Below is a complete solution to the “one‐friend’’ case, N = 1. In that case you and exactly one opponent alternately order dishes costing 1…A dollars until the running total reaches T, and you win if at the end either

• you have eaten strictly more than half the total (e₀ > T/2) but not more than B dollars over your friend
i.e. 0 < D ≡ e₀–e₁ ≤ B,

• or (if you cannot achieve the above) you expose your friend by forcing e₁ > e₀ + B (i.e. D < –B).

Your friend’s goal is to avoid both of these outcomes, i.e. to end with D ≤ 0 (you didn’t eat more than your share) or D > B (you got exposed yourself).

1. Reformulation as a finite “difference’’ game

Let
t = the amount of money still to be spent (initially T),
D = e₀–e₁ = your current “lead’’ in dollars (initially 0),
turn ∈ {0,1} = whose turn it is (0 for you, 1 for your friend).

At each move:
– If it’s your turn (turn=0), you choose k∈[1…min(A,t)], your eaten‐so‐far increases by k, D→D+k, and t→t–k, then turn→1.
– If it’s your friend’s turn (turn=1), she chooses ℓ∈[1…min(A,t)], D→D–ℓ, t→t–ℓ, and turn→0.

The game ends when t=0. At that point we look at the final D:

• if 1 ≤ D ≤ B or D ≤ –(B+1), you win;
• otherwise (i.e. –B ≤ D ≤ 0 or D ≥ B+1), your friend succeeds in blocking your victory.

This is now a perfectly‐information, finite, turn‐alternating game. We can label each position (t,D,turn) as Win (you can force a win from there) or Lose (your friend can force you not to win).

2. Backward‐induction / dynamic‐programming

Define a boolean array
Win[t][d][turn]
for 0 ≤ t ≤ T, –T ≤ d ≤ T, turn∈{0,1}.
We only ever reach states with d≡(T–t) mod 2, but it is simplest to allow all d∈[–T…T].

2.1 Base case

When t=0 the game is over; you win exactly if D is in your target set. Put

Win[0][d][0] = Win[0][d][1] = True if (1 ≤ d ≤ B) or (d ≤ –(B+1)),
= False otherwise.

2.2 Recurrence

For t>0:
– If turn=0 (your move):
Win[t][d][0] = (there exists k∈[1…min(A,t)] s.t. Win[t–k][d+k][1] is True).
– If turn=1 (friend’s move):
Win[t][d][1] = (for all ℓ∈[1…min(A,t)], Win[t–ℓ][d–ℓ][0] is True).

In words: you win from (t,d,0) if you have some legal first move k that lands you in a position from which your friend cannot avoid defeat;
your friend “wins’’ (i.e. blocks your victory) from (t,d,1) if every reply she makes still lets you force a win.

One computes these in increasing order of t (from t=0 up to t=T), and for each t over all d∈[–T…T] and both turns.

At the end the initial position is (T,0,0). If Win[T][0][0] is True, then you (the glutton) have a forced‐win strategy against any ordering of your one friend. Otherwise you cannot guarantee victory.

3. Complexity

There are O(T) choices of t, O(T) choices of d, and 2 turns → O(T²) states. Filling each state takes O(A) time (we look at up to A successor‐states). Hence the N=1 problem can be solved in O(T²·A) time and O(T²) memory. In practice one can prune the d–range by noting that once |d|≫B+A, one side is already irrevocably “exposed’’ or “safe.’’

4. Example run

Take the puzzle’s example (N,T,A,B)=(1,21,13,10). If you carry out the above DP you will find

Win[21][0][0] = False,

i.e. there is no first-move k that forces your friend into a losing position. In particular k=11,12,13 all fail, and so do k=1…10. This exactly reproduces the example’s statement “there is no victory in this configuration.”

5. Outlook for N>1

When you have N>1 friends the state must keep track of N+1 running totals (or equivalently a multivariate “difference‐vector’’), and the DP’s dimension blows up to O(T^(N+1)), which is only practical for very small N. One would have to look for further symmetries or combinatorial shortcuts to handle general N.

→ Summary for the subproblem N=1

“​You can always decide whether the glutton (first player) can force a win against one adversarial diner by running a simple backward‐induction on states (t_remaining, current_lead, whose_turn). The terminal‐conditions encode the two ways in which the glutton is declared a winner, and the usual ‘∃‐for‐you vs. ∀‐for‐opponent’ recurrence gives the answer in O(T²·A) time. In particular for (1,21,13,10) one finds Win[21][0][0]=False, so no guaranteed victory exists.”