MASTERMIND

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## MASTERMIND

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**MASTERMIND**• Did anyone play the game over the weekend? • Any thoughts on strategy?**Combinatorics Review**CSC 172 SPRING 2004 LECTURE 12**Assignments With Replacement**• Example: Passwords • Are there more strings of length 5 built from three symbols or strings of length 3 built from 5 symbols? • What would make a better password? • 4 digit “pins” • 3 letter initials**In general**We are given n “items”, to each we must assign one of k “values” Each value may be used any number of times Let W(n,k) be the number of ways How many different ways may we assign values to the items? “Different ways” means that one or more of the items get different values.**Inductive Definition**Basis: W(1,k) == k Induction: If we have n+1 items, we assign the first one in one of k ways and the remaining n in W(n,k) ways**Recurrence**W(1,k) = k W(n+1,k) = k * W(n,k) T(1) = k T(n) = k * T(n-1) Easy expansion T(n) = kn**Example: Are there more strings of length 5 built from three**symbols or strings of length 3 built from 5 symbols? Length 5, from 3 “values” = {0,1,2} “items” are the 5 positions N = 35 = 243 Length 3, from 5 “values” = {0,1,2,3,4} “items” are the 3 positions N = 53 = 125**Example: Are there more 4-digit pins, or 3 letter initials?**Length 4, from 10 “values” = {0,1,2,3,4,5,6,7,8,9} “items” are the 4 positions N = 104 = 10000 Length 3, from 26 “values” = {a,b,c,…,x,y,z} “items” are the 3 positions N = 263 = 17576**MASTERMIND**• How many codes? • N colors • M positions Can you build an array of unknown Dimensionality?**Exercise (aside)**Can you write public int blackPegs( String [] correct, String[] guess){ // return the number of correct colors // in the correct positions }**Exercise (aside)**• Can you write public int whitePegs( String [] correct, String[] guess){ // return the number of correct colors // in the incorrect positions }**Permutations**Example: “SCRABBLE” - start with 7 letters (tiles) - how many different ways can you arrange them - we don’t care about the word’s legality**Scrabble**We can pick the first letter to be any of the 7 tiles For each possible 1st letter, there are 6 choices of second letters Or, 7*6 = 42 possible two letter prefixes Similarly, for each of the 42, there are 5 choices of the third letter. 42 * 5 = 210, and so on Total choices = 7*6*5*…*1 = 7! = 5040 In general, there are n! permutations of n items.**Ordered Selections**Suppose we want to begin Scrabble with a 4 letter word? How many ways might we form the word from our 7 distinct tiles? For each possible 1st letter, there are 6 choices of second letters 7*6 = 42 possible two letter prefixes For each of the 42, there are 5 choices of the third letter. 42 * 5 = 210 For each of the 120, there are 4 choices of the third letter. 210 * 4 = 840**In general**P(n,m), the number of ways to pick a sequence of m things out of n == n*(n-1)*(n-2)*…*(n-m+1) == n!/(n-m)!**Combinations**Suppose we give up trying to make a word and want to throw 4 of our 7 tiles back in the pile? How many different ways can we get rid of 4 tiles? Ordered selection 7!/(7-4!) = 840 However, we don’t care about order. So, how many ways are there to order 4 items? 4! = 24 840/24==35 == 7!/((7-4)!4!) = 35**In general**“n choose m”**Recursive Definition for n choose m**We want to choose m things out of n, we can either take or reject the first item. If we take the first, then we can take the rest by choosing m-1 of the remaining n-1 We can do this in (n-1) choose (m-1) ways OTOH, if we reject the first item, then we can get the rest by choosing m of the remaining n-1 We can do this in (n-1) choose m ways**Inductive Definition**Basis: for all n there is only one way to choose all or none of the elements Induction: for 0 < m < n**Proving Inductive Definition = Direct Definition**What is the induction parameter? Zero for the basis case decreases in the inductive step Complete induction on m(n-m) Prove: c(n,m) = n!/((n-m)!m!)**Basis**If m(n-m) == 0, then either m == 0 or m == n If m == 0, then n!/(n-m)!m! == n!/n! = 1 = c(n,0) If m == n, then n!/(n-m)!m! == n!/n! = 1 = c(n,n)**Induction**By definition c(n,m) = c(n-1,m-1) + c(n-1,m) Assume, c(n-1,m-1) = (n-1)!/((n-m)!(m-1)!) c(n-1,m) = (n-1)!/((n-m-1)!m!) Add the left sides == c(n,m), by definition Add the right sides == n!/(n-m)!m! , clearly**Reminders**• Midterm is Tuesday – in class • Project is due before you go on Spring break • For workshop read Weiss Section 13.1 • The Josephus Problem • I will be out, most of next week • Thursday’s guest lecture will involve “game AI” • 7.7 & 10.2 (alpha-beta pruning)**Midterm**• Closed book, notes • Calculators will not be necessary – no laptops • Proof by induction • Solve recurrence relations • Big-Oh proof • Big-Oh analysis of some code • Linked list programming • Stacks, queues, arrays • Recursion/backtracking programming analysis • Sorting (merge, quick, insertion, shell) • Combinatorics**Orders With Some Equivalent Items**In real life, we play Scrabble with duplicate letters Suppose you draw {S,T,A,A,E,E,E} at star, how many 7-letter “words” can you make. Similar to permutations, but now, some are indistinguishable, because of duplicates Trick: we can mark the letters to make them distinguishable S,T,A1,A2,E1,E2,E3 Then we get 7!=5040 ways**How much “sameness”**But some order are the same E3TA1E1SA2E2 == E3TA2E1SA1E2 The two As can be ordered in 2! = 2 ways The three Es can be ordered in 3! = 6 ways So, the number of different words is 7!/2!3! = 540/(2*6) = 420**In general**The orders of n items with groups i1,i2,…,ik equivalent items is n!/(i1!i2!..ik!)**Items into Bins**Suppose we throw 7 dice (6 sided). How many outcomes are there? Place each of 7 items into one of 6 bins The tokes are the dice The bins are then number of dice Putting the second token into bin 3 means that the second die shows 3**Trick**Imagine 5 markers, denoted “*” that represent separation between bins, and 7 tokens “T” that represent dice The string “*TT**TTT*T*T” corresponds to no 1s, two 2s, no 3s, three 4s, a 5 and a 6 How many such strings? “orders with identical items” 12 items, 5 of type “*”, 7 of type “T” 12!/5!7! = 792**In general**The number of ways to assign n items to m bins The number of orders of n-1 markers and n tokens We are picking n out of the possible n+m-1 positions for the tokens**Several kinds of items into Bins**Order in bin can be either important (queues) or not**Several kinds of items, order within bin unimportant**If we have items of k colors, with ij items of the jth color, then the number of distinguishable assignments into m bins is: Example: 4 red dice, 3 blue dice, 2 green dice 6 bins**Order important**If there are m bins into which ij items are placed and order within the bin matters Imagine m-1 markers separating the bins and ij tokens of the jth type By “orders with identical items”