CS 350 Algorithms and Complexity

Transcription

1 CS 350 Algorithms and Complexity Spring 2015 Lecture 8: Decrease & Conquer (continued) Andrew P. Black Department of Computer Science Portland State University

2 Ferrying Soldiers!A detachment of n soldiers must cross a wide and deep river with no bridge in sight. They notice two 12-year-old boys playing in a rowboat by the shore. The boat is so tiny, that it can hold just two boys or one soldier. " How can the soldiers get across the river and leave the boys in joint possession of the boat? " How many times need the boat pass from shore to shore? 2

3 Ferrying Soldiers!Apply decrease-by-1 process: " Ferry one soldier to the far side, leaving boat and boys back at their initial positions 3

4 Ferrying Soldiers!Apply decrease-by-1 process: " Ferry one soldier to the far side, leaving boat and boys back at their initial positions " If no soldiers remain, we have finished, 3

5 Ferrying Soldiers!Apply decrease-by-1 process: " Ferry one soldier to the far side, leaving boat and boys back at their initial positions " If no soldiers remain, we have finished, " otherwise, ferry remaining n 1 soldiers 3

6 Ferrying Soldiers!Apply decrease-by-1 process: " Ferry one soldier to the far side, leaving boat and boys back at their initial positions " If no soldiers remain, we have finished, " otherwise, ferry remaining n 1 soldiers!how many (one way) boat trips will it take to ferry one soldier? 3

7 Ferrying Soldiers!Apply decrease-by-1 process: " Ferry one soldier to the far side, leaving boat and boys back at their initial positions " If no soldiers remain, we have finished, " otherwise, ferry remaining n 1 soldiers!how many (one way) boat trips will it take to ferry one soldier? A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 3

8 Alternating Glasses! There are 2n glasses standing in a row, the first n of them filled with beer, while the remaining n glasses are empty. Make the glasses alternate in a filled-empty-filled-empty pattern in the minimum number of moves. " Interchanging (any) two glasses is one move 4

9 Alternating Glasses! There are 2n glasses standing in a row, the first n of them filled with beer, while the remaining n glasses are empty. Make the glasses alternate in a filled-empty-filled-empty pattern in the minimum number of moves. " Interchanging (any) two glasses is one move 4

10 Alternating Glasses! There are 2n glasses standing in a row, the first n of them filled with beer, while the remaining n glasses are empty. Make the glasses alternate in a filled-empty-filled-empty pattern in the minimum number of moves. " Interchanging (any) two glasses is one move 4

11 Alternating Glasses! There are 2n glasses standing in a row, the first n of them filled with beer, while the remaining n glasses are empty. Make the glasses alternate in a filled-empty-filled-empty pattern in the minimum number of moves. " Interchanging (any) two glasses is one move! Apply decrease by-a-constant: " What smaller problem can we solve that will help? 5

12 Depth-first Search!Levitin says: " Depth-first Search uses a Stack, while " Breadth-first search uses a queue 7

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15 Example: BFS traversal of undirected graph a b c d e f g h BFS traversal queue: BFS tree: 10

16 Depth-first Search!Levitin says: Depth-first Search uses a Stack, Breadth-first search uses a queue!where s the stack? 11

17 DFS with explicit stack!dfs(r) S stack.empty S.push(r) while {S.notEmpty} do { } u S.pop if (u.hasbeenvisited.not) then { } u.markvisited u.adjacentverticesdo { v S.push(v) } 12

18 Example: DFS traversal of undirected graph a b c d e f g h DFS traversal stack: DFS tree: 13

19 Topological Sorting Example Order the following items in a food chain tiger human fish shrimp sheep plankton wheat 14

20 Topological Sort using decrease-by-one!basic idea: " topsort a graph with one less vertex " combine the additional vertex with the sorted graph!problem: " How to choose a vertex that can be easily recombined? 15

21 Which vertex should we remove? fish human shrimp tiger sheep A. fish B. shrimp C. plankton D. wheat E. sheep F. human G. tiger plankton wheat 16

22 Decrease by a Constant Factor!binary search and bisection method ( 12.4)!exponentiation by squaring!multiplication à la russe 17

23 Variable-size decrease!euclid s algorithm!median (or percentile) by partition!nim-like games 18

24 Finding the Median!The Median of an array of numbers is the middle number, when sorted.!we can obviously find the median by sorting j the array, and then picking the n k th element 2!How much work is that (in average case)? A. O(n) B. O(n lg n) C. O(n 2 ) D. something else 19

25 Median in Linear Time?! Can we do better?! After all, sorting the whole array is more work than is needed to find the median! Key insight: generalize the problem!! Rather than seeking an algorithm for the th element, lets look for the k th element, k [1..n] Suppose that we have a way of partitioning the array at element with value p: How can this help? l p p p r j n 2 k 20

26 l r A = p p p!suppose that we are looking for the 10 th element, and: # Alo = 5 # Ap = 1 Alo Ap Ahi " Then we can seek the 4 th element of Ahi instead " We have reduced the problem size by a variable amount, in this case Alo + Ap = 6 21

27 l r A = p p p Alo Ap Ahi!Suppose that we are looking for the 10 th element, and: # Alo = 28 " Then we can seek the 10 th element of Alo instead " We have reduced the problem size by a variable amount, in this case Ap + Ahi 22

28 l r A = p p p!suppose that we are looking for the 8 th element, and: # Alo = 6 # Ap = 2 Alo Ap Ahi " Then we can seek the 2 nd element of Ap instead. " We have now solved the problem, because all the elements of Ap are p 23

29 Variable-size decrease?!what s the connection? " suppose that we have A[1:20] and are looking for the 7 th -smallest element: " run partition, find s = 9, say " Where do we look for the 7 th -smallest element? A: A[1..20] B: A[1..8] C: A[1..9] D: A[10..20] 24

30 Variable-size decrease?!what s the connection? " suppose that we have A[1:20] and are looking for the 7 th -smallest element: " run partition, find s = 3, say " Where do we look for the 7 th -smallest element? A: A[1..3] B: A[1..4] C: A[3..20] D: A[4..20] 25

31 What s the Efficiency?!Dasgupta s analysis shows that: if we can do the partition in O(n) time, then we can select the k th element in O(n) time!how can we do partition in O(n) time? Lomuto Partition Hoare Partition 26

32 Lomuto Partition!While algorithm is running: l s i r p < p p?!invariant: " A[l] = p A[l+1.. s] < p A[s+1.. i 1] p l s < i r!establish invariant initially: " p A[l]; s l; i s+1 // makes < p interval and p intervals both empty 27

33 I don't like Lumuto Partition!It does more swaps than necessary l s i r p < p p? 28

34 I don't like Lumuto Partition!It does more swaps than necessary l s i r p < p p? " half of the swap is wasted!it confuses students! " Quicksort does not use the Lumuto Partition 28

35 Hoare Partition!Classic algorithm of computing!developed in 1959, published in 1961.!Not only linear, but peculiarly efficient!!tony Hoare won the Turing Award for Quicksort, which is based on this algorithm... and some other things! 29

36 Partition: CACM (Vol 4) July 1961 ALGORITHM 63 PARTITION C. A. R. HOARE Elliott Brothers Ltd., Borehamwood, Hertfordshire, En procedure partition (A,M,N,I,J); value M,N; array A; integer M,N,1,J; conunent I and J are output variables, and A is the array (wi subscript bounds M:N) which is operated upon by this procedur Partition takes the value X of a random element of the array and rearranges the values of the elements of the array in such way that there exist integers I and J with the following propertie M _-< J < I =< NprovidedM < N A[R] =< XforM =< R _-< J A[R] = XforJ < R < I A[R] ~ Xfor I =< R ~ N The procedure uses an integer procedure random (M,N) whi chooses equiprobably a random integer F between M and N, a also a procedure exchange, which exchanges the values of its tw parameters ; begin real X; integer F; F := random (M,N); I:=M; J:=N; up: X := A[F]; for I : = I step 1 until N do if X < A [I] then go to down; I:=N; down: forj := J step --1 until M do if A[J]<X then go to change; J:=M; change: else if I < J then begin exchange (A[IL A[J]); I := I+ 1;J:= J - 1; go to up end if [ < F then begin exchange (A[IL A[F]) i I:=I+l end else if F < J tllen begin exchange (A[F], A[J]) ; J:=J-1 end ; end partition 30

37 Partition: CACM (Vol 4) July 1961 ALGORITHM 63 PARTITION C. A. R. HOARE Elliott Brothers Ltd., Borehamwood, Hertfordshire, En procedure partition (A,M,N,I,J); value M,N; array A; integer M,N,1,J; conunent I and J are output variables, and A is the array (wi subscript bounds M:N) which is operated upon by this procedur Partition takes the value X of a random element of the array and rearranges the values of the elements of the array in such way that there exist integers I and J with the following propertie M _-< J < I =< NprovidedM < N A[R] =< XforM =< R _-< J A[R] = XforJ < R < I A[R] ~ Xfor I =< R ~ N The procedure uses an integer procedure random (M,N) whi chooses equiprobably a random integer F between M and N, a also a procedure exchange, which exchanges the values of its tw parameters ; begin real X; integer F; F := random (M,N); I:=M; J:=N; up: X := A[F]; for I : = I step 1 until N do if X < A [I] then go to down; I:=N; down: forj := J step --1 until M do if A[J]<X then go to change; J:=M; change: else if I < J then begin exchange (A[IL A[J]); I := I+ 1;J:= J - 1; go to up end if [ < F then begin exchange (A[IL A[F]) i I:=I+l end else if F < J tllen begin exchange (A[F], A[J]) ; J:=J-1 end ; end partition Important features: random pivot double-ended search works in place two outputs 30

38 Hoare Partition method partition(a, lo, hi) { def pivotindex = randombetween(lo)and(hi) def pivot = A[pivotIndex] var i := lo 1 var j := hi+1 while { do { i := i + 1 } while { (i <= hi).andalso {A[i] <= pivot} } do { j := j 1 } while { (j >= lo).andalso {A[j] >= pivot} } i < j } do { exchange(a, i, j) } if (i < pivotindex) then { exchange(a, i, pivotindex) ; i := i + 1 } elseif (j > pivotindex) then { exchange(a, pivotindex, j) ; j := j 1 } list.with(i, j) } 31

39 Before partition begins: i lo p hi j 7

40 Before partition begins: i lo p hi j Leave elements that are already in the right place: i j lo hi 7

41 Before partition begins: i lo p hi j Leave elements that are already in the right place: i j lo hi Now a[i] p a[j], so swap a[i] and a[j]: i j lo hi 7

42 Before partition begins: i lo p hi j Leave elements that are already in the right place: i j lo hi Now a[i] p a[j], so swap a[i] and a[j]: i j lo p hi 7

43 Before partition begins: i lo p hi j Leave elements that are already in the right place: i j lo hi Now a[i] p a[j], so swap a[i] and a[j]: i j lo p p hi 7

44 Before partition begins: i lo p hi j Leave elements that are already in the right place: i j lo hi Now a[i] p a[j], so swap a[i] and a[j]: i j lo p And continue... i j p hi lo hi 7

45 when do we stop? And continue... i j lo hi until i and j cross! j i lo p p hi 33

46 when do we stop? And continue... i j lo hi until i and j cross! j i lo p p hi is this possible? j i lo hi 33

47 12 Coins " This problem is originally stated as:! You have a balance scale and 12 coins, 1 of which is counterfeit. The counterfeit weigh less or more than the other coins. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter? " A harder and more general problem is:! For some given n > 1, there are (3 n 3)/2 coins, 1 of which is counterfeit. The counterfeit weigh less or more than the other coins. Can you state a priori n weighting experiments with a balance, with which you determine the counterfeit coin, and tell if it is heavier or lighter? 34

48 Problem: Mazes One can model a maze by having a vertex for a starting point, a finishing point, dead ends, and all the points in the maze where more than one path can be taken, and then connecting the vertices according to the paths in the maze. a. Construct such a graph for the following maze. b. Which traversal DFS or BFS would you use if you found yourself in a maze and why? 35

49 Problem: Mazes One can model a maze by having a vertex for a starting point, a finishing point, dead ends, and all the points in the maze where more than one path can be taken, and then connecting the vertices according to the paths in the maze. a. Construct such a graph for the following maze. b. Which traversal DFS or BFS would you use if you found yourself in a maze and why? 35

50 Problem: Mazes One can model a maze by having a vertex for a starting point, a finishing point, dead ends, and all the points in the maze where more than one path can be taken, and then connecting the vertices according to the paths in the maze. a. Construct such a graph for the following maze. b. Which traversal DFS or BFS would you use if you found yourself in a maze and why? A: DFS B: BFS 35

51 Problem: Gray Code Use the decrease-by-one technique (Algorithm BRGC) to generate the binary reflected Gray code for n = 4. 36

52 Problem: Gray Code Use the decrease-by-one technique (Algorithm BRGC) to generate the binary reflected Gray code for n = n = 1 n = 2 n = 3 36

53 Problem: Gray Code Algorithm! Trace the following algorithm for generating the Binary Gray Code of order 4. Start with code = 0000 output code for i = 1 to 15 do: b position of least significant 1 in binary rep of i code code XOR (bit b) output code 37

54 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 38

55 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 38

56 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 38

57 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 39

58 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 39

59 Nim!1 pile of n chips!players take turns removing 1 k m chips!the player removing the last chip wins m = 4 39

60 Multiplication à la russe n m = ( n 2 2m if n is even n 1 2 2m + m if n is odd 40

61 You try it!!multiply n m