Outline. Computer programming. Cross product algorithm. Cross product. Set every element of vector p to 1. Find next element as follows:

Transcription

1 Outline Computer programming "He who loves practice without theory is like the sailor who boards ship without a ruder and compass and never knows where he may cast." Leonardo da Vinci Applications Combinatorial generation: generating subsets Cross product (Cartesian product) Combinations Permutations Arrangements Power set Simple sorting algorithms Counting sort Insertion sort Shell sort Bubble sort Selection sort T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2 Cross product Considern sets of positive integers: A, A 2,..., A n. Set A i, for i=, 2,..., n has n i elements The cross product (Cartesian product, set direct product, product set) is required. i.e. A A2... A n = A i n i= having elements which will i= n i be generated in a vector, p=[ p p 2... p n ] n Cross product algorithm Set every element of vector p to. This is the first element of the cross product Find next element as follows: Find the highest index i for which p i < n i. If such an index cannot be found then all elements have been generated. Stop Find next element of the cross product as p, p 2, p i -, p i +,,,..., T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş

2 Cross product implementation. Non recursive Cross product implementation. Non recursive #include <stdio.h> #define MAXN void listproduct(int n, int prodnb, int p[]) int i; printf("\n%d ", prodnb); for (i = ; i <=n; i++) printf(" %2d", p[i]); if ( prodnb % 2 == ) getch(); void crossprodnonrec(int n, int nelem[]) /* n = number of sets; nelem = vector with number of elements per set */ int i, prodnb, p[maxn]; prodnb = ; for (i = ; i <= n; i++) p[i] = ; listproduct(n, prodnb, p); i = n; while ( i > ) p[i]++; // find highest such as p[i] > nelem[i] if (p[i] > nelem[i]) p[i] = ; i--; prodnb++; listproduct(n, prodnb, p); i = n; int main(int argc, char *argv[]) int i, n, nelem[maxn]; printf("\nnumber of sets [<%d]=", MAXN); scanf("%d", &n); for (i = ; i <= n; i++) printf("number of elements in set %d=", i); scanf("%d", &nelem[i]); printf("\nthe members of the cross product"); printf("\nno. Elements"); crossprodnonrec(n, nelem); getchar(); return ; T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 5 T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 6 Cross product implementation. Recursive #include <stdio.h> #include <conio.h> #define MAXN int prodnb, p[maxn], nelem[maxn]; void listproduct(int n) int i; printf("\n%d ", prodnb); for (i = ; i <=n; i++) printf(" %2d", p[i]); if ( prodnb % 2 == ) getch(); void crossprodrec(int n, int i) /* n = number of sets; nelem = vector with number of elements per set */ int j; for (j = ; j <=nelem[i]; j++) p[i] = j; if (i< n) crossprodrec(n, i+); prodnb++; listproduct(n); T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 7 Cross product implementation. Recursive int main(int argc, char *argv[]) int i, n; printf("\nnumber of sets [<%d]=", MAXN); scanf("%d", &n); for (i = ; i <= n; i++) printf("number of elements in set %d=", i); scanf("%d", &nelem[i]); printf("\nthe members of the cross product"); printf("\nno. Elements"); crossprodrec(n, ); getch(); return ; T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 8

3 Combinations Let P be a set of n elements All ways of picking k unordered elements of the n elements = generating all subsets with k n elements of P such as any two subsets are distinct The number of subsets is the binomial coefficient or choice number and read "n choose k " T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 9 Combinations algorithm First subset is p=, 2,..., k Given a subset, its successor is found as follows: Going from k down to find index i which satisfies the relationships pi < n k + i pi+ = n k + i +... pk = n p = n k Successor set is: p, p2,..., pi +, pi + 2,..., pi + n k + The last subset is: n k +, n k + 2,..., n, n T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş Combinations algorithm implementation. Non recursive void combinnonrec(int n, int k) int p[maxn]; int i, j, combinnb; for (i=; i <=k; i++) p[i]=i; /* first combination */ listcombin(k, combinnb, p); i = k; while (i > ) /* generate the next combinations */ p[i]++; // find index satisfying relation set if (p[i] > n k + i) i--; for (j = i + ; ; j <= k; j++) p[j]=p[j-] + ; combinnb++; listcombin(k, combinnb, p); i = k; T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş Combinations algorithm implementation. Recursive void combinrec(int n, int k, int i) int j; for (j = p[i-]+; j <= n-k+i; j++) p[i] = j; if (i < k) combinrec(n, k, i+); combinnb++; listcombin(k, combinnb, p); Notes Array p, and combinnb must be global Invocation is: combinrec(n, k, ) T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2

4 Generating Permutations For instance 52 is the permutation that maps to, 2 to 5, to 2, to, and 5 to. We know that there are 7! permutations on,2,,,5,6,7. Suppose we want to list them all. Is there an efficient way to do so? It turns out to be fairly simple to list them lexicographically. The only hard question is, given one permutation, how do we find the next one? The lexicographically first permutation is 2... n and the last is n n -... Generating Permutations It is intuitively reasonable that if the final digits of a permutation are in descending order, then no rearrangement will make them larger. For instance in 2576 we cannot produce a larger number by rearranging the 76. Instead we must increase the next most significant digit (the 5) by the next larger digit in 76 (the 6). Then the remaining digits (the 5 and the 7) must be arranged to form the smallest number possible. Thus the next permutation in lexicographic order is T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş Generating permutations in lexicographical order a. First permutation is p =, 2,... n b. Given vector p=[p p 2... p n ] the next permutation is found as follows:. Look from n down to for the highest valued index which satisfies the relationships: p i <p i + p i + > p i +2 >...> p n 2. Find the maximum element, p k > p i of p i +, p i +2,..., p n. Swap p k with p i. Revert p i +, p i +2,..., p n by swapping p i+ and p n, p i+2 and p n-, a.s.o. Permutations implementation. Non recursive void permnonrec(int n) int p[maxn]; int i, k, permnb = ; /* first permutation, step a */ for (i = ; i <= n; i++) p[i] = i; listperm(n, ++permnb, p); do /* generate the next permutations */ i = n - ; while (p[i] > p[i+] && i > ) i--; /* step b */ if (i > ) for (k = n; p[i] > p[k]; k--); /* step b2 */ swap( &p[i], &p[k] ); /* step b */ revert(p, i, n, (n - i ) / 2); listperm(n, ++permnb, p); while (i > ); void swap(int *i, int *j) int temp; temp = *i; *i = *j; *j = temp; void revert(int p[], int i, int n, int k) int j; for (j = ; j <= k; j++) swap(p[i+j], p[n+-j]); T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 5 T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 6

5 A note on swapping Swapping integers in place void swap(int *i, int *j) *i += *j; // i == i + j *j = *i - *j;// j == i + j j == i *i -= *j; // i == i + j i == j void swap(int *i, int *j) *i ^= j; // a == a^b; *j ^= *i;// b == b^(a^b) *i ^= *j;// a == (a^b)^(b^(a^b)) Second version based on proof on the nearby table, where a and b are bit positions a b a^b final b b^(a^b) T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 7 final a (a^b)^(b^(a^b)) Permutations implementation. Recursive void permrec(int nb) int i, j; if ( nb == ) permnb++; listperm(); permrec(nb ); for (i = ; i <= nb ; i++) swap( p[i], p[nb] ); permrec(nb ); swap( p[i], p[nb] ); Notes Generates recursively permutations of elements p p 2... p n- with p n in position n Then swaps p i with p n for i =..n-, and generates all permutations Array p, n, and permnb must be globals First permutation must be initialized separately Invocation: permrec(n) Order is not loxicographic T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 8 Generating arrangements For set p=, 2,..., n and k n, positive generate all k-subsets such any two subsets must differ either in the composing elements or in their order Algorithm idea: generate all combinations of n elements taken k, and, for each combination generate the k! permutations T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 9 Generating arrangements implementation void arrange(int n, int m, int i) int j, k, r; for (j = p[i-] + ; j <= n - m + i; j++) // recursively generate combinations p[i] = j; if (i < m) arrange(n, m, i + ); arrnb++; listarrang(m, p); for (k = ; k <= m; k++) v[k] = p[k]; // save combination do // nonrecursively permute combination k = m - ; while (v[k] > v[k + ] && k > ) k--; if (k > ) for (r = m; v[k] > v[r]; r--); swap(&v[k], &v[r]); revert(k, m, (m - k) / 2); arrnb++; listarrang(m, v); while (k > ); T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2

6 Generating a power set We wish to generate all 2 n subsets of set A=a,...,a n (power set) All subsets of A=a,...,a n can be divided to two groups Ones that contain a n Ones that does not contain a n Ones that does not contain a n are all subsets of a,...,a n- When we have all subsets of a,...,a n- we can create all subsets of a,...,a n by adding all elements with a n inserted T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2 Generating a power set Again, we try to generate all subsets without generating power sets of smaller sets We can associate 2 n subsets of n elements set A=a,...,a n with 2 n bit strings b...b n b i = if a i is element of set b i = if a i is not element of set For elements set a, a 2, a the empty set the set itself subset a, a 2 We can create bit strings by generating binary numbers from to 2 n - Example: Bit strings Subsets a a 2 a 2,a a a,a a,a 2 a,a 2,a T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 22 Generating a power set. Non recursive Generating a power set. Recursive void allsubsetsnr(int n) int i, setnb, p[maxn]; // p[i]= if element i is a member, // i.e p is a characteristic vector setnb=; for (i=; i<=n; i++) p[i]=; /*empty set */ listset(n, setnb, p); for (i=n; i>; ) /* generate next subsets */ if (p[i] == ) p[i] = ; setnb++; listset(n, setnb, p); i = n; p[i] = ; i--; void allsubsetsrec(int n, int i) int j; for (j=; j <= ; j++) p[i] = j; if (i < n) allsubsetsrec(n, i+); setnb++; listset(n, setnb); void listset(int n, int setnb) int i; printf("\n%d ", setnb); for (i = ; i <=n; i++) if (p[i] == ) printf(" %2d,", i); printf("\b "); if ( setnb % 2 == ) getch(); Note that array p and variable setnb must be global T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2 T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2

7 Sorting We assume that objects to be sorted are records consisting of one or more fields. One of the fields, called the key is of a type for which a linear ordering relationship < is defined. The sorting problem: to arrange a sequence of records so that the values of their key fields form a nondecreasing sequence, i.e. for records r, r 2,..., r n with key values k, k 2,..., k n respectively, we must produce the same records in an order such that T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 25 Sort. Auxiliary routines void swap( void *px, void *py, size_t size) /* swap exchanges the values of x and y */ void *ptemp; if (NULL==(ptemp=malloc(size))) fatalerror("no memory"); memcpy(ptemp, px, size); memcpy(px, py, size); memcpy(py, ptemp, size); free(ptemp); /* swap */ int cmpkey(keyt pk, keyt pk2, size_t size) // example for objects stored in high to low value order return(memcmp(pk, pk2, size)); T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 26 Counting Sort CountingSort is an algorithm which requires that the keys positive integers in the range of to k and exploits the fact that keys can be used for indexing an array, hence it is not based on comparison. First we count the occurrences of each key of the array A and store the count in array C: A = C = 2 Then we determine the number of elements which are less than or equal to each of the keys by calculating the prefix sums: C = 6 Counting Sort We produce the output in sequence B by going through A and determining for each element to which position in B goes by looking that up in C. Then we decrease the entry for that element in C. A = B = C = B = C = B = C = T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 27 T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 28

8 Counting sort implementation variant void countsort(int n) // array A is an array of integer keys int i, j; int b[maxrec]; int c[maxrec]; for (i=; i < n; i++) c[i]=; //init counting vector b[i] = A[i]; // duplicate A // counting for (j = ; j < n; j++) for (i = ; i < j; i++) if (A[i] < A[j])) c[j]++; c[i]++; // rearange vector A for (i = ; i < n; i++) A[c[i]] = b[i]; /* countsort */ Insertion Sort In place sorting algorithm using a table A[..n ] At the beginning of each step i, for i < n, the table consists of: left, sorted part A[..i ] marked element x A[i] right, unsorted part A[i+..n ] The marked element x is inserted in the sorted part; the rest of the sorted part is shifted to the right T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 29 T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş Insertion Sort Insertion sort operation void inssort() int i, j; memcpy(a[].key, MINUS_INFINITY, sizeof(keyt)); for (i = 2; i <= n; i++) j = i; while (cmpkey(a[j].key, A[j-].key, sizeof(keyt)) < ) swap( &A[j], &A[j-], sizeof(recordt)); j--; /* inssort */ T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 2

9 Shell sort idea Sort in place, using a table A[..n-] Idea InsertionSort only elements at positions i, i+h, i+2h Repeat for i=,,..., h- InsertionSort the result Improvement: iterate on k, decreasing hk Motivation: reduce the maximal shifting distance in each phase. void shellsort() int h, i, j; Shell Sort h = n/2; // set increment value while (h > ) /* sort-by-insertion in increments of h */ for (i = h + ; i <= n; i++) j = i - h; while ( j > ) if ( cmpkey( A[j].key, A[j+h].key,sizeof(keyT) ) > ) swap(&a[j+h], &A[j], sizeof(recordt)); j -= h; j = ; h /= 2; /* shellsort */ T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş Bubble Sort enum FALSE, TRUE; void bubblesort() int i, j, done = FALSE; for (i = ; i < n &&!done; i++) for (j = n; j > i &&!done; j--) done = TRUE; if (cmpkey(a[j].key, A[j-].key, sizeof(keyt)) < ) swap(&a[j], &A[j-], sizeof(recordt)); done = FALSE; /* bubblesort */ T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 5 Selection Sort void selsort() static keyt lowkey; // the currently smallest key found on a pass // through A[i],..., A[n] int lowindex ; // the position of lowkey int i, j; for (i = ; i < n; i++) /* select the lowest among A[i],..., A[n] and swap it with A[i] */ lowindex = i; memcpy( &lowkey, &A[i].key, sizeof( keyt )); for ( j = i+; j <= n; j++) /* compare each key with current lowkey */ if ( cmpkey( A[j].key, lowkey, sizeof( keyt )) < ) memcpy( &lowkey, &A[j].key, sizeof( keyt )); lowindex = j; swap( &A[i], &A[lowindex], sizeof( recordt )); /* selsort */ T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 6

10 Selection sort operation...and one step more T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 7 Summary Applications Combinatorial generation: generating subsets Cross product (Cartesian product) Combinations Permutations Arrangements Power set Simple sorting algorithms Counting sort Insertion sort Shell sort Bubble sort Selection sort T.U. Cluj-Napoca - Computer Programming - lecture - M. Joldoş 8