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Exact number of flips required to sort a burnt stack of pancakes

Gerold Jäger, Nacim Oijid

TL;DR

The paper tackles the exact minimum number of flips needed to sort a burnt stack of pancakes, focusing on the worst-case burnt instance and modeling the problem on the burnt pancake graph. It introduces an algorithmic search approach to generate flipping sequences for base cases and then generalizes these patterns to all odd sizes $n$ with $n \equiv 1 \bmod 4$, proving $T(n)=\frac{3n+3}{2}$ for $n\ge 29$; together with existing results for $n\equiv 3 \pmod{4}$, this resolves the odd-$n$ case completely and reduces the even-$n$ case to two possibilities. The method hinges on decomposing sequences into three phases (W,A,B) that create clans, convert them into blocks, and finally merge blocks to form the sorted stack, with explicit flipping families $S_1,S_5,S_9$ depending on $n\bmod 12$. This work advances the understanding of exact bounds in burnt pancake sorting and demonstrates how targeted state-space search can produce constructive, scalable solutions for large combinatorial problems.

Abstract

For the buffet, the waiter of a restaurant gets a large stack of pancakes from the overworked cook. As usual, one side is burnt, and as the level of batter decreases, the pancakes became smaller and smaller. Hence, the waiter ends up with a stack of one-sided burnt pancakes sorted by size, with the larger at the bottom and burnt side up. However, the waiter cannot serve them this way. He needs to turn all the burnt sides down, without changing the order. Having only a spatula, he can only perform flips to the top of the stack. How can he perform this transformation in a minimum number of flips? Having n pancakes, this problem can be modeled in the burnt pancake graph, having 2^n*n! vertices, where each possible stack of pancakes corresponds to a vertex expressed by a permutation of size n, where the pancakes are ordered by size, and the pancake numbers are multiplied by -1, if the corresponding pancake has the burnt side side up. An edge exists in this graph, if the corresponding stacks can be reached from each other by one flip. Let T(n) be the minimum number of flips to sort the stack of n pancakes (-1,...,-n). General burnt pancake sorting has been introduced by Bill Gates and Papadimitriou. The instance (-1,...,-n) has strong relevance because of its easy structure and as it has been shown to be a worst-case instance for several small n. Heydari and Sudborough gave the currently best upper bound of T(n), namely (3n+3)/2 for n = 3 mod 4, which later has been shown to be exact by a work of Cibulka. Except these two works, no progress regarding lower and upper bounds has been made until now. In our work, we present that (3n+3)/2 is also an upper bound of T(n) for n = 1 mod 4, which again matches the lower bound of Cibulka and thus is exact. The case of even n keeps an open problem, where two possible values for T(n) are possible, namely (3/2)n + 1 or (3/2)n + 2.

Exact number of flips required to sort a burnt stack of pancakes

TL;DR

The paper tackles the exact minimum number of flips needed to sort a burnt stack of pancakes, focusing on the worst-case burnt instance and modeling the problem on the burnt pancake graph. It introduces an algorithmic search approach to generate flipping sequences for base cases and then generalizes these patterns to all odd sizes with , proving for ; together with existing results for , this resolves the odd- case completely and reduces the even- case to two possibilities. The method hinges on decomposing sequences into three phases (W,A,B) that create clans, convert them into blocks, and finally merge blocks to form the sorted stack, with explicit flipping families depending on . This work advances the understanding of exact bounds in burnt pancake sorting and demonstrates how targeted state-space search can produce constructive, scalable solutions for large combinatorial problems.

Abstract

For the buffet, the waiter of a restaurant gets a large stack of pancakes from the overworked cook. As usual, one side is burnt, and as the level of batter decreases, the pancakes became smaller and smaller. Hence, the waiter ends up with a stack of one-sided burnt pancakes sorted by size, with the larger at the bottom and burnt side up. However, the waiter cannot serve them this way. He needs to turn all the burnt sides down, without changing the order. Having only a spatula, he can only perform flips to the top of the stack. How can he perform this transformation in a minimum number of flips? Having n pancakes, this problem can be modeled in the burnt pancake graph, having 2^n*n! vertices, where each possible stack of pancakes corresponds to a vertex expressed by a permutation of size n, where the pancakes are ordered by size, and the pancake numbers are multiplied by -1, if the corresponding pancake has the burnt side side up. An edge exists in this graph, if the corresponding stacks can be reached from each other by one flip. Let T(n) be the minimum number of flips to sort the stack of n pancakes (-1,...,-n). General burnt pancake sorting has been introduced by Bill Gates and Papadimitriou. The instance (-1,...,-n) has strong relevance because of its easy structure and as it has been shown to be a worst-case instance for several small n. Heydari and Sudborough gave the currently best upper bound of T(n), namely (3n+3)/2 for n = 3 mod 4, which later has been shown to be exact by a work of Cibulka. Except these two works, no progress regarding lower and upper bounds has been made until now. In our work, we present that (3n+3)/2 is also an upper bound of T(n) for n = 1 mod 4, which again matches the lower bound of Cibulka and thus is exact. The case of even n keeps an open problem, where two possible values for T(n) are possible, namely (3/2)n + 1 or (3/2)n + 2.
Paper Structure (19 sections, 10 theorems, 96 equations)

This paper contains 19 sections, 10 theorems, 96 equations.

Key Result

Theorem 1

Let $n \ge 23$ be an integer with $n \equiv 3 \bmod 4$. Then it holds that $T(n) = \frac{3n+3}{2}$.

Theorems & Definitions (10)

  • Theorem 1: Heydari and Sudborough HS97
  • Theorem 2: Cohen and Blum CB95
  • Theorem 3: Cohen and Blum CB95
  • Theorem 4
  • Lemma 5
  • Lemma 6
  • Lemma 7
  • Lemma 8
  • Corollary 9
  • Corollary 10