Finding Order Statistics: The Selection Problem A more interesting problem is selection : finding the i th smallest element of a set We will show: A practical randomized algorithm with O(n) expected running time A cool algorithm of theoretical interest only with O(n) worst-case running time

Randomized Selection Key idea: use partition() from quicksort But, only need to examine one subarray This savings shows up in running time: O(n) We will again use a slightly different partition than the book: q = RandomizedPartition(A, p, r)  A[q]  A[q] q p r

Randomized Selection RandomizedSelect(A, p, r, i) if (p == r) then return A[p]; q = RandomizedPartition(A, p, r) k = q - p + 1; if (i == k) then return A[q]; // not in book if (i < k) then return RandomizedSelect(A, p, q-1, i); else return RandomizedSelect(A, q+1, r, i-k);  A[q]  A[q] k q p r

Randomized Selection Analyzing RandomizedSelect() Worst case: partition always 0:n-1 T(n) = T(n-1) + O(n) = ??? = O(n 2 ) (arithmetic series) No better than sorting! “ Best” case: suppose a 9:1 partition T(n) = T(9 n /10) + O(n) = ??? = O(n) (Master Theorem, case 3) Better than sorting! What if this had been a 99:1 split?

Randomized Selection Average case For upper bound, assume i th element always falls in larger side of partition: Let’s show that T( n ) = O( n ) by substitution What happened here?

Randomized Selection Assume T( n )  cn for sufficiently large c : What happened here? “ Split” the recurrence What happened here? What happened here? What happened here? The recurrence we started with Substitute T(n)  cn for T(k) Expand arithmetic series Multiply it out

Randomized Selection Assume T( n )  cn for sufficiently large c : What happened here? Subtract c/2 What happened here? What happened here? What happened here? The recurrence so far Multiply it out Rearrange the arithmetic What we set out to prove

Worst-Case Linear-Time Selection Randomized algorithm works well in practice What follows is a worst-case linear time algorithm, really of theoretical interest only Basic idea: Generate a good partitioning element Call this element x

Worst-Case Linear-Time Selection The algorithm in words: 1. Divide n elements into groups of 5 2. Find median of each group ( How? How long? ) 3. Use Select() recursively to find median x of the  n/5  medians 4. Partition the n elements around x . Let k = rank( x ) 5. if (i == k) then return x if (i < k) then use Select() recursively to find i th smallest element in first partition else (i > k) use Select() recursively to find ( i-k )th smallest element in last partition

Worst-Case Linear-Time Selection (Sketch situation on the board) How many of the 5-element medians are  x? At least 1/2 of the medians =  n/5  / 2  =  n/10  How many elements are  x? At least 3  n/10  elements For large n , 3  n/10   n/4 (How large?) So at least n /4 elements  x Similarly: at least n /4 elements  x

Worst-Case Linear-Time Selection Thus after partitioning around x , step 5 will call Select() on at most 3 n /4 elements The recurrence is therefore: ??? ??? ??? ??? ???  n/5   n/5 Substitute T(n) = cn Combine fractions Express in desired form What we set out to prove

Worst-Case Linear-Time Selection Intuitively: Work at each level is a constant fraction (19/20) smaller Geometric progression! Thus the O(n) work at the root dominates

Linear-Time Median Selection Given a “black box” O(n) median algorithm, what can we do? i th order statistic: Find median x Partition input around x if ( i  (n+1)/2) recursively find i th element of first half else find ( i - (n+1)/2)th element in second half T(n) = T(n/2) + O(n) = O(n) Can you think of an application to sorting?

Linear-Time Median Selection Worst-case O(n lg n) quicksort Find median x and partition around it Recursively quicksort two halves T(n) = 2T(n/2) + O(n) = O(n lg n)

lecture 10

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lecture 10