A276128 - OEIS

%I #22 Jul 12 2025 08:39:05

%S 0,1,2,4,6,8,10,12,14,16,18,21,24,27,30,34,38,42,46,49,52,55,58,61,64,

%T 67,70,73,76,79,82,86,90,94,98,102,106,110,114,119,124,129,134,139,

%U 144,149,154,160,166,172,178,182,186,190,194,198,202,206,210,214,218,222,226,230,234,238,242,246,250,254,258,262,266,270,274,278,282,286,290

%N a(n) is the minimum cost to determine a number among 1...n by multiple guesses, where the cost of each guess is the guessed value and the answer is whether the guess is low, high, or correct.

%C Definition of a(n): For a positive integer n, let the single-player game G(n) be as follows: x is a number in {1, 2, ..., n}, but unknown to the player. The player can guess as many times as he wants to determine the value of x. For each guess, the player can propose a possible value c in {1, 2, ..., n}, but such guess will cost the player c dollars. After each guess, the player will get response to show whether c<x, c=x, or c>x.

%C A guess strategy will consist a series of guesses to determine x. The cost of multiple guesses is defined to be the sum of the cost of each guess. The cost of guess strategy is defined to be the worse case of the cost of the guess series. The optimal guess strategy for the game G(n) is the guess strategy that has the minimum cost. a(n) is the cost of the optimal guess strategy.

%H Robert Israel, <a href="/A276128/b276128.txt">Table of n, a(n) for n = 1..1000</a>

%e For n = 3, the best strategy is to use only one guess with value 2. So the cost is 2.

%e For n = 4, the best strategy is first make a guess with value 3. If the target value is less than 3, then make another guess 1. So the cost is 3+1=4.

%p f:= proc(a,b) option remember;

%p local t;

%p      if b-a <=0 then return 0

%p      elif b-a = 1 then return a

%p      else min(seq(t + max(procname(a,t-1), procname(t+1,b)), t=a..b))

%p      fi

%p end proc:

%p map2(f,1,[$1..100]); # _Robert Israel_, Sep 08 2016

%t A276128[n_] := (

%t    (*Allocate Memory*)

%t    dp = Table[0, {first, 1, n + 1}, {last, 1, n + 1}];

%t    (*Bottom-up Dynamic Programming*)

%t    For[length = 2, length <= n, length++,

%t        For[first = 1, first <= n + 1 - length, first++,

%t            (*[First,Last) contains n elements in total*)

%t            last = first + length;

%t            dp[[first, last]] = Min[Table[mid + Max[dp[[first, mid]], dp[[mid + 1, last]]], {mid, first, last - 1}]];]];

%t    (*Achieve Results*)

%t    Table[dp[[1, idx + 1]], {idx, 1, n}]);

%t A276128[79]

%o (C++)

%o // O(n^2) dynamic-programming:

%o // Let k0[first,last] = max{k: dp[first,k] <= dp[k+1,last]}, then

%o // max{dp[first,k],dp[k+1,last]} = dp[first, k] if k0[first,last]<k<=last

%o //                            else dp[k+1, last]

%o // So dp[first,last]

%o //    = min{max{dp[first,k],dp[k+1,last]}+k : first<=k<last}}

%o //    = min{f1[first,last], f2[first,last]}

%o // where f1[first,last] = min{first<=k<=k0[first,last]: dp[k+1,last]+k}

%o //       f2[first,last] = min{k0[first,last]<k<last: dp[first,last]+k}

%o //                      = dp[first,k0] + k0 + 1

%o int A276128(int n) {

%o     vector<vector<int>> dp(n + 1, vector<int>(n + 1));

%o     for (int last = 2; last <= n; ++last) { // for each last (included)

%o         int k0 = last - 1; // init k0

%o         deque<pair<int, int>> f1Cands; // candidates of f1

%o         for (int first = last - 1; first > 0; --first) { // for each first

%o             while (dp[first][k0 - 1] > dp[k0 + 1][last]) {

%o                 if ((!f1Cands.empty()) && (f1Cands.front().second == k0)) {

%o                      f1Cands.pop_front(); // out of range, remove

%o                 }

%o                 --k0; // update k0

%o             }

%o             int vn = first + dp[first + 1][last]; // new entry into the range

%o             while ((!f1Cands.empty()) && (vn < f1Cands.back().first)) {

%o                 f1Cands.pop_back();

%o             }

%o             f1Cands.emplace_back(vn, first);

%o             // update dp[first][last]

%o             int f1 = f1Cands.front().first;

%o             int f2 = dp[first][k0] + k0 + 1;

%o             dp[first][last] = min(f1, f2);

%o         }

%o     }

%o     return dp[1][n];

%o }

%K nice,nonn

%O 1,3

%A _Zhiqing Xiao_, Aug 21 2016