AbstractWe consider schemes of discrete search in the set {1,…,n} using only comparison queries, i.e. those of the form x<a? for aϵ{1,…,n}. Such a scheme is called i-adaptive if questions may be stated in i series, so that answers are obtained after each series but not between particular questions in it. We describe a worst-case optimal algorithm of i-adaptive search and determine its complexity. The latter turns out to be i·n1i where i may be a function of n. We also consider the closely related version of continuous search with a given error bound

Pelc, Andrzej

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Weakly adaptive comparison searching 

Theoretical Computer Science 66 (1989) 105-l 11 
North-Holland 
105 
NOTE 
WEAKLY ADAPTIVE COMPARISON SEARCHING* 
Andrzej PELC 
Dgpartement d’lnformatique, Universite’ du Qutfbec ci Hull, Case Postale 1250, Succursale “B”‘, Hull, 
Que’bec, Canada JSX 3X? 
Communicated by N. Nivat 
Received June 1986 
Abstract. We consider schemes of discrete search in the set { 1,. . . , n} using only comparison 
queries, i.e. those of the form x < a? for a E (1,. . . , n}. Such a scheme is called i-adaptive if 
questions may be stated in i series, so that answers are obtained after each series but not between 
particular questions in it. We describe a worst-case optimal algorithm of i-adaptive search and 
determine its complexity. The latter turns out to be i l n*ji where i may be a function of n. We 
also consider the closely related version of continuous search with a given error bound. 
We consider the problem of identifying an unknown number x E (1,. . . , n} using 
only comparison queries, i.e. questions of the form x c a ? for a E { 1, . . . , n}. This 
search can be viewed as a two-person game between the Questioner and the 
Responder. The latter thinks of a number x E { 1, . . . , n} which the Questioner has 
to determine using comparison queries. The game is called an i-adaptive search if 
the Questioner has to state all his questions in i series: questions from a given series 
are asked all at once, without waiting for answers. Those are disclosed only at the 
end of the series and are known at the time of constructing the next series of 
questions. The aim of this paper is to provide a worst-case optimal algorithm of 
i-adaptive search and to determine its complexity. By optimal we mean of course 
an algorithm using the minimal number of comparisons. 
Let us first remark that if arbitrary yes-no queries are allowed, the length of the 
Questioner’s optimal strategy is the same in the i-adaptive search for an 
equals [log n 1. (!og is the logarithm with base 2 and [xl denotes the least integer 
greater or equal than x.) This optimal strategy is then clearly asking about consecutive 
digits of the binary expansion of X, regardless of the answers. 
However, if only comparison queries are allowed, the degree of adaptability of 
the Questioner’s trategy turns out to have a major influence on the least possible 
length of the search. We shall denote by F(n) the minimal number of comparisons 
which enable the Questioner to win an i-adaptive searching game in (1,. . . , n), 
independently of the Responder’s choice and play. It is clear Gat T’(;j) s ?;-(n) for 
* This work was partially supported by the Natural Sciences and Engineering Research Council of 
Canada, Grant No. A8136. 
0304-3975/89/$3.50 @ 1989, Elsevier Science Publishers B.V. (North-Holland) 
106 A. Pelt 
i > j and Ti(n) G Ti(m) for n c m. Obviously, Ti( n) = [log n 1 for any i 2 [log n 1 
because [log n 1 comparisons are sufficient if the game is played in a fully interactive 
way. Then each series consists of a unique question which means that at the time 
of asking a query all previous answers are known. 
We shall now describe an optimal algorithm of i-adaptive search in { 1, _ . . , n}. 
If i = 1, i.e. all questions must be asked at once, it is easily seen that one needs to 
ask all queries XC~?, x<3?,..., x < n? Those n - 1 comparisons are clearly 
sufficient; on the other hand, omitting the comparison x < k? prohibits to determine 
x in case when answers to all queries x < m? for m < k are “no” and answers to 
all queries x < m ? for m > k are “yes”: we would not be able to decide whether 
x=k or x=&l. Hence T,(n)=n-1 for any natural n. 
Consider the i-adaptive search for i > 1. Let x < al?, x c a2?, . . . , x < ak._,?, for 
a,<-* 
answers 
Since al 
< ak._i j be the questions asked in the first series. Any consistent set of 
restricts the possible location of x to exactly one of the sets 
S()={I,...,al-I}, S~={a~,...,a~-]L},...,Sk_~={ak_~,...,n}. 
least one of them has [n/k] elements we get 
q(n)amin{(k-l)+&_,([n/kl): lskcn} , 
it is this set Sj of size 3 [n/k 1 which may have to be further searched using because 
(i - 1).adaptive search. 
On the other hand, if a1 < a2< l l l C a&1 are chos.\rl so that the sets 
S&S I,***, Sk_1 have sizes differing by at most 1 (in this case we say that 
a, 9 42, . . . , ak.+ form and even splitting of (1, . . . , n}), then all sets S’ have size at 
most [n/k]. It follows that stating queries x < a, ?, . . . , x < ak+ ? in the first series 
and then searching any set Sj using (i - l)-adaptive search takes (k L 1) + Ti-,( [n/ kl) 
comparisons. Thus 
;rl-(n)smin{(k-l)+T,_,([n/kl): lsksn). 
This gives the recursive formula for T(n): 
T,(n) = n - 1, 
z(n)=min{(k-l)+T_,([n/kl): lsksn} fori>l. 
The above considerations show that the following is a worst-case optimal algorithm 
of i-adaptive search in (1,. . . , n}. It is assumed that, for j6 i and m s n, the integer 
q(m) defined above and the integer tj( m) such that 
T(m) =@b) - O+ q-d [mlti(m)l) 
are known. (G(m) is equal to the value of k at which the minimum in the recursive 
formula for q(m) is taken.) 
Algorithm. i-adaptive comparison search in the set { 1, . . . , n}. 
Input: n: size of the search space 
i: number of series 
Weakly adaptive comparison searching 107 
Output: x: unknown element of (1, . . . , n}. 
begin 
set: ={l, . . . , n}; size:= n; min:= 1; max:= n 
{set keeps the value of the set presently searched, min is its smallest, max its 
largest element and size keeps its number of elements} 
for j:= i downto 2 do 
begin 
find an even splitting al, . . . , atj(size) of set; 
ask queries x < a,?, . . . , x c atjtsize)?; 
get answers; 
if (answer to x C a,? is “yes”) then 
{the leftmost interval was chosen} 
begin 
set:= {min, . . . , a, - 1); 
size := al - min; 
max:=a,-1 
end 
else 
if (answer to x C aritsizej? is “no”) then 
{the rightmost interval was chosen} 
begin 
set := { arj(sizej, . . . , max}; 
size := max- uti(size) +1; 
. 
mm :== Qti( size) 
end 
else 
begin 
find I such that the answer to x C a,? is “no” and the answer to x < al+,? 
is yes; 
set:={al,aI+l,...,al+,-1) 
size := aI+ 1 - al ; 
min := al; max := al+1 -1 
end 
end; 
{Now the unknown x must be in the current value of set which is {min, min+ 1, 
. . . , max} and has to be found in one series of comparisons}. 
ask queries x c min + 1 ?, . . . , x < max?; 
get answers; 
if (answer to x c min + 1 ? is “yes”) then 
X := min 
else 
if (answer to x c max? is “no”) then 
X := max 
108 A. Pelt 
else 
begin 
find c such that the answer to x < c? is “no” and the answer to x < c + 1 ? is 
“yes”; 
X 
.- 
l - c 
end 
end. 
Let us now consider the continuous version of the search problem: the task is 
now to find an unknown point x in an interval I of length a stating comparison 
queries of the form x c c? for c E 1, in an i-adaptive way. Clearly, it is now impossible 
to detect the exact position of the point. We rather try to minimize the length of 
the interval which is known to contain x. 
Denote by L(a, S i) this minimal length if the initial interval has length a and k 
questions can be stated in an i-adaptive way. Since after the first series of m questions 
the worst-case minimal length of inter-al to which search may be restricted is 
af (m + l), we get 
Lb, S 1) = aI@+ l), 
L(a,k,i)=min{L(a/(m+l),k-m,i-l):m<k-i+l} forkai>l. 
(m may take only positive integer values. It must be at most k - i + 1 to allow at 
least one query in each of the remaining i - 1 series.) 
This formula yields a worst-case optimal algorithm similarly as in the discrete 
case. In fact, both the continuous and discrete versions are closely related. The 
discrete search in the set { 1, . . . , n} can be interpreted as the search in [0, 1) with 
error bound l/n: all we have to do is locate x in one of the intervals 
I,=[OJ); **=[$f) ,..., k=[y,l). 
We shall now estimate the value of L(a, k, i). 
Lemma I. For any positive a and any positive integers k3 i we have 
Proof. The lemma is proved by inducGn on i. It is obvious for i = 1. (For k = i the 
search is fully interactive and the lower bound is taken: L(a, k, k) = a/2’.) 
Suppose that k> i and a/(k/i+l)‘s L(a, k, i). For i+l we get 
L(a, k, i+l)a min 
a 
msk-i(m+l)((k-m)/ii-1)’ 
2 inf- a 
m (m+l)((k-m)/i+l)” 
This lower bound is taken at the critical point of the function f(m) = 
(m+ l)((k-m)/i+l)‘; the derivative off is 
fi 1 ‘m = ( k-m . -+l ‘+(m+l)=i 
i ) 
(y*l)i-‘. (A); 
Weakly adaptive comparison searching 109 
hence,J’(m)=Oiff m=k/(i+l). Thus we obtain 
L(a, k, i+l)a 
(i:l+l)( a 
k-k/(i+l) 
i 
+l)‘=(~:l)i+19 
which proves the first inequality by induction. 
The second inequality is also true for i = 1. Assume 
L(a, k i)s(k/4.,‘. 
Let [r] denote the largest integer which does not exceed r. We have 
L(a, k, i+l)s min 
msk-i+l 
a 
s k 
i+l 
which concludes the proof by induction. Cl 
Let N(n, i, E) denote the least number k of questions ufficient o find a point x 
in an interval of length a with error < C, using i-adaptive comparison search. The 
Lemma 1 enables to estimate N(a, i, E). As a consequence we also estimate the 
number K(n) corresponding to the disc&e i-adaptive search in (1, . . . , n}. 
Thearem 1. (a) i(a/e)‘/‘-ii N(a, i, E)S i(a/&)“‘. 
(b) in”‘- i < Ti( n) G in”‘. 
Proof. (a): By Lemma 1 we have 
a(N(a,i,c)/i+l)-i~e~a(N(a, 
hence, 
N(a, i, E)+i> a ‘Ii, N(a, i, e) 
i (3 - E i 
i, E)/ i)-‘; 
and finally, 
i(a/ @Ii - is N(a, i, E) s i(a/e)‘/‘. 
110 A. Pelt 
(b): As remarked before, the i-adaptive discrete comparison search in { 1, . . . , n} 
can be interpreted as the i-adaptive continuous comparison search in [0, 1) with 
error bound l/n. Hence Ti( n) = N( 1, i, l/n), which gives 
inI/’ -is Ti(n)S in”‘. [zl 
Let US now discuss the complexity of i(n)-adaptive discrete search in the set 
{I ¶...9 n} where i(n) is a function of n. We need to evaluate the order of magnitude 
of Titn,( n). Clearly, i(n) s [log n 1 can be assumed because 
?;-(n) = TrlOg.l(n) = [log nl for anyja [log nj. 
For large n we have 3’Og n= n’Og3 s $n2, hence, for i(n) G [log n 1 s log n + 1, we get 
which implies $i( n) s i( n)n”““‘. Finally, we obtain 
7;:,,,(n)ai(n) l n”““‘-i(n)>fi(z) l n”““‘. 
Together with zfn,( n) S i(n) l n”““‘, this shows that 
magnitude of i(n) l n”““‘. 
T(,)(n) is of the order of 
In particular, if i is constant hen the worst-case optimal algorithm of i-adaptive 
comparison search in (1,. . . ) n} uses @(n”‘) queries. This is in sharp contrast to 
the case of arbitrary questions where increasing adaptability of the Questioner’s 
strategy does not shorten the scheme. 
Let us remark that, in order to perform our algorithm of weakly adaptive com- 
parison search, we have to supply the integers tj( m) for many values j s i and m e n. 
Though in terms of the number of comparisons the algorithm is optimal, the actual 
computation of those integers indicating the length of each series may be quite 
tedious. Below we show a simple modification of the main strategy for which the 
total number of comparisons may be larger but is of the same order of magnitude. 
The advantage of the modified algorithm is that the number of questions in each 
series is very easy to find (and thus the actual queries are also easy to construct). 
We put tF( m) = [rn”jl and replace the function fi by ~7 everywhere in our 
searching algorithm. Thus, if j series remain, the current one consists now of [m ‘“1 
questions corresponding to an even splitting of the set of size m which is currently 
searched. Let T”(n) denote the total number of comparisons used by this modified 
algorithm of i-adaptive search in { 1,. . . , n}. We shall show that the order of 
magnitude of T&,(n) is : , same as of Ticn,( n), where i(n) is a function of n. First 
we need the following inequality: for any positive integer i and any positive real X, 
w I/is ix’/‘+ 1. (I) 
Indeed, we have 
ii[x]s r;(x+ I) = iix+ iid i’x+ ii(x*/i)i-* 
= (ix’/‘)‘+ i(ixlli)i-l 6 (ix’li+ 1)‘; 
hence, i[xl*% ix’/‘+ 1. 
Weakly adaptive comparison searching 111 
The following lemma is proved by induction on i. 
Lemma2. T~(n)+n”‘+iforanyn, ial. 
Proof. Since T:(n) = T,(n) = n - 1, our lemma holds for i = 1. Suppose that T?(n) s 
in”‘+ i. Dividing the set { 1, . . . , n) using an even splitting into [n”‘] parts gives 
all parts of size at most [n/ [n"'l 1. Hence applying the inductive hypothesis and 
(I) we get 
Tr+,(n)S([n”““‘I-l)+i (1 Inl,~+*‘ll)l’i+ i 
s [n l/(i+l) 
1 (I 
+ i n 
l/i 
nl/(i+l) 1 ) 
+i 
n ( > 
I/i 
s (n’/(‘+“+ I)+ i - 
nl/“+l’ +i=(r+l)n*‘(‘+*‘+(i+l), 
which concludes the proof by induction. cl 
Now we are ready to prove the final theorem. 
Theorem 2. T(,, ,( n) and Tz,,,( n) have the same order of magnitude. 
Proof. As remarked before we may assume i(n) s [log n 1 and hence the inequalities 
i(n) G J”i( n)n’l’(“’ and z&n) 3 fi( n)n 1/i(n) 
hold for large n. Thus, using the lemma we obtain 
fi( n)n”““‘S Kcnl (n)c T:,,(n)si(n)n l/i(n) + qn) 
< $i(n)nlli(n) 
which concludes the proof. Cl 
The above result shows that for large values of i and n the modified 
may turn out more suitable because the queries are easier to construct 
number does not increase significantly compared to the former strategy. 
References 
algorithm 
and their 
[I] E. Horowitz and S. Sahni, Fundamentals ofcomputer Algorithms (Computer Science Press, Rockville, 
MD, 1978). 
[2] D.E. Knuth, The Art of Computer Programming Vol. III: Sorting and Searching (Addison-Wesley, 
Reading, MA, 1973). 


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Weakly adaptive comparison searching

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