ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI
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1 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI MICHAEL RAND 1. Introduction The Tower of Hnoi puzzle ws creted over century go by the number theorist Edourd Lucs [, 4], nd it nd its vrints hve chllenged children s well s professionl mthemticins nd computer scientists ever since. The puzzle consists of three verticl pegs ttched to bse, nd number of disks of different dimeters stcked on them. In the simplest version of the puzzle, ll the disks re plced on one peg in order of decresing size, so tht the lrgest disk is on the bottom nd the smllest t the top of the pile: The gol is to move the entire tower to nother peg, while obeying the following restrictions: Only one disk my be moved t time. No disk my be plced on smller disk. The min questions regrding this puzzle hve been: Wht is the minimum number of moves required to trnsfer the entire tower from one peg to nother? Wht is the move sequence tht will chieve this? The minimum number of moves for tower of n disks ws quickly shown to be n 1, with the simple recursive solution s follows: Lbel the three pegs strt, gol, nd temp. To move n pegs from the strt peg to the gol peg vi the temp peg, (1 If n 1, move the one disk from strt to gol. ( If n > 1, (i Recursively move the top n 1 disks from strt to temp vi gol. (ii Move the n-th disk from strt to gol. (iii Recursively move the n 1 disks from temp to gol vi strt. This solution tkes n 1 moves: (1 If n 1, f(n 1 n 1 ( If n > 1, f(n ( n n + 1 n 1 Dte: 4/6/009. 1
2 MICHAEL RAND In 1908, Henry Ernest Dudeney published book of puzzles which included vrition on the stndrd Tower of Hnoi [1]. Dudeney replced the disks with wheels of cheese, but more importntly dded fourth peg, gretly complicting the problem. He described this problem s The Reve s Puzzle. Then in 1939 the Americn Mthemticl Monthly published generlized Tower of Hnoi problem, sking for solution for n disks on ny number k of pegs [5]. Two yers lter, the sme journl published pir of solutions, by J. S. Frme nd the puzzle s proposer, B. M. Stewrt [6]. Frme nd Stewrt offered lgorithms for solving the problem in minimum number of moves, but n editoril note pointed out tht neither hd proved their lgorithm correct. This is where we stnd tody: no one hs yet offered proven solution to the generlized Tower of Hnoi problem, but most ssume tht the solutions of Frme nd Stewrt re correct. Since Frme nd Stewrt s solutions were lter proven to be equivlent (see [3], we will describe the simpler of the two, nd refer to it s the Frme-Stewrt lgorithm. In this description we will use the following definition: Definition 1.1. For k N, k 3, let H k : N N be function which returns the minimum number of moves to solve the Tower of Hnoi problem for n disks on k pegs ccording to the Frme-Stewrt lgorithm. For exmple, H 3 (n n 1 nd H 4 (5 13. The Frme-Stewrt lgorithm is s follows: For n disks on k pegs, if k 3, use the lgorithm given bove. If n 0, obviously no moves re required. For k > 3, n > 0, choose n integer l stisfying 0 l < n tht minimizes the steps used in the following formul: Move the top l disks from the strt peg to n intermedite peg; this cn be ccomplished in H k (l moves (since the bottom n l disks do not interfere with movements t ll. Move the bottom n l disks from the strt peg to the gol peg using H k 1 (n l moves. (Since one peg is occupied by tower of smller disks, it cnnot be used in this stge. Move the originl l disks from the intermedite peg to the gol peg in H k (l moves. Therefore, the recursive formul for solutions to the Tower of Hnoi problem ccording to the Frme-Stewrt lgorithm is s follows: 0 if n 0 H k (n x 1 if k 3, n > 0 min H k(l + H k 1 (n l if k > 3, n > 0 0 l<n Remrk 1.. It is interesting to note tht we could derive the results for 3-peg Tower of Hnoi from the Frme-Stewrt lgorithm using the degenerte cse of k. If, following our logicl intuition, we set H (0 0, H (1 1, H (n for ll n > 1 then for k 3, n > 1 we would hve to set l n 1 (in order tht our second term H (n l be finite, nd our formul becomes H 3 (n H 3 (n 1 + H (1 H 3 (n which ws our originl recursive solution for k 3.
3 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 3. Frme-Stewrt Differences In this section we will prove some interesting numericl results regrding the solutions produced by the Frme-Stewrt lgorithm. Although these results re probbly well-known to experts in the subject, the literture in the field is somewht scttered. We hope tht the cler nd explicit presenttion of these results nd their proofs provided in this pper will be of interest to both newcomers nd those fmilir with the problem. The min thrust of this section will be Theorem.3; the rest of the section will be dedicted to proof of this Theorem. (In wht follows, ll vribles represent elements of Z 0 : {0, 1,,...} unless stted otherwise. Remrk.1. It my be useful to mention the following identity regrding binomil coefficients, s it will come up repetedly in this pper. ( ( ( b b + 1 b + 1 A proof is provided in ppendix A. Definition.. Let D k : N N be the function defined by D k (i H k (i H k (i 1. Thus, D k (i represents the number of dditionl moves required by the Frme-Stewrt lgorithm s the number of disks is incremented from n 1 to n. Theorem.3. Let n, k, x N, n 1, k 3. Then for the Tower of Hnoi problem on k pegs, ( ( + x k + x x 0, D k (n x < n k k where ( ( b is the usul binomil coefficient nd < b b 0. Equivlently, if we view D k s sequence {D k (i} i N, the sequence will consist of ( k 3 of the vlue 0, followed by ( k 3+1 k 3 of the vlue 1, ( k 3+ k 3 of, etc. Exmple.4. For k 5 the sequence {D 5 } is s follows: {( ( ( {D 5 } of 0, of 1, { 1 of 0, 3 of 1, 6 of, 10 of 3,... } { 0, 1, 1, 1,,,,,,, 3, 3... } {1,,,, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8,...} } of,... Proof of Theorem.3. The proof of Theorem.3 will use double strong induction on the number of disks nd the number of pegs. It will follow the following generl outline: Prove tht Theorem.3 holds for ll n if k 3. Prove tht Theorem.3 holds for n 1 for ll k > 3. Prove tht, ssuming Theorem.3 holds for ll k < k 0 nd n < n 0, then Theorem.3 holds for k k 0, n n 0. Bse Cse. Theorem.3 holds for ll n if k 3.
4 4 MICHAEL RAND Proof. We need to show tht D 3 (n x But H 3 (n n 1, so we hve ( ( x 1 < n 1+x 1, i.e. D3 (n n 1. D 3 (n H 3 (n H 3 (n 1 n 1 ( n 1 1 Bse Cse. Theorem.3 holds for n 1 for ll k > 3. n 1 Proof. The sttement of Theorem.3 reduces to showing tht D k (1 1. This is trivilly true, since for ll k it is esy to show H k (1 1. Inductive Step. Assuming Theorem.3 for ll k < k nd n < n, then Theorem.3 holds for k, n. In order to prove the inductive step, we will need the following definitions: Definition.5. For the Tower of Hnoi problem on k pegs, we will cll number n of disks perfect if n cn be represented s n ( k +x k for some x 0. Definition.6. For k pegs nd n disks, we will cll number l with 0 l < n n optiml prtition number if H k (n H k (l + H k 1 (n l Note tht this mens l is n optiml prtition number if nd only if l such tht 0 l < n, H k (l + H k 1 (n l H k (l + H k 1 (n l In other words, l is prtition number relizing the minimum number of moves using the Frme-Stewrt lgorithm. The proof of the inductive step will consist of the following prts: Assuming tht Theorem.3 holds for ll numbers of disks less thn n nd for ll numbers of pegs less thn k, We will determine some upper nd lower bounds for the optiml prtition number, l, of n. We will show tht within these bounds, ll l re optiml prtition numbers. Using this informtion, we will be ble to directly clculte D k (n for the relevnt n using the inductive hypothesis. Lemm.7. Assuming Theorem.3 up to but not including some number of disks n, then for n stisfying ( ( k < n k +x k, we cn stte the following bout ny optiml prtition number l: ( ( + x l k k Note ( tht our proof of the bse cse n 1 implies tht there does exist some ( k < n stisfying Theorem.3, specificlly k 3+1 k 1. Proof. If l is n optiml prtition number, then, by definition, l stisfying 0 l < n, H k (l + H k 1 (n l H k (l + H k 1 (n l We will use proof by contrdiction: If n optiml prtition number l is outside of the forementioned bounds, then there exists n l with H k (l + H k 1 (n l < H k (l + H k 1 (n l.
5 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 5 Assume first tht l < ( k. We will prove tht l l + 1 contrdicts the bove inequlity. Since l < n, our inductive hypothesis, combined with the fct tht l ( k, implies tht Dk (l x. Additionlly, l < ( ( k nd n > k imply tht n l > ( ( k ( k k 3. Therefore, by our inductive hypothesis, D k 1 (n l x. Now, reclling tht D k (n H k (n H k (n 1, we get H k (l + H k 1 (n l (H k (l + D k (l + (H k 1 (n l D k 1 (n l H k (l + ( x + H k 1 (n l x < H k (l + H k 1 (n l contrdicting our definition of l s n optiml prtition number. Next ssume l > ( k nd compre it to l l 1. (Recll tht we require l < n, so we cn use the inductive hypothesis on l nd l. Since l > ( k nd n ( ( k +x k, we hve n l < k +x ( k ( k ( k 3 nd n l k 3. Now the inductive hypothesis long with l > ( k implies tht Dk (l x, nd together with n l ( k 3 implies tht Dk 1 (n l x. Therefore H k (l + H k 1 (n l (H k (l D k (l + (H k 1 (n l + D k 1 (n l H k (l ( x + H k 1 (n l + x < H k (l + H k 1 (n l once gin contrdicting the definition of l s n optiml prtition number. Lemm.8. Assuming Theorem.3 up to but not including some number of disks n, then for n stisfying ( ( k < n k +x k nd ny optiml prtition number l, ( ( + x n l Proof. The proof is similr to tht of Lemm.7: First ssume n l < ( k 3. Then l > n ( ( k 3 > ( k ( k 3 k so tht Dk (l x 1, nd if we let l l 1 we get D k 1 (n l x 1. Then H k (l + H k 1 (n l (H k (l D k (l + (H k 1 (n l + D k 1 (n l H k (l ( x 1 + H k 1 (n l + x 1 < H k (l + H k 1 (n l ( so l cnnot be n optiml prtition number. Therefore we know tht n l k 3. Now ssume n l > ( k 3. Then Dk 1 (n l x+1, nd l < n ( k 3 ( k +x ( k ( k 3 k so tht if we let l l+1 then the inductive hypothesis implies D k (l x 1. Now we cn clculte H k (l + H k 1 (n l (H k (l + D k (l + (H k 1 (n l D k 1 (n l H k (l + ( x 1 + H k 1 (n l x+1 < H k (l + H k 1 (n l contrdicting the ssumption tht l is n optiml prtition number, so we conclude tht n l ( k 3.
6 6 MICHAEL RAND Remrk.9. Note tht for ny perfect n ( k +x k, Lemm.7 sttes tht ll optiml prtition numbers l re ( k, nd Lemm.8 sttes tht n l ( ( k 3 l k, so we hve proved tht Theorem.3 lso implies ( ( k + x + x n l k k In other words, when n is perfect the optiml prtition number l is both unique nd perfect, s well s simple to clculte. We will return to this interesting point lter. Lemm.10. Assuming Theorem.3 up to but not including some number of disks n, then for n stisfying ( ( k < n k +x k, if n integer l stisfies the conditions stted in Lemms.7 nd.8, then l is n optiml prtition number. Proof. We will show tht ny choice of l in the given rnge results in the sme expression for H k (l + H k 1 (n l. Since the definition of n optiml prtition number sttes tht the vlue of H k (l + H k 1 (n l is minimized, this implies tht ll l in the given rnge re optiml. Let n ( ( k +, where 0 < ( k 3. Now set l k + b nd n l ( k 3 + c, where b + c. By Lemms.7 nd.8 we hve ( ( k l ( k, nd ( k 3 n l k 3. By the bounds on l (Lemm.7, nd the inductive hypothesis we see tht (( ( ( H k (l H k + l x 1 k k H k (l b + (b x 1 while by the sme rgument the bounds on n l (Lemm.8 imply (( ( ( H k 1 (n l H k 1 + n l x H k 1 (n l c + (c x ( Therefore, H k (l + H k 1 (n l (H k (l b + (b x 1 + (H k 1 (n l c + (c x H k (l b + H k 1 (n l c + (b + c x (( (( H k + H k 1 + x k H k (n + x This lst line relies on the fct tht by the inductive hypothesis we know tht the (unique optiml prtition number for n ( ( k is l k. Therefore ( ( ( ( H k (n H k k + H k 1 k 3. Since we ttin the sme expression no mtter wht we choose for b nd c (provided they fulfill the conditions stted in Lemms.7 nd.8, ll such l re optiml prtition numbers. Finlly, ( lso completes the proof of Theorem.3
7 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 7 3. Consequences of Theorem An itertive method for computing H k. Now tht we hve more informtion bout the vlues of H k, we cn clculte H k (n without using recursive formul: Theorem 3.1. For k 3, nd ( k x 1 ( + i H k (n i0 ( < n k +x for some x 0, i + ( n Proof. This is direct result of Theorem.3, since n H k (n H k (0 + D k (i x i0 i1 ( k +i k j( k 3+i k +1 x 1 (( k + i k i0 x 1 ( + i i0 i + ( i + k n i( k +1 ( + i i + k ( ( + x n ( + x ( x ( n x ( + x x k x 3.. Clculting l directly. It would be useful to be ble to clculte n optiml prtition number l directly from the number of disks n. In the following Theorem nd Corollry, we provide function tht clcultes l directly from n for the 4-peg Tower of Hnoi. Theorem 3.. For the Tower of Hnoi puzzle on k 4 pegs, for ny number of disks n, if ( ( x+1 < n x+, then is n optiml prtition number. l n (x + 1 Proof. Recll the bounds we defined for n optiml prtition number l in Lemms.7 nd.8: For n stisfying ( ( k < n k +x k, we require ( ( ( ( + x + x l nd n l k k For k 4 this reduces to the following: For ( ( x+1 < n x+, ( ( ( ( x x + 1 x x + 1 l nd n l 1 1 The second condition is obviously stisfied by our choice of l, since n l x + 1. The first condition is lso stisfied: ( ( ( x + x + 1 x + 1 l n (x + 1 1
8 8 MICHAEL RAND nd ( ( x + 1 x + 1 l n (x + 1 > 1 ( ( ( x + 1 x x l n (x (The lst eqution is implied becuse we re deling exclusively with integers. Since l is within the bounds specified by Lemms.7 nd.8, Lemm.10 implies tht l is n optiml prtition number. Corollry 3.3. For the Tower of Hnoi puzzle on k 4 pegs, for ny number of disks n, n 1 l n + is n optiml prtition number. Proof. First we crete function g : N N which inverts the binomil function f(x ( ( x in the following mnner: g(y x x 1 ( < y x. Then g(y x ( ( x+1 < y x+, nd by Theorem 3., we cn find n optiml prtition number l defined by l n (x + 1 n ((g(n + 1 As proven in Appendix B, g(y y + 3 inverts the binomil function in the mnner described, nd therefore n 3 n 1 l n (g(n + 1 n + 1 n + is n optiml prtition number A simple recursive solution for perfect numbers. Recll from Remrk.9 tht when n is some perfect number n ( k +x k, we know tht there is unique optiml l ( ( k, nd therefore unique optiml n l k 3. In prticulr, l is perfect on k pegs, nd n l on k 1 pegs, so tht ech prt of the recursive solution H k (n H k (l + H k 1 (n l is lso perfect nd esy to clculte. In fct, the recursive solution bers strong resemblnce ( to the binomil eqution tht forms the bsis for Pscl s tringle: +1 ( b+1 ( b+1 + b. In this cse the eqution is (( (( (( + 1 H k H k + H k 1 b + 1 b + 1 b Using two vrints of Pscl s tringle, we cn quickly clculte H k (n for ny perfect number n of disks s follows: Recll tht the entries in Pscl s tringle, which we will denote P, hve the form P i,j ( i j 1 if j 0 0 if i 0, j > 0 P i 1,j + P i 1,j 1 if i > 0, j > 0 where P i,j is the entry in row i, column j of P, indexed from 0.
9 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 9 First we build version of Pscl s tringle, P, with the column indices offset by two, i.e. P i,j is only defined on i 0, j nd P i,j P i,j so tht P i, 1 for ll i 0. This llows us to mp x nd k to perfect n ( x k s follows: P x,k ( x k n. Next we build second tringle P with the sme column offset, but insted of Pscl s P i,j P i 1,j + P i 1,j 1 we set P i,j P i 1,j + P ( ( x k i 1,j 1 Then we cn mp x nd k to P x,k H ( (x 1 ( (x 1 k H k k + H k 1 k 3. Now, using P to clculte x bsed on n, nd P to clculte H k (n bsed on x, we cn quickly nd conveniently clculte H k (n for ny perfect number of disks n on k pegs: P k x P k x In order to find H k (n for some perfect number n, first we look up n under column k in P. Then we cn discover x by looking t the index of the row contining n. Finlly, we look up x, k in P to get P x,k H k(n. Exmple 3.4. If we wnt to look up H k (n for the perfect number of disks n 15 on k 4 pegs, first we look for n in tble P under column k 4. We discover tht P 6,4 15, so x 6. Then we look up P x,k P 6,4 to discover tht H 4 ( Exmple 3.5. To find H 5 (10, we first look up 10 in tble P under column k 5. Then we see tht x 5, nd looking up P 5,5 shows tht H 5 (10 31.
10 10 MICHAEL RAND Figure 1. The set of ll optiml prtition numbers l of n on 5 pegs Appendix A. Proof tht ( ( b + ( b+1 +1 b+1 Proof. By the definition of ( b! ( b!b!, we hve ( (! + b b + 1 ( b!b! +! ( b 1!(b + 1!!(b + 1 ( b!(b + 1! +!( b ( b!(b + 1!!( b + b + 1 ( b!(b + 1! ( + 1! (( + 1 (b + 1!(b + 1! ( + 1 b + 1 Appendix B. Inverse of ( x Proposition B.1. The function g : N N defined by y 3 g(y + fulfills the condition g(y x ( x 1 < y We will find the following lemms useful in our proof: ( x
11 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 11 b Lemm B.. For, b N,, b > 0, < b Lemm B.3. For, b N,, b > 0, b b < +b Proof. ( b > 0 + b > b ( + b > 4b ( + b > b Lemm B.4. Let f(x x(x 1 N, g(f(x x. ( x nd g(y y + 3. Then x Proof. Exmine g(f(x: ( x g(f(x + 3 x(x (x x (by Lemm B. nd by Lemm B.3 we hve x + (x 1 x(x 1 < 3 x(x 1 + < x x + 1 x(x < x + 1 x(x So g(f(x is n integer greter thn or equl to x nd less thn x + 1, i.e. g(f(x x. We cn now complete the proof tht g(y y + 3 fulfills the necessry condition: Proof. For x, y N, if y stisfies ( x 1 (x (x 1 < y (x 1(x ( x, then g(y x. First, we will prove tht y > (x (x 1 implies g(y x: Since both y
12 1 MICHAEL RAND nd (x (x 1 re integers, we hve (x (x 1 y + 1 x 3x + 4 ( x 3 > 3 y + > x y 3 + x (since x is n integer Second, y (x 1x f(x implies g(y x: Since g(y is n incresing function, nd by Lemm B.4 we hve g(f(x x, then y f(x implies g(y g(f(x x. This completes our proof tht ( ( x 1 x g(y x < y
13 ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI 13 References [1] H. E. Dudeney. The Cnterbury puzzles, nd other curious problems. E.P. Dutton nd co., New York, [] A. M. Hinz. An itertive lgorithm for the Tower of Hnoi with four pegs. Computing, 4(- 3: , [3] Sndi Klvžr, Uroš Milutinović, nd Ciril Petr. On the Frme-Stewrt lgorithm for the multi-peg Tower of Hnoi problem. Discrete Appl. Mth., 10(1-3: , 00. [4] Dvid G. Poole. The towers nd tringles of Professor Clus (or, Pscl knows Hnoi. Mth. Mg., 67(5:33 344, [5] B. M. Stewrt, Richrd Bellmn, F. A. Lewis, nd V. Thebult. Problems for Solution: Amer. Mth. Monthly, 46(6: , [6] B. M. Stewrt nd J. S. Frme. Problems nd Solutions: Advnced Problems: Solutions: Amer. Mth. Monthly, 48(3:16 19, 1941.
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