- Email: [email protected]
On the random construction of heaps
Information Processing North-Holland
29 February
Letters 27 (1988) 103-109
ON THE RANDOM
CONSTRUCTION
1988
OF HEAPS
Alan M. FRIEZE Deportment of Computer Science and Statistics. United Kingdom El 4NS
Queen MOT College, University
of London, Mile End Rood, London,
Communicated by P. Henderson Received 1 April 1987 Revised 1 June 1987
We give a simple proof of the result of Bollobb and Simon (1985) on the linear expected time construction heap. We also give bounds on the probability that the construction time is much more than this and generalise d-heaps, da 2.
Keywork
Binary
heap,
d-heap,
upper
of a binary the result to
bound
1. Introduction The binary
heap is one of the basic data structures
of Computer
Science.
An array A[O] = - co, A[l], [3]) that an originally unsorted array A can be put into heap order in O(n) time. On the other hand, it is also known that the ‘repeated insertion’ Algorithm RI below requires S2(n log n) time in the worst case.
A[n] is in heap order if A[i] a A[[+i]] for i z 1. It is well known (see, for example. API,...,
Algorithm RI. The input, the bottom. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
real array A[l..n],
is transformed
into a binary
heap by repeated
insertion
from
begin A[O]:= -00; for t = 1 to n do begin p := t; q := [it]; a := A[t]; while A[ q] > a do begin A[p]:=a[q]; p:=q; end A[ p] := a end
q:=[+pJ
end
Bollobds and Simon [l] analysed the average case time complexity of Algorithm RI under the assumption that the input array A[l..n] contains a random permutation of { 1, 2,. . . , n}. If Tm( A) denotes the (random) number of exchanges (= executions of line 8) for input A, then they proved the following theorem. 0020-0190/88/%3.50
6 1988, Elsevier Science
Publishers
B.V. (North-Holland)
103
INFORMATION
Volume 27, Number 2
PROCESSING
LETTERS
29 February 1988
Theorem A
E(T,,(A))d(l+++o(l))n,
where+=
E - 1 ,=I 1+2’
= 0.7645.
(I)
(Strictly speaking they prove the result for n + 1 a power of 2. We do the same. For arbitrary n one can only double the bound from what is proued.) This is a nice result but its proof is rather involved and the main purpose of this paper is to simplify the proof. (It also removes the o(1) term in inequality (1)). By using a different model of randomness we are able to cut out most of the ‘combinatorial computation’. We are also able to prove some related results. For example, we prove that Algorithm RI is unlikely to require much more than (1 + $)n exchanges (n + 1 = 29. Theorem B. Let 0 < E < 1 be fixed.
Then, for some constant c > 0,
Pr(T,,(A)>,(l+++&)n)
=o(e-C’Z”“‘O).
We also consider d-heaps, d > 2, where all we require is A[i] 2 A[[(i + d - 2)/d J] and analyse the obvious generalisation of Algorithm RI,, in which [itj in line 5 is replaced by ]( t + d - 2)/d ] and ]ip] in line 8 is replaced by ]( p + d - 2)/d 1. We prove the following. Theorem C
2. Model of randomness and notation The main device for simplifying the proof of Bollobas and Simon is a change of model. Instead of considering A[l..n] to contain a random permutation of (1, 2,..., n} we assume that A[l], A[2],..., A[n] are independent uniform [0, 1) random variables. All we have to observe is that the number of exchanges usedbyAlgorithmRIonlydependsoni,, i,,...,i,,whereA[i,]cA[i,]< ...
Ai,k =
C jEL,
104
“j.a
ad
Bi,k =
c aj.k j-0
forO
INFORMATION
Volume 27, Number 2 3. Proof of Theorem
29 February 1988
PROCESSING LETTERS
A
Fix k > 0 and consider the insertion of the elements in level k. For t E L, let P(t) = {[t/2’] : j = k - l} be the vertices on the path from vertex t to the root of T,. Let
1, 2,...,
X,= ]{iEP(t):u,,,_l>ut}
1
= the number of exchanges in the insertion of u, by Algorithm RI d I{ i E
p(t) I u,,,~_,>
u,}I = Y,
say,
since ~~,~+tQ uiTSfor s > i. We prove that E
c i
and
then
r,
<(l++)lLkl,
fGL,
1
inequality
(1)
(2)
k=l,2,...,
follows from X, G x, as observed in [l].
Now, E(Y, I U’.7t-,’ i E P(t)) =
C Pr( 4 < ui,sk_,I qTk_, i E p(t)) = C iaP(r) iSP(f)
ui.7k_1.
Hence, E
c
k-l y,lU,,,k_,~l
=
tGL,
IEL,
C ui.s,_,= ,Co2*-‘Aj,k-l
C
iCP(t)
‘2s
k-2
= 2k
c
2-(J+‘)BJ,k_, + 2-(k-1’Bk_,,k-,
j-0
by straightforward algebra. Removing the conditioning yields E( ,zkx)=2’(
x2-‘/i”h(Bj,,_,)
+2-“-“E(B~-~*k-*)).
(3)
All that is needed are upper bounds to E( Bj,i) for 0
(4)
E( Bo,i) = l/2’+’
since Bo_i is the minimum of 2’+’ - 1 independent
uniform [0, l] random variables.
(5)
E(B,,,)=2’-f since B, i is the sum of 2’+’ - 1 independent uniform [0, l] random variables. To estimate E( Bj,;) for j < i we observe (see [l, Lemma 51) that
Bj,i G Bj,i-1 -
C
max{O, u,,T,_, - VI}9
tcL,
where u, = mm{ U, : s E Li and t E P(S)}. 105
INFORMATION
Volume 27, Number 2
PROCESSING
29 February 1988
LETTERS
value u, will rise to level j at time s and To see this, note that if u, < Us.,,_,. then the corresponding push down the contents of t. Thus, by this time, when looking at the sum of the contents of the first j levels, u, has replaced u ,.,, _, and some other values may have been reduced. Hence, BJ.iGBJ,i-t-
C
(ur,T,_,-ut)=BJ-,.i-,
+
C
(6)
01
IEL,
IGL,
and so
Bj.i) < E(B,_,,,_,)+ 2’/(2’-J+
E(
since each u, is the minimum
1)
of 2’-J independent 2’+2’-‘+
E(Bj.i)
G E(BO,i-j)
+
uniform
variables.
Thus,
2J+‘-1
0.1 +2
2’-’
[0, l] random
G
+ 1
2’-‘+I
(7)
’
on using (4). Using (5) and (7) in (3) yields k-2 C
2-(i+l)
2if’-1 p-1-j
j=O
As 1L, 1= 2k we have (2) and Theorem
+
+ 1 - 1/2k
< 2”(1+
$8).
1
A.
4. Proof of Theorem B
8,
Again we consider n + 1 = 2k and the insertion of level k and the random We shall prove that be the event {Z, E LkY, > , (1 + e)(l + +) 1L, I}.
variables
K for t E I!+,. Let
(8)
Pr( 8,) = o(e-r’n”‘/25). Since C, E L, Y, dj I LjIalways, we then will have Pr j=lk/21
=o
(ke
tCL,
- e*n”‘“/25
1
and this will prove the theorem. Let g2 be the event k-2
c 2-"+1)Bjek_1 +2-(k-1)Bk_ -l.k-1
<
(1
+
t&)(1
+
d’)
. i
j-0
Then, Pr(8,)
d Pr(gl
l&2;)
+
P@J
We shall show later on that pr( z2) =
and we quickly 106
o(e-‘2n”‘/25)
dispose
of Pr(&t I d$).
(9)
INFORMATION
Volume 27, Number 2
PROCESSING
LE-ITERS
29 February 1988
Consider the following part of Theorem 1 by Hoeffding [2, inequality (2.3)]: let 2,. Z,,. . . , Z,,, be independent random variables where 0 < Z, d 1 for i = 1, 2,. . . , m and E(( Z, + Z, + . . . + Z,,,)/nz) = /-L. Then, for t > 0, Pr((Z,+Z,+ We now observe Also,
that, given
E 2-k i Applying yields
inequality
c
+ Z,)/m
...
2 p + t) Q e-2mr’.
ui,?_ ,,...,
r, 1g2
(10) Y,, t E L,, are independent
u,~_,+,~_,, the variables
d (1 + &)(l
and 0 G Y, G k.
++).
i
teL,
(10) with
m = 2k, Z, = y/k,
i E L,,
p G (1 + ie)(l
+ 9)/k
and
t = ~(1 + +)/(2k)
pr( g1 , g*;) Q e-n(4++v*b3~)* and the theorem follows once we have dealt with equation Consider r E Lj and j < i. We show that 24I.7, d (j + 1)st smallest
24,: s E
of
()
,
L,andrEP(s)
u-j+1
i i.e., u, 7 is at most the (j + 1)st smallest of 2’-j+’ To prove this we inductively show that u,.,,<(j+l)stsmallestof
(9).
1
- 2 2 2*-j independent
{u,:rEP(s)},
jz0,
rEL,,
uniform
[0, l] random
t>,O,
variables.
(11)
where u, = u, if t z s and u, = cc otherwise. Now, clearly (11) holds for j = 0 and t z 0; assume it holds for all r E Lj_, and t z 0 for some j > 0. Now, (11) trivially holds while 1{s : r E P(s) and u, < cc} 1 -c j + 1 and so we have a basis for assuming (11) true up to some value of t. Let S = {s : r E P(s) and u, < cc}. We must confirm (11) for the next case t’~tsuchthatr~P(t’).LetS’=S~{t’}.Wemustshowthatu,,~~(j+l)stsmallestof{u,:sES’}. This is clearly true if u,, 2 u,,~ and if u,, < u,,,, then u,.,~ = u,~,~):, which is at most the jth smallest of a set which includes S. Hence, (11) continues to hold and the inductive step is complete. It then follows that Pr( A,,; >, a2*j-‘)
Q Pr(3r G
2j
e-(a-n]2J-‘-j(i-j)logc2)
Since Aj,i < 2j always, we can remove Pr( Bjai 2 $a2*jmi)
E Lj: u,,,, 2 a2j-j)
< 2j(*‘T)(l G 2j
the restriction
- a2j-i)2’-‘-i
e-(a-aj2J-‘-kz)
if a d 2’-j,
if a < 2’-j.
on a in the above inequality.
Hence,
< Pr( gJ< j : A;,, >, a2*/-‘) j Q
e-(a-mj2’-‘-kZ)
c
2/<
2j+l
e-(a-aj2’-‘-k2).
(12)
J%
Putting
a = 2k’5 we see that (i+l)Bj,k_l
2 $2-k/s
< Pr(3j
Q +k : 2-(j+‘)Bj+k_l
Q zn3/5
e-2k’5-‘_
> $ X 2j-4k’5)
(13) 107
Volume 27, Number 2
INFORMATION
PROCESSING
LETTERS
29 February 1988
Next we give a simple consequence of Hoeffding’s Theorem 1 [2] (easily derivable from inequality (2.1) of that paper): if Z,, Z,,..., Z, are as in (10) and 0 < E < 1, then 2 (1 + e)~) < e-EZmp/3.
Pr(( Z, + Z, + . . . + Z,)/m
(14)
Now suppose that i > [:kJ. It follows from (6) and (14) that 2i-r
Pr Bi_,,k-l-r ( forOgr
-e222’-‘-‘/3
(1 + E) 2/c-l-i+1
- Bi-r_l,k_2_ra
(15)
‘e i
Thus, s-l
Pr 2-(i+1)Bi,k_1 - 2-(i+‘)B,_S,k_,_s
2 (1 + E) c
,x’J”,+T
1
r-0
s-l d
If [;k] <
_r222t-k-‘,3
Ce r=O
i G [$I?],
-,=2”=/3
(16)
then we can use (12) with a = 2k/5 to show that
Pr( Bj_s,k_l_s/2i+1 If
>, $2-k/‘)
< 2i-S+1e-2k’5-‘.
[jk], then we use Bi_s,k_l_s/2i+’ Q 2-” Q 2-k/5. In conjunction with (16) we get that, for i > ]ik],
i >
Pr( B,,,_,/2’+’
> (1 + ~)/(2~-‘-’
+ 1) + 4
x
Zmk/‘)
=
o( n e-e’n”‘/6)_
(17)
It follows next from (14) that Pr(2- +l)B,,
k_l a 1 + E) s
e&2&-‘-v2)/3
as BO,k_ 1 is the sum of 2k - 1 independent (18) then imply that
(18)
uniform [0, l] random variables. (1n)equalities (13), (17), and
k-2
Pr
C2(
(i+l)Bj,k_l +
2-(k-1)Bk_l,k_l
2 (1 + .e)(l + +) + 4
X
j-0
2-k/5
= o(n e-e2n”J/6). i 09)
Replacing E in (19) by E’ satisfying (1 + e’)(l + +) + 4 X 2-k’5 = (1 + $e>(l + 9) yields (9).
5. Proof of Theorem C The idea of the proof is the same as that of Theorem A. This time 7” denotes an infinite d-ary tree where vertex i has children (i - l)d+j, i= 2, 3, d + 1. The level sets L, and the Ai,k, Bi,k are defined analogously, as are the variables X,, Y,, t E L,. In place of (3) we have
E(,z,K) =dk( xd-“+“(d108
l)E(B&,)
+ d+-%(Bk_l,k_l)
(20)
Volume
27. Number
INFORMATION
2
PROCESSING
LE-ITERS
29 February
1988
In place of (4) we have E(&;)
= (d-
l)/(d’+’
2).
(21)
d - l)] .
(22)
+ d-
In place of (5) we have E( B,,,) = (d’+’ - 1)/[2( In place of (6) we have
E( B,,~)< E(
B,_,,,_,) + d/Cd’-‘+
It then follows from (20)-(23)
and Theorem
1) <
d’+’
(d-
_
1
l)(d'-'+
1) ’
(23)
that
C follows.
6. Final comments We do not believe that the upper bound in Theorem A is tight, although it is not clear how to improve it significantly. We have not mentioned lower bounds. We can obtain a rather weak bound of approximately $n on the expected number of exchanges as follows: the number of exchanges needed to insert u,, t E L,, is at least what would the required assuming that levels L,, L,, . . . , L,_, have their contents at time TV. Thus, (24)
Bj,k is at least the sum of the 2’+’ - 1 smallest values in ui, ZQ,. . . , z+~+I_,, and so E(B, k) 2 2i+1(2/+1_ 1)/2k+2. Substitution of this into (24) yields E(ZE, E LIX,) 2 ($ - 0(1))2~.
Now,
Acknowledgment I am particularly Dyer for his helpful
grateful to Trevor comments.
Fenner
for his careful
reading
of the manuscript,
and to Martin
References [I] B. Bollobb and I. Simon, Repeated random insertion into a priority queue, J. Algorithms 6 (1985) 466-477. [2] W. Hoeffding, Probability inequalities for sums of bounded random variables, J. Amer. Statist. Assoc. 58 (1963) 13-30.
[3] D.E. Knuth, The Art of Computer Programming, Vol. 3, Sorting and Searching (Addison-Wesley, Reading, MA, 1973).
109
29 February
Letters 27 (1988) 103-109
ON THE RANDOM
CONSTRUCTION
1988
OF HEAPS
Alan M. FRIEZE Deportment of Computer Science and Statistics. United Kingdom El 4NS
Queen MOT College, University
of London, Mile End Rood, London,
Communicated by P. Henderson Received 1 April 1987 Revised 1 June 1987
We give a simple proof of the result of Bollobb and Simon (1985) on the linear expected time construction heap. We also give bounds on the probability that the construction time is much more than this and generalise d-heaps, da 2.
Keywork
Binary
heap,
d-heap,
upper
of a binary the result to
bound
1. Introduction The binary
heap is one of the basic data structures
of Computer
Science.
An array A[O] = - co, A[l], [3]) that an originally unsorted array A can be put into heap order in O(n) time. On the other hand, it is also known that the ‘repeated insertion’ Algorithm RI below requires S2(n log n) time in the worst case.
A[n] is in heap order if A[i] a A[[+i]] for i z 1. It is well known (see, for example. API,...,
Algorithm RI. The input, the bottom. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
real array A[l..n],
is transformed
into a binary
heap by repeated
insertion
from
begin A[O]:= -00; for t = 1 to n do begin p := t; q := [it]; a := A[t]; while A[ q] > a do begin A[p]:=a[q]; p:=q; end A[ p] := a end
q:=[+pJ
end
Bollobds and Simon [l] analysed the average case time complexity of Algorithm RI under the assumption that the input array A[l..n] contains a random permutation of { 1, 2,. . . , n}. If Tm( A) denotes the (random) number of exchanges (= executions of line 8) for input A, then they proved the following theorem. 0020-0190/88/%3.50
6 1988, Elsevier Science
Publishers
B.V. (North-Holland)
103
INFORMATION
Volume 27, Number 2
PROCESSING
LETTERS
29 February 1988
Theorem A
E(T,,(A))d(l+++o(l))n,
where+=
E - 1 ,=I 1+2’
= 0.7645.
(I)
(Strictly speaking they prove the result for n + 1 a power of 2. We do the same. For arbitrary n one can only double the bound from what is proued.) This is a nice result but its proof is rather involved and the main purpose of this paper is to simplify the proof. (It also removes the o(1) term in inequality (1)). By using a different model of randomness we are able to cut out most of the ‘combinatorial computation’. We are also able to prove some related results. For example, we prove that Algorithm RI is unlikely to require much more than (1 + $)n exchanges (n + 1 = 29. Theorem B. Let 0 < E < 1 be fixed.
Then, for some constant c > 0,
Pr(T,,(A)>,(l+++&)n)
=o(e-C’Z”“‘O).
We also consider d-heaps, d > 2, where all we require is A[i] 2 A[[(i + d - 2)/d J] and analyse the obvious generalisation of Algorithm RI,, in which [itj in line 5 is replaced by ]( t + d - 2)/d ] and ]ip] in line 8 is replaced by ]( p + d - 2)/d 1. We prove the following. Theorem C
2. Model of randomness and notation The main device for simplifying the proof of Bollobas and Simon is a change of model. Instead of considering A[l..n] to contain a random permutation of (1, 2,..., n} we assume that A[l], A[2],..., A[n] are independent uniform [0, 1) random variables. All we have to observe is that the number of exchanges usedbyAlgorithmRIonlydependsoni,, i,,...,i,,whereA[i,]cA[i,]< ...
Ai,k =
C jEL,
104
“j.a
ad
Bi,k =
c aj.k j-0
forO
INFORMATION
Volume 27, Number 2 3. Proof of Theorem
29 February 1988
PROCESSING LETTERS
A
Fix k > 0 and consider the insertion of the elements in level k. For t E L, let P(t) = {[t/2’] : j = k - l} be the vertices on the path from vertex t to the root of T,. Let
1, 2,...,
X,= ]{iEP(t):u,,,_l>ut}
1
= the number of exchanges in the insertion of u, by Algorithm RI d I{ i E
p(t) I u,,,~_,>
u,}I = Y,
say,
since ~~,~+tQ uiTSfor s > i. We prove that E
c i
and
then
r,
<(l++)lLkl,
fGL,
1
inequality
(1)
(2)
k=l,2,...,
follows from X, G x, as observed in [l].
Now, E(Y, I U’.7t-,’ i E P(t)) =
C Pr( 4 < ui,sk_,I qTk_, i E p(t)) = C iaP(r) iSP(f)
ui.7k_1.
Hence, E
c
k-l y,lU,,,k_,~l
=
tGL,
IEL,
C ui.s,_,= ,Co2*-‘Aj,k-l
C
iCP(t)
‘2s
k-2
= 2k
c
2-(J+‘)BJ,k_, + 2-(k-1’Bk_,,k-,
j-0
by straightforward algebra. Removing the conditioning yields E( ,zkx)=2’(
x2-‘/i”h(Bj,,_,)
+2-“-“E(B~-~*k-*)).
(3)
All that is needed are upper bounds to E( Bj,i) for 0
(4)
E( Bo,i) = l/2’+’
since Bo_i is the minimum of 2’+’ - 1 independent
uniform [0, l] random variables.
(5)
E(B,,,)=2’-f since B, i is the sum of 2’+’ - 1 independent uniform [0, l] random variables. To estimate E( Bj,;) for j < i we observe (see [l, Lemma 51) that
Bj,i G Bj,i-1 -
C
max{O, u,,T,_, - VI}9
tcL,
where u, = mm{ U, : s E Li and t E P(S)}. 105
INFORMATION
Volume 27, Number 2
PROCESSING
29 February 1988
LETTERS
value u, will rise to level j at time s and To see this, note that if u, < Us.,,_,. then the corresponding push down the contents of t. Thus, by this time, when looking at the sum of the contents of the first j levels, u, has replaced u ,.,, _, and some other values may have been reduced. Hence, BJ.iGBJ,i-t-
C
(ur,T,_,-ut)=BJ-,.i-,
+
C
(6)
01
IEL,
IGL,
and so
Bj.i) < E(B,_,,,_,)+ 2’/(2’-J+
E(
since each u, is the minimum
1)
of 2’-J independent 2’+2’-‘+
E(Bj.i)
G E(BO,i-j)
+
uniform
variables.
Thus,
2J+‘-1
0.1 +2
2’-’
[0, l] random
G
+ 1
2’-‘+I
(7)
’
on using (4). Using (5) and (7) in (3) yields k-2 C
2-(i+l)
2if’-1 p-1-j
j=O
As 1L, 1= 2k we have (2) and Theorem
+
+ 1 - 1/2k
< 2”(1+
$8).
1
A.
4. Proof of Theorem B
8,
Again we consider n + 1 = 2k and the insertion of level k and the random We shall prove that be the event {Z, E LkY, > , (1 + e)(l + +) 1L, I}.
variables
K for t E I!+,. Let
(8)
Pr( 8,) = o(e-r’n”‘/25). Since C, E L, Y, dj I LjIalways, we then will have Pr j=lk/21
=o
(ke
tCL,
- e*n”‘“/25
1
and this will prove the theorem. Let g2 be the event k-2
c 2-"+1)Bjek_1 +2-(k-1)Bk_ -l.k-1
<
(1
+
t&)(1
+
d’)
. i
j-0
Then, Pr(8,)
d Pr(gl
l&2;)
+
P@J
We shall show later on that pr( z2) =
and we quickly 106
o(e-‘2n”‘/25)
dispose
of Pr(&t I d$).
(9)
INFORMATION
Volume 27, Number 2
PROCESSING
LE-ITERS
29 February 1988
Consider the following part of Theorem 1 by Hoeffding [2, inequality (2.3)]: let 2,. Z,,. . . , Z,,, be independent random variables where 0 < Z, d 1 for i = 1, 2,. . . , m and E(( Z, + Z, + . . . + Z,,,)/nz) = /-L. Then, for t > 0, Pr((Z,+Z,+ We now observe Also,
that, given
E 2-k i Applying yields
inequality
c
+ Z,)/m
...
2 p + t) Q e-2mr’.
ui,?_ ,,...,
r, 1g2
(10) Y,, t E L,, are independent
u,~_,+,~_,, the variables
d (1 + &)(l
and 0 G Y, G k.
++).
i
teL,
(10) with
m = 2k, Z, = y/k,
i E L,,
p G (1 + ie)(l
+ 9)/k
and
t = ~(1 + +)/(2k)
pr( g1 , g*;) Q e-n(4++v*b3~)* and the theorem follows once we have dealt with equation Consider r E Lj and j < i. We show that 24I.7, d (j + 1)st smallest
24,: s E
of
()
,
L,andrEP(s)
u-j+1
i i.e., u, 7 is at most the (j + 1)st smallest of 2’-j+’ To prove this we inductively show that u,.,,<(j+l)stsmallestof
(9).
1
- 2 2 2*-j independent
{u,:rEP(s)},
jz0,
rEL,,
uniform
[0, l] random
t>,O,
variables.
(11)
where u, = u, if t z s and u, = cc otherwise. Now, clearly (11) holds for j = 0 and t z 0; assume it holds for all r E Lj_, and t z 0 for some j > 0. Now, (11) trivially holds while 1{s : r E P(s) and u, < cc} 1 -c j + 1 and so we have a basis for assuming (11) true up to some value of t. Let S = {s : r E P(s) and u, < cc}. We must confirm (11) for the next case t’~tsuchthatr~P(t’).LetS’=S~{t’}.Wemustshowthatu,,~~(j+l)stsmallestof{u,:sES’}. This is clearly true if u,, 2 u,,~ and if u,, < u,,,, then u,.,~ = u,~,~):, which is at most the jth smallest of a set which includes S. Hence, (11) continues to hold and the inductive step is complete. It then follows that Pr( A,,; >, a2*j-‘)
Q Pr(3r G
2j
e-(a-n]2J-‘-j(i-j)logc2)
Since Aj,i < 2j always, we can remove Pr( Bjai 2 $a2*jmi)
E Lj: u,,,, 2 a2j-j)
< 2j(*‘T)(l G 2j
the restriction
- a2j-i)2’-‘-i
e-(a-aj2J-‘-kz)
if a d 2’-j,
if a < 2’-j.
on a in the above inequality.
Hence,
< Pr( gJ< j : A;,, >, a2*/-‘) j Q
e-(a-mj2’-‘-kZ)
c
2/<
2j+l
e-(a-aj2’-‘-k2).
(12)
J%
Putting
a = 2k’5 we see that (i+l)Bj,k_l
2 $2-k/s
< Pr(3j
Q +k : 2-(j+‘)Bj+k_l
Q zn3/5
e-2k’5-‘_
> $ X 2j-4k’5)
(13) 107
Volume 27, Number 2
INFORMATION
PROCESSING
LETTERS
29 February 1988
Next we give a simple consequence of Hoeffding’s Theorem 1 [2] (easily derivable from inequality (2.1) of that paper): if Z,, Z,,..., Z, are as in (10) and 0 < E < 1, then 2 (1 + e)~) < e-EZmp/3.
Pr(( Z, + Z, + . . . + Z,)/m
(14)
Now suppose that i > [:kJ. It follows from (6) and (14) that 2i-r
Pr Bi_,,k-l-r ( forOgr
-e222’-‘-‘/3
(1 + E) 2/c-l-i+1
- Bi-r_l,k_2_ra
(15)
‘e i
Thus, s-l
Pr 2-(i+1)Bi,k_1 - 2-(i+‘)B,_S,k_,_s
2 (1 + E) c
,x’J”,+T
1
r-0
s-l d
If [;k] <
_r222t-k-‘,3
Ce r=O
i G [$I?],
-,=2”=/3
(16)
then we can use (12) with a = 2k/5 to show that
Pr( Bj_s,k_l_s/2i+1 If
>, $2-k/‘)
< 2i-S+1e-2k’5-‘.
[jk], then we use Bi_s,k_l_s/2i+’ Q 2-” Q 2-k/5. In conjunction with (16) we get that, for i > ]ik],
i >
Pr( B,,,_,/2’+’
> (1 + ~)/(2~-‘-’
+ 1) + 4
x
Zmk/‘)
=
o( n e-e’n”‘/6)_
(17)
It follows next from (14) that Pr(2- +l)B,,
k_l a 1 + E) s
e&2&-‘-v2)/3
as BO,k_ 1 is the sum of 2k - 1 independent (18) then imply that
(18)
uniform [0, l] random variables. (1n)equalities (13), (17), and
k-2
Pr
C2(
(i+l)Bj,k_l +
2-(k-1)Bk_l,k_l
2 (1 + .e)(l + +) + 4
X
j-0
2-k/5
= o(n e-e2n”J/6). i 09)
Replacing E in (19) by E’ satisfying (1 + e’)(l + +) + 4 X 2-k’5 = (1 + $e>(l + 9) yields (9).
5. Proof of Theorem C The idea of the proof is the same as that of Theorem A. This time 7” denotes an infinite d-ary tree where vertex i has children (i - l)d+j, i= 2, 3, d + 1. The level sets L, and the Ai,k, Bi,k are defined analogously, as are the variables X,, Y,, t E L,. In place of (3) we have
E(,z,K) =dk( xd-“+“(d108
l)E(B&,)
+ d+-%(Bk_l,k_l)
(20)
Volume
27. Number
INFORMATION
2
PROCESSING
LE-ITERS
29 February
1988
In place of (4) we have E(&;)
= (d-
l)/(d’+’
2).
(21)
d - l)] .
(22)
+ d-
In place of (5) we have E( B,,,) = (d’+’ - 1)/[2( In place of (6) we have
E( B,,~)< E(
B,_,,,_,) + d/Cd’-‘+
It then follows from (20)-(23)
and Theorem
1) <
d’+’
(d-
_
1
l)(d'-'+
1) ’
(23)
that
C follows.
6. Final comments We do not believe that the upper bound in Theorem A is tight, although it is not clear how to improve it significantly. We have not mentioned lower bounds. We can obtain a rather weak bound of approximately $n on the expected number of exchanges as follows: the number of exchanges needed to insert u,, t E L,, is at least what would the required assuming that levels L,, L,, . . . , L,_, have their contents at time TV. Thus, (24)
Bj,k is at least the sum of the 2’+’ - 1 smallest values in ui, ZQ,. . . , z+~+I_,, and so E(B, k) 2 2i+1(2/+1_ 1)/2k+2. Substitution of this into (24) yields E(ZE, E LIX,) 2 ($ - 0(1))2~.
Now,
Acknowledgment I am particularly Dyer for his helpful
grateful to Trevor comments.
Fenner
for his careful
reading
of the manuscript,
and to Martin
References [I] B. Bollobb and I. Simon, Repeated random insertion into a priority queue, J. Algorithms 6 (1985) 466-477. [2] W. Hoeffding, Probability inequalities for sums of bounded random variables, J. Amer. Statist. Assoc. 58 (1963) 13-30.
[3] D.E. Knuth, The Art of Computer Programming, Vol. 3, Sorting and Searching (Addison-Wesley, Reading, MA, 1973).
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