ACM notes Why… ? Not always an easy question! Entropy problem -- Huffman Codes InputOutput A..AB..BC..CD..DE..EF compression ratio, to 1 place of precision bits used in ASCII bits used in an optimal “prefix-free” encoding chars 1

Prefix-free Codes via binary trees A..AB..BC..CD..DE..EF D CD ABCDEF BCDF BF A E C 16 AE B F *2 + 49*3 = 245

Prefix-free Codes via binary trees A..AB..BC..CD..DE..EF D CD ABCDEF BCDF BF A E C 16 AE B F *2 + 49*3 = codewords are read down the paths

Huffman Codes Building the tree from the bottom up… A..AB..BC..CD..DE..EF EF 10 E F A..AB..BC..CD..DE..F

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

Huffman Codes Building the tree from the bottom up… EF 10 E F 9 1 DEFBC B C D A..AB..CD..E..F A..AD..E..FB..BC..C

Huffman Codes Building the tree from the bottom up… EF 10 E F 9 1 DEF BCDEF BC B C D 12 A A..AB..CD..E..F A..AB..C..D..E..F 4058

Huffman Codes Building the tree from the bottom up… EF 10 E F 9 1 DEF ABCDEF BCDEF BC B C D 12 A A..AB..C..D..E..F 4058 A..B..C..D..E..F 58

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

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

Graphs A B E D C A B C D E FROM TO Matrix representation ABCDE [A,E], [A,B], [A,C], [B,D], [B,E], [C,B], [D,A], [D,C], [E,D] (2d array) Adjacency List Representation (vector or array of pairs ?)

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

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

All-pairs shortest paths A B C D E D 0 = (d ij ) A B C D E D 1 = (d ij ) 1 d ij = shortest distance from i to j through {1, …, k} k d ij = k A B C D E “Floyd-Warshall algorithm”

All-pairs shortest paths A B C D E D 0 = (d ij ) A B C D E D 1 = (d ij ) 1 d ij = shortest distance from i to j through {1, …, k} k d ij = k A B C D E “Floyd-Warshall algorithm”

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

All-pairs shortest paths A B C D E D 2 = (d ij ) A B C D E D 3 = (d ij ) A B C D E D 4 = (d ij ) 4 A B C D E D 5 = (d ij ) 5 to store the path, another matrix can track the last intermediate vertex

Graphs Representations InputOutput AAAAABCD compression ratio, to 1 place of precision bits used in ASCII bits used in an optimal “prefix-free” encoding

Graphs Representations InputOutput AAAAABCD compression ratio, to 1 place of precision bits used in ASCII bits used in an optimal “prefix-free” encoding

Problems Entropy InputOutput AAAAABCD compression ratio, to 1 place of precision bits used in ASCII bits used in an optimal “prefix-free” encoding A 0 B 10 C 110 D 111

Problems N-Credible Mazes InputOutput Maze #1 can be travelled startend dimensions edge start edge end (or not…)

C++ STL set 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 ::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 m; // holds multiple copies multiset #include set #include

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

C++ STL vector 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 d; // double-ended queue d.push_front(42); // add to front d.front(42); // return front element d.pop_front(42); // remove from front sort #include vector #include deque #include last time

Other problems Change counting input: output: There are 292 ways to make $1.00 There are 2 ways to make $0.06 Sigma series input: output: Shortest sequences from 1 to N such that each element is the sum of two previous elements.

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

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:

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

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

Key Skill #2: anxiety Anxiety!

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

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

Dynamic programs can be short #include vector v(10000); vector m(100); // old mods vector 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:

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 for (int j=1 ; j<N ; ++j) { cin >> Array[i]; } Table[i + n % k] = 1; Table[i - n % k] = 1; filling in the table in the “divis” problem:getting the input in the “pea” problem: Best bugs from last week:

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

Knapsack Problem V(n,w) = max value stealable w/ ‘n’ objects & ‘w’ weight V(n,w) = object wt. val Maximize loot w/ weight limit of 4. 4 Number of objects considered Weight available for use n w

C Output printf, fprintf, sprintf(char* s, const char* format, …) the destinationthe format string the values h.412-#0%d start character flags - left-justify 0 pad w/ zeros + use sign (+ or -) (space) use sign ( or -) # deviant operation minimum field width precision size modifier h short l long (lowercase L) L long double type 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 ‘%’ allowed size modifiers possible format strings

C Output %10.4d value = 42value = %-#12x 0x2a0xffffffd6 %+10.4g value = 42value = %- 10.4g %-#10.4g value = “forty-two” %10.5s forty

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++; } Change Brute Force Dynamic Programming 1¢ 1¢, 5¢ 1¢, 5¢, 10¢ 1 using total 1¢ 2¢3¢ 4¢ 5¢ 6¢7¢ 8¢ 9¢ 10¢11¢ 12¢ ¢