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ACM notes

ACM notes. Why… ? Not always an easy question!. Entropy problem -- Huffman Codes. bits used in ASCII. Input. Output. A..AB..BC..CD..DE..EF. 784 224 3.5. 1. 9. 40. 20. 16. 12. compression ratio, to 1 place of precision. bits used in an optimal “prefix-free” encoding. 98 chars.

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ACM notes

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  1. ACM notes • Why… ? Not always an easy question! Entropy problem -- Huffman Codes bits used in ASCII Input Output A..AB..BC..CD..DE..EF 784 224 3.5 1 9 40 20 16 12 compression ratio, to 1 place of precision bits used in an optimal “prefix-free” encoding 98 chars

  2. Prefix-free Codes via binary trees A..AB..BC..CD..DE..EF 1 9 40 20 16 12 ABCDEF 98 AE 49 BCDF 49 BF CD 21 E A 28 40 9 C D F B 12 20 16 1 49*2 + 49*3 = 245

  3. Prefix-free Codes via binary trees A..AB..BC..CD..DE..EF 1 9 40 20 16 12 ABCDEF 98 0 1 AE 49 BCDF 49 1 1 0 0 BF CD 21 E A 28 40 9 1 0 1 0 01 00 C D F B 12 20 16 1 111 110 101 100 codewords are read down the paths 49*2 + 49*3 = 245

  4. Huffman Codes Building the tree from the bottom up… A..AB..BC..CD..DE..EF 1 9 40 20 16 12 EF 10 1 0 A..AB..BC..CD..DE..F F E 10 40 20 16 12 9 1

  5. Huffman Codes Building the tree from the bottom up… A..AB..BC..CD..DE..F 10 40 20 16 12 DEF 22 0 D EF 10 12 1 A..AD..E..FB..BC..C 0 40 22 20 16 F E 9 1

  6. Huffman Codes Building the tree from the bottom up… A..AD..E..FB..BC..C 40 22 20 16 BC DEF 36 22 1 0 1 0 D EF C 10 B 12 20 16 1 A..AB..CD..E..F 0 40 36 22 F E 9 1

  7. Huffman Codes Building the tree from the bottom up… A..AB..CD..E..F 40 36 22 A 40 BCDEF 58 1 0 BC DEF 36 22 1 0 1 0 D EF C 10 B 12 20 16 1 0 A..AB..C..D..E..F F E 40 58 9 1

  8. Huffman Codes Building the tree from the bottom up… A..AB..C..D..E..F 40 58 ABCDEF 98 1 0 A 40 BCDEF 58 1 0 BC DEF 36 22 1 0 1 0 D EF C 10 B 12 20 16 1 0 A..B..C..D..E..F 58 F E 9 1

  9. Huffman Codes Total number of bits needed... 40*1 + 36*3 + 12*3 + 10*4 = 224 ABCDEF 98 1 0 A 40 BCDEF 58 1 0 BC DEF 36 22 1 0 1 0 D EF C 10 B 12 20 16 1 0 F E 9 1

  10. Graphs [A,E], [A,B], [A,C], [B,D], [B,E], [C,B], [D,A], [D,C], [E,D] B Adjacency List Representation A (vector or array of pairs) C D E

  11. 0 8 13 - 1 - 0 - 6 12 - 9 0 - - 7 - 0 0 - - - - 11 0 Graphs [A,E], [A,B], [A,C], [B,D], [B,E], [C,B], [D,A], [D,C], [E,D] B Adjacency List Representation 8 9 A (vector or array of pairs ?) 13 C 6 0 7 1 TO 12 D A B C D E E A 11 B C FROM D Matrix representation E (2d array)

  12. Problems Roman Forts (forts.cc) J C A D I F G B E K H fortify the most vulnerable...

  13. Problems Single Points of Failure (spf.cc) 1 2 5 4 3 1 3 2 3 4 3 5 0 Network #1 SPF node 3 leaves 2 subnets Output 1 Input 1 Graph 1 Graph 2 Output 2 Network #2 No SPF nodes

  14. 0 0 8 8 13 13 - - 1 1 - - 0 0 - - 6 6 12 12 - - 9 9 0 0 - - - - 7 7 - - 0 0 0 0 - - - - - - - - 11 11 0 0 All-pairs shortest paths... “Floyd-Warshall algorithm” A k B dij = C D E 0 D0 = (dij ) A A 0 8 13 - 1 B B - 0 - 6 12 C C - 9 0 - - D D 7 15 0 0 8 E E - - - 11 0 1 D1 = (dij ) k dij = shortest distance from i to j through {1, …, k}

  15. 0 0 8 8 13 13 - - 1 1 - - 0 0 - - 6 6 12 12 - - 9 9 0 0 - - - - 7 7 - - 0 0 0 0 - - - - - - - - 11 11 0 0 All-pairs shortest paths... “Floyd-Warshall algorithm” A k B dij = C D E 0 D0 = (dij ) A A 0 8 13 - 1 B B - 0 - 6 12 C C - 9 0 - - D D 7 15 0 0 8 E E - - - 11 0 1 D1 = (dij ) k dij = shortest distance from i to j through {1, …, k}

  16. Geometric Problems [A,E], [A,B], [A,C], [B,D], [B,E], [C,B], [D,A], [D,C], [E,D] B Adjacency List Representation A (vector or array of pairs) C D code resources to keep in mind E

  17. All-pairs shortest paths... A 0 8 13 14 1 A 0 8 13 14 1 B 13 0 6 6 12 B - 0 - 6 12 C 22 9 0 15 21 C - 9 0 15 21 D 7 9 0 0 8 D 7 15 0 0 8 E 18 20 11 11 0 E - - - 11 0 4 D4 = (dij ) 2 D2 = (dij ) A 0 8 12 12 1 A 0 8 13 14 1 B 13 0 6 6 12 B - 0 - 6 12 C 22 9 0 15 21 C - 9 0 15 21 D 7 9 0 0 8 D 7 9 0 0 8 E 18 20 11 11 0 E - - - 11 0 5 3 D5 = (dij ) D3 = (dij ) to store the path, another matrix can track the last intermediate vertex

  18. Graphs Representations bits used in ASCII Input Output AAAAABCD 64 13 4.9 compression ratio, to 1 place of precision bits used in an optimal “prefix-free” encoding

  19. Graphs Representations bits used in ASCII Input Output AAAAABCD 64 13 4.9 compression ratio, to 1 place of precision bits used in an optimal “prefix-free” encoding

  20. Problems Entropy bits used in ASCII Input Output AAAAABCD 64 13 4.9 compression ratio, to 1 place of precision bits used in an optimal “prefix-free” encoding A 0 B 10 C 110 D 111

  21. Problems N-Credible Mazes dimensions Input Output 2 0 0 2 2 0 0 0 1 0 1 0 2 0 2 1 2 0 2 0 3 1 2 2 2 -1 start end Maze #1 can be travelled (or not…) edge start edge end

  22. C++ STL www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/ set<int> s; // basically a bin. tree s.size(); // returns an int s.insert(14); // adds 14 s.insert(-9); // adds -9 s.insert(42); // adds 42 set<int>::iterator i; // may want to typedef // think of an iterator as a pointer i = s.find(42); // return 42’s iterator cout << (*i) << endl; // prints 42 cout << (*--i)); // prints ... i = s.find(43); // not there ! // at this point ( i == s.end() ) is true s.erase(-9); // removing elements multiset<int> m; // holds multiple copies set #include <set> multiset #include <set>

  23. Breadth-first search • algorithm data structures queue, deque, hashtable (map)

  24. C++ STL www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/ vector<int> v; // basically an int array v.reserve(10); // assure 10 spots v.push_back(42); // adds 42 to the end v.back(); // returns 42 v.pop_back(); // removes 42 v.size(); // # of elements v[i]; // ith element sort( v.begin(), v.end() ); // default sort sort( v.begin(), v.end(), mycompare ); deque<int> d; // double-ended queue d.push_front(42); // add to front d.front(42); // return front element d.pop_front(42); // remove from front vector #include <vector> sort #include <algorithm> last time deque #include <deque>

  25. Other problems • Change counting input: output: 1.00 0.06 0 There are 292 ways to make $1.00 There are 2 ways to make $0.06 • Sigma series Shortest sequences from 1 to N such that each element is the sum of two previous elements. input: output: 3 4 87 99 -1 1 2 3 1 2 4 1 2 4 8 16 24 28 29 58 87 1 2 4 8 16 32 33 66 99

  26. Useful C functions int atoi(char* s); double atof(char* s); int strcasecomp(char* s1, char* s2); long strtol(char* s, NULL, int base) converts C strings to ints atoi(“100”) == 100 converts C strings to doubles atoi(“100.0”) == 100.0 case-insensitive C string comparison strcasecmp(“aCm”,“ACm”) == 0 arbitrary conversion from a string in bases (2-36) to a long int strtol(“Charlie”, NULL, 36)== 2147483647L use man for more...

  27. sprintf int sprintf(char* str, char* format, ...); prints anything to the string str char str[100]; sprintf(str,“%d”,42); // str is “42” sprintf(str,“%f”,42.0); // str is “42.0” sprintf(str,“%10d”,42); // str is “ 42” sprintf(str,“%-10d”,42); // str is “42 ” flexible formatting: right/left justify:

  28. Two ACM programming skills A chance to “improve” your C/C++ … Preparation for the ACM competition ... Problem Insight and Execution ... 2 1 Anxiety! Get into the minds of the judges

  29. Get into the minds of the judges Key Skill #1: mindreading 100% 0% “What cases should I handle?” spectrum

  30. Key Skill #2: anxiety Anxiety!

  31. Dynamic Programming Strategy: create a table of partial results & build on it. divis.cc T(n) = number of steps yet to go T(n) = T(3n+1) + 1 if n odd T(n) = T(n/2) + 1 if n even

  32. Dynamic Programming Keys: create a table of partial results, articulate what each table cell means, then build it up... divis.cc j = items considered so far Table T 0 1 2 3 4 5 6 1 1 6 2 -3 0 the list 1 4 i = possible remainder the divisor 2 3 T[i][j] is 1 if i is a possible remainder using the first j items in the list.

  33. Dynamic programs can be short #include <cstdio> #include <iostream> #include <vector> vector<int> v(10000); vector<bool> m(100); // old mods vector<bool> m2(100); // new mods int n, k; bool divisible() { fill(m.begin(),m.end(),false); m[0] = true; for (int i=0; i<n; i++) { /* not giving away all of the code */ /* here the table is built (6 lines) */ } return m[0]; } int main() { cin >> n; // garbage while (cin >> n) { cin >> k; for (int i=0; i<n; i++) { cin >> v[i]; v[i] = abs(v[i]); v[i] %= k; } cout << (divisible() ? "D" : "Not d") << "ivisible\n"; } cout << endl; } acknowledgment: Matt Brubeck STL: http://www.sgi.com/Technology/STL

  34. General ACM Programming Try brute force first (or at least consider it) -- sometimes it will work fine… -- sometimes it will take a _bit_ too long -- sometimes it will take _way_ too long Best bugs from last week: getting the input in the “pea” problem: filling in the table in the “divis” problem: for (int j=1 ; j<N ; ++j) { cin >> Array[i]; } Table[i + n % k] = 1; Table[i - n % k] = 1;

  35. New Problem Word Chains hertz jazz hajj zeroth doze aplomb ceded dozen envy ballistic yearn Input A list of words Output yes or no -- can these words be chained together such that the last letter of one is the first letter of the next… ?

  36. Knapsack Problem objectwt.val. 1 3 8 2 2 5 3 1 1 4 2 5 Maximize loot w/ weight limit of 4. w Weight available for use Number of objects considered 0 1 2 3 4 1 V(n,w) = max value stealable w/ ‘n’ objects & ‘w’ weight 2 3 4 n V(n,w) =

  37. C Output printf, fprintf, sprintf(char* s, const char* format, …) the destination the format string the values possible format strings h % -#0 12 .4 d d decimal integers u unsigned (decimal) ints o octal integers x hexadecimal integers f doubles (floats are cast) e doubles (exp. notation) g f or e, if exp < -3 or -4 c character s string n outputs # of chars written !! % two of these print a ‘%’ type size modifier minimum field width precision allowed size modifiers flags - left-justify 0 pad w/ zeros + use sign (+ or -) (space) use sign ( or -) # deviant operation h short l long (lowercase L) L long double start character

  38. C Output value = 42 value = -42 %10.4d 0042 -0042 %-#12x 0x2a 0xffffffd6 value = 42 value = -42.419 %+10.4g +42 -42.42 %- 10.4g 42 -42.42 %-#10.4g 42.00 -42.42 value = “forty-two” %10.5s forty

  39. Change const int p=1,n=5,d=10,q=25,h=50; int counter=0; int pn, nn, dn, qn, hn; for (hn = 0; hn*h <= num; hn++) for (qn = 0; hn*h + qn*q <= num; qn++) for (dn = 0; hn*h + qn*q + dn*d <= num; dn++) for (nn = 0; hn*h + qn*q + dn*d + nn*n <= num; nn++) for (pn = 0; hn*h + qn*q + dn*d + nn*n + pn*p <= num; pn++) { if (hn*h + qn*q + dn*d + nn*n + pn*p == num) counter++; } Brute Force Dynamic Programming total 5¢ 9¢ 0¢ 1¢ 6¢ 7¢ 10¢ 11¢ 2¢ 3¢ 8¢ 12¢ 4¢ using 1¢ 1 1 1 1 1 1 1 1 1 1 1 1 1 1¢, 5¢ 1 2 1 1 2 2 1 2 3 1 2 3 3 1 2 1 1 2 2 1 2 4 1¢, 5¢, 10¢ 1 2 4 4

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