Searching for quicksand ideals in partially ordered sets
aa r X i v : . [ m a t h . C O ] S e p SEARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS
ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTHAbstract.
We consider a combinatorial question about searching for an unknown ideal µ within a known poset λ . Elements of λ may be queried for membership in µ , but at most k positive query results are permitted. The goal is to find a search strategy which guarantees asolution in a minimal total number q k ( λ ) of queries. We provide tight bounds for q k ( λ ), andconstruct optimal search strategies for the case where k = 2 and λ is the product poset of totallyordered finite sets, one of which has cardinality not more than six. Introduction
Quicksand puzzle.
A surveyor stands in the northeast corner of a rectangular field λ ofdimension m × n . In the southwest corner of the field there may exist a rectangular quicksandpit µ of unknown dimension m ′ × n ′ . The surveyor has k stones available to toss into the fieldin order to identify safe and unsafe regions of the field. × × In order to gain information, the surveyor tosses a stone into some location x in the field. Ifthe stone does not sink, it follows that the region northeast of x is safe; the surveyor can ventureinto the field to retrieve the stone and use it again. If the stone does sink, the surveyor knowsthat the quicksand pit extends at least as far as x , but they now have one less stone with whichto work. How can the surveyor identify the location of the quicksand pit, and do so in a minimalnumber of tosses?1.2. Quicksand ideals in posets.
As we explain in § λ be a finite poset and k ∈ N . We seek to identify a (possibly empty)‘quicksand’ ideal µ contained in λ by sequentially querying elements of λ for membership in µ ,under the restriction that at most k positive query results are permitted. Letting q k ( λ ) representthe minimum total number of queries needed to guarantee identification of µ , our goal is to solve: Problem . Find the value q k ( λ ), and identify a search strategy which realizes this value. ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH
For all k ∈ N , the value q k ( λ ) has a recursive combinatorial description, as explained in § q k ( λ ) = λ = ∅ ; | λ | if k = 1;min { max { q k ( λ u ) , q k − ( λ ≻ u ) } | u ∈ λ } + 1 if k > , λ = ∅ , For any x ∈ Z ≥ , let T k ( x ) = P ki =1 (cid:0) xk (cid:1) , and let τ k ( x ) be the smallest integer such that x ≤ T k ( τ k ( x )). Our first main result provides bounds for q k ( λ ): Theorem A.
For all k ∈ N and posets λ , we have τ k ( | λ | ) ≤ q k ( λ ) ≤ | λ | .This appears as Theorem 4.2 in the text. These bounds are tight, in that q k ( λ ) = | λ | when λ has the trivial partial order, and q k ( λ ) = τ k ( | λ | ) when λ is totally ordered. In fact, when λ istotally ordered, Problem 1 is related to the ‘ k -egg’ or ‘ k -marble’ problem [ , – ], which appearsin numerous texts on dynamic programming and optimization, and perhaps apocryphally, as aninterview question for certain coding positions in big tech.1.3. Quicksand ideals in the product order, k = 2 case. After investigating general resultsdescribed in § κ, ν are totallyordered sets, we consider κ × ν to be a poset under the product partial order ; i.e.,( x , y ) (cid:23) ( x , y ) ⇐⇒ x ≥ x and y ≥ y , for x , x ∈ κ and y , y ∈ ν . We consider the k = 2 case, where T ( x ) is the triangular number1 + 2 + · · · + x = x ( x + 1) /
2, and τ ( x ) = ⌈ ( √ x + 1 − / ⌉ . Our second main result, whichappears as Corollary 6.5 in the text, provides a partial solution to Problem 1 in this setting: Theorem B.
Let κ, ν be finite totally ordered sets, with | κ | ≤ | ν | ≤
6. Then q ( κ × ν ) = ( | κ | = | ν | = 6; τ ( | κ || ν | ) otherwise . In Algorithm 6.3 we describe an explicit strategy, for any such κ, ν , which realizes the value q ( κ × ν ) above. In general, this strategy—and hence the proof of Theorem B—is rather deli-cately connected to the congruence class of τ ( | κ || ν | ) modulo | κ | and | ν | , and relies heavily onsome interesting number theoretic facts about triangular numbers proved in § q ( κ × ν ) in general, see § Solving the quicksand puzzle.
Theorem B offers a solution to the puzzle in § k = 2 and one dimension of the field is not more than six. Indeed, we mayconsider the field λ as the poset [1 , m ] × [1 , n ], depicted as a rectangular array of boxes in thefirst quadrant of the Cartesian plane. The quicksand pit is then an unknown ideal in λ , sinceany ideal µ ⊆ λ is either empty or equal to [1 , m ′ ] × [1 , n ′ ] for some m ′ ≤ m , n ′ ≤ n .Take the k = 2, λ = [1 , × [1 ,
7] example from § (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Strategy for 5 × The surveyor tosses their first stone into the locations marked (cid:13) , (cid:13) , (cid:13) , . . . , in sequence. If thisstone never sinks, then µ = ∅ . If the stone sinks on say, the i th toss, the remaining unclearedarea weakly northeast of this location (belonging to the same colored region as i (cid:13) ), is checkedsequentially with the remaining stone, in a southwesterly fashion. When the second stone sinksit will determine the northeast corner of µ , and if it never sinks, the northeast corner of µ is at i (cid:13) . This strategy identifies the quicksand pit µ in at most τ (5 ·
7) = 8 total tosses.2.
Partially ordered sets
In this section we give a brief primer on partially ordered sets and provide some preliminarydefinitions. See [ , ] for a complete treatment of the subject. We introduce the q k -functionwhich is the central topic of this paper, and explain how it relates to Problem 1.2.1. Posets. A partially ordered set (or poset ) is a set λ together with a binary relation (cid:23) , whichsatisfies the following conditions for all u, v, w ∈ λ :(i) u (cid:23) u ( reflexivity );(ii) u (cid:23) v and v (cid:23) u imply u = v ( antisymmetricity );(iii) u (cid:23) v and v (cid:23) w imply u (cid:23) w ( transitivity ).We use a ≻ b to indicate a (cid:23) b and a = b . The order (cid:23) is a total order if either u (cid:23) v or v (cid:23) u for all u ∈ v . An order-preserving map of posets λ, ν is a set map f : λ → ν such that f ( u ) (cid:23) f ( v ) whenever u (cid:23) v . We say two posets λ, ν are isomorphic and write λ ∼ = ν if thereexist mutually inverse order-preserving maps λ ⇄ ν .If κ, ν are posets, then κ × ν is a poset under the product partial order :( x , y ) (cid:23) ( x , y ) ⇐⇒ x (cid:23) κ x and y (cid:23) ν y , for all x , x ∈ κ and y , y ∈ ν . Our main examples of posets in this paper are the following: Example 2.1.
The trivial partial order on a set λ has u (cid:23) v if and only if u = v for all u, v ∈ λ . Example 2.2.
The natural numbers N = { , , . . . } are totally ordered under the usual ≥ relation, as is any interval [ a, b ] = { a, a + 1 , . . . , b } ⊂ N . In fact, if λ is any finite totally orderedset of cardinality m , then λ ∼ = [1 , m ]. Example 2.3.
Let m, n ∈ N . Then [1 , m ] , [1 , n ] are totally ordered sets as in Example 2.2. Wewrite J m, n K as shorthand for the poset [1 , m ] × [1 , n ] under the product partial order. If κ, ν are totally ordered sets of cardinality m, n respectively, then κ × ν ∼ = J m, n K .We represent elements of J m, n K as boxes situated in the first quadrant of the plane, arrangedso that ( a, b ) is a box in the a th row from the bottom, and in the b th column from the left.In this scheme, we have u (cid:23) v for u, v ∈ J m, n K if and only if the v box is weakly below andto the left (i.e. ‘southwest’) of the u box. For example, in the figure below we show the poset J , K , with the elements x = (4 , , y = (2 , , z = (2 , x (cid:23) z, y (cid:23) z , with x, y incomparable. x yz The poset J , K , with elements x, y, z ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH
Lower sets and ideals.
Let U be a subset of a poset λ . Then U is itself a poset underthe partial order inherited from λ , and we always assume we take this partial order on U . Wesay U is a lower set in λ provided that for all u ∈ U , v ∈ λ , u (cid:23) v implies v ∈ U . We say U is a directed set in λ provided that for all u, v ∈ U , there exists w ∈ U such that w (cid:23) u, v . We say U is an ideal in λ if it is a lower set and a directed set. In particular, we allow ideals to be empty.Let S, U ⊆ λ . We define subsets: S (cid:23) U = { v ∈ S | v (cid:23) u for some u ∈ U } S ≻ U = { v ∈ S | v ≻ u for some u ∈ U } S (cid:22) U = { v ∈ S | v (cid:22) u for some u ∈ U } S U = { v ∈ S | v u for all u ∈ U } = S \ S (cid:23) U . When U = { u } , we will write S (cid:23) u in place of S (cid:23){ u } , and so on. For any ordered sequence u = ( u , . . . , u r ) of elements of λ , we will also write S (cid:23) u in place of S (cid:23){ u ,...,u r } , and so on. Wewill often apply these definitions with S = λ .We will focus primarily on finite posets λ . In this setting every ideal is either empty or principal ; i.e. of the form λ (cid:22) u for some u ∈ λ , and every lower set is equal to λ (cid:22) U for some U ⊆ λ .2.3. The q k -function and Problem 1.Definition 2.4. Let k ∈ N , and let λ be a finite poset. We define the value q k ( λ ) ∈ Z ≥ recursively by setting: q k ( λ ) = λ = ∅ ; | λ | if k = 1;min { max { q k ( λ u ) , q k − ( λ ≻ u ) } | u ∈ λ } + 1 if k > , λ = ∅ , where we implicitly take the partial orders on λ u , λ ≻ u to be those inherited from λ . Example 2.5.
It is easy to check from Definition 2.4 that q k ( λ ) = | λ | when | λ | ≤
2. Let A = { a, b, c } , and consider the posets λ , λ , λ , λ with underlying set A , where the strictcomparisons in these posets are given as follows: λ : ∅ λ : { b ≻ a } λ : { b, c ≻ a } λ : { c ≻ b ≻ a } . Note that λ is the trivial poset on A and λ is a totally ordered set on A . We have q ( λ i ) = 3for i = 0 , , , q k ( λ ) = 3 q k ( λ ) = 2 q k ( λ ) = 3 q k ( λ ) = 2for all k ≥ µ is an unknown ideal in λ we wish to identify, and we may sequentially query elementsof λ for membership in µ , with the restriction that we must stop after the k th positive query.Note that since µ is an ideal in a finite set, we have that µ = ∅ or µ = λ (cid:22) x for some x ∈ λ . Let q ′ k ( λ ) represent the minimum total number of queries needed to guarantee identification of µ .We explain now that q ′ k ( λ ) = q k ( λ ).2.3.1. The λ = ∅ case. In this case we must have µ = ∅ , so no queries are needed to identify µ . Thus q ′ k ( ∅ ) = 0 = q k ( ∅ ). EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 5
The k = 1 case. With only one positive search query available, the search strategy isvery limited. Assume that u ∈ λ and we know v / ∈ µ for all v ≻ u by previous queries. Thena positive query at u will identify µ to be the ideal λ (cid:22) u . On the other hand, if there exists aelement v ≻ u whose membership in µ is unknown, a positive query result at u would result infailure, as µ could potentially be λ (cid:22) v or λ (cid:22) u , and we would be left with no further queries todistinguish these possibilities.We see then that the only permissible search strategy is to query all of the elements of λ insome non-increasing sequence, where the first positive query result will identify the generatorof the ideal µ . If µ = ∅ , the ideal will only be identified after the final (negative) query, so wehave q ′ k ( λ ) = | λ | = q k ( λ ).2.3.3. The general k > , | λ | > case. By induction, assume that q ′ ℓ ( ν ) = q ℓ ( ν ) for all ℓ < k or | ν | < | λ | . Assume the first query is at some element u ∈ λ . If the query is negative, this impliesthat µ ⊆ λ u , and we still have k positive queries to work with. By induction, the minimal totalnumber of queries necessary to guarantee identification of µ in λ u is q k ( λ u ).On the other hand, assume the query at u ∈ λ is positive. This implies that the ideal generator x could be any element in λ (cid:23) u , and we now have k − µ ′ be the ideal µ ∩ λ ≻ u in λ ≻ u . Then we have µ ′ = ∅ if and only if x = u , and µ ′ is nonempty ifand only if x ∈ λ ≻ u and µ ′ = ( λ ≻ u ) (cid:22) x . Therefore, identifying µ is equivalent to identifying theideal µ ′ in λ ≻ u . By induction, q k − ( λ ≻ u ) is the minimal total number of queries necessary toguarantee success in this search.Therefore if we begin by querying u , the minimal number of queries that will be necessary toguarantee identification of µ in λ is q k ( λ u ) + 1 if u / ∈ µ , and q k − ( λ ≻ u ) + 1 if u ∈ µ . Thus, byfirst querying u , the minimal number of queries necessary ismax { q k ( λ u ) , q k − ( λ ≻ u ) } + 1 . Therefore, taking the minimum over all possible choices of the initial query u , we have that q ′ k ( λ ) = q k ( λ ), as desired.3. Binomial sums and triangular numbers
Bounds for the q k -function will be shown to be directly related to binomial sums, and, in the k = 2 case, triangular numbers. In preparation for establishing this fact, we investigate someproperties of binomial sums, and triangular numbers in particular.3.1. Binomial sums.
Throughout this section, we fix k ∈ N . Definition 3.1.
Define the function T k : Z ≥ → Z ≥ via: T k ( x ) = k X i =1 (cid:18) xi (cid:19) . Notably, when k = 1 we have T ( x ) = x , and when k = 2 we have T ( x ) = x ( x + 1)2 = 1 + 2 + · · · + x, (3.2)the x th triangular number . The following function is key in describing lower bounds for the q k -function. Definition 3.3.
Define the function τ k : Z ≥ → Z ≥ by setting τ k ( x ) to be the unique non-negative integer such that T k ( τ k ( x ) − < x ≤ T k ( τ k ( x )) . ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH
The following two lemmas are clear from definitions.
Lemma 3.4.
For any x ∈ Z ≥ , we have τ k ( T k ( x )) = x . Lemma 3.5.
For any x ≤ y , we have τ k ( x ) ≤ τ k ( y ) . We now prove some additional useful technical lemmas on T k and τ k . Lemma 3.6.
For all x > , we have T k ( x ) = T k ( x −
1) + T k − ( x −
1) + 1 .Proof.
We have1 + T k ( x −
1) + T k − ( x −
1) = (cid:18) x − (cid:19) + k X i =1 (cid:18) x − i (cid:19) + k − X i =1 (cid:18) x − i (cid:19) = k X i =1 (cid:20)(cid:18) x − i (cid:19) + (cid:18) x − i − (cid:19)(cid:21) = k X i =1 (cid:18) xi (cid:19) = T k ( x ) , where the third equality follows from the binomial recurrence relation. (cid:3) Lemma 3.7.
Let x, y ∈ Z ≥ be such that y < T k − ( τ k ( x ) −
2) + 2 . Then τ k ( x ) − ≤ τ k ( x − y ) .Proof. We have by Lemma 3.6 that T k ( τ k ( x ) −
2) = T k ( τ k ( x ) − − T k − ( τ k ( x ) − − < T k ( τ k ( x ) − − y + 1 . By the definition of τ k ( x ) we have T k ( τ k ( x ) − < x , so T k ( τ k ( x ) − ≤ T k ( τ k ( x ) − − y < x − y. so by the definition of τ k ( x − y ), we have τ k ( x − y ) > τ k ( x ) −
2. Thus τ k ( x − y ) ≥ τ k ( x ) − (cid:3) Lemma 3.8. If x, y ∈ Z ≥ and n ∈ N are such that x ≡ n ) and T k ( τ k ( x )) ≡ y (mod n ) ,where ≤ y < n , then T k ( τ k ( x )) − x ≥ y .Proof. By definition, T k ( τ k ( x )) ≥ x . Then T k ( τ k ( x )) − x ≡ y (mod n ) , and T k ( τ k ( x )) − x ≥ T k ( τ k ( x )) − x = y + nt for some t ∈ Z ≥ , so the result follows. (cid:3) Triangular numbers.
Now we prove some technical lemmas in the case k = 2, recallingthat T ( x ) is the triangular number 1 + · · · + x . The next lemma is just a special case ofLemma 3.7. Lemma 3.9. If x, y ∈ Z ≥ , with y < τ ( x ) , then τ ( x ) − ≤ τ ( x − y ) . Lemma 3.10.
Let x, y ∈ Z ≥ , ℓ ∈ N , with ≤ y ≤ x − ℓτ ( x ) + T ( ℓ − . Then we have τ ( y ) ≤ τ ( x ) − ℓ. Proof.
By Definition 3.3, we have y ≤ x − ℓτ ( x ) + T ( ℓ − ≤ T ( τ ( x )) − ℓτ ( x ) + T ( ℓ − · · · + τ ( x )] − ℓτ ( x ) + [1 + 2 + · · · + ( ℓ − · · · + τ ( x )] − [( τ ( x ) − ( ℓ − · · · + ( τ ( x ) −
1) + τ ( x )]= 1 + 2 + · · · + ( τ ( x ) − ℓ ) = T ( τ ( x ) − ℓ ) . Then, applying τ to both sides of the inequality, we have by Lemmas 3.4 and 3.5 that τ ( y ) ≤ τ ( T ( τ ( x ) − ℓ )) = τ ( x ) − ℓ, as desired. (cid:3) EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 7
Lemma 3.11.
Let y, r, n ∈ N , with y ≡ r (mod n ) . Then: T ( y ) ≡ ( T ( r ) + n (mod n ) if n ≡ , y − rn ≡ T ( r ) (mod n ) otherwise . Proof.
We may assume without loss of generality that y ≥ r . Note that since y ≡ r (mod n ) ,we have y − r = nℓ for some ℓ ∈ Z ≥ . We prove the claim by induction on ℓ . Let ℓ = 0. Then y = r and y − rn ≡ T ( y ) = T ( r ) ≡ T ( r ) (mod n ) so the base case holds.Now assume ℓ > ℓ ′ < ℓ . Then T ( y ) = T ( r + nℓ )= ( r + nℓ ) + ( r + nℓ −
1) + · · · + ( r + nℓ − ( n − T ( r + n ( ℓ − nr − (0 + · · · + ( n − T ( r + n ( ℓ − nr − T ( n −
1) + T ( r + n ( ℓ − nr − ( n − n T ( r + n ( ℓ − ≡ − ( n − n T ( r + n ( ℓ − n )We consider three separate cases, based on the parity of n and ℓ . Case 1.
Suppose n is odd. Then we have that T ( r + n ( ℓ − ≡ T ( r ) (mod n ) by theinduction assumption. Therefore, T ( y ) ≡ − ( n − n T ( r + n ( ℓ − ≡ − n · ( n − T ( r ) ≡ T ( r ) (mod n ) . Case 2.
Suppose n is even and ℓ is odd. Then ℓ − T ( r + n ( ℓ − ≡ T ( r ) (mod n ) by the induction assumption. Then T ( y ) ≡ − ( n − n T ( r + n ( ℓ − ≡ − n n −
1) + T ( r ) (mod n ) ≡ − n −
1) + T ( r ) ≡ T ( r ) + n n ) . Case 3.
Suppose n is even and ℓ is even. Then ℓ − T ( r + n ( ℓ − ≡ T ( r ) + n (mod n ) by the induction assumption. Then T ( y ) ≡ − ( n − n T ( r + n ( ℓ − n ) ≡ − n n −
1) + T ( r ) + n ≡ n + T ( r ) ≡ T ( r ) (mod n ) . Thus in any case, the claim holds for ℓ , completing the induction step and the proof. (cid:3) Bounds on the q k -function Now we establish bounds on the q k -function. The following lemma is clear from Definition 2.4. Lemma 4.1. If λ ∼ = ν , then q k ( λ ) = q k ( ν ) . Theorem 4.2.
For all λ, k , we have τ k ( | λ | ) ≤ q k ( λ ) ≤ | λ | .Proof. We first prove that q k ( λ ) ≤ | λ | . The claim holds for k = 1 and λ = ∅ by Definition 2.4.Now let k > | λ | >
0, and assume q k ′ ( λ ′ ) ≤ | λ ′ | for all k ′ < k , | λ ′ | < | λ | . Let v be any maximal ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH element in λ . Then we have λ ≻ v = ∅ and | λ v | = | λ | −
1, so: q k ( λ ) = min { max { q k ( λ u ) , q k − ( λ ≻ u ) } | u ∈ λ } + 1 ≤ max { q k ( λ v ) , q k − ( λ ≻ v ) } + 1 ≤ max {| λ v | , } + 1 = ( | λ | −
1) + 1 = | λ | , as desired.Now we prove that τ k ( | λ | ) ≤ q k ( λ ). The claim holds for k = 1, as q ( λ ) = | λ | = (cid:0) | λ | (cid:1) = T ( | λ | ) , and the claim holds for λ = ∅ , as we have q k ( ∅ ) = 0 = P ki =1 (cid:0) i (cid:1) = T k (0) . Now let k > | λ | > τ k ′ ( | λ ′ | ) ≤ q k ′ ( λ ′ ) for all k ′ < k , | λ ′ | < | λ | . For some u ∈ λ , we have q k ( λ ) = max { q k ( λ u ) , q k − ( λ ≻ u ) } + 1 . Then by the induction assumption we have q k ( | λ | ) ≥ q k ( | λ u | ) + 1 ≥ τ k ( | λ u | ) + 1 (4.3)and q k ( | λ | ) ≥ q k − ( | λ ≻ u | ) + 1 ≥ τ k − ( | λ ≻ u | ) + 1 . (4.4)Assume by way of contradiction that q k ( | λ | ) < τ k ( | λ | ). First we claim that | λ ≻ u | < T k − ( τ k ( | λ | ) − | λ ≻ u | ≥ T k − ( τ k ( | λ | ) − τ k − ( | λ ≻ u | ) >τ k ( | λ | ) −
2, so τ k − ( | λ ≻ u | ) ≥ τ k ( | λ | ) −
1. But then τ k − ( | λ ≻ u | ) + 1 ≥ τ k ( | λ | ) > q k ( | λ | ) , a contradiction of (4.4). Thus | λ ≻ u | < T k − ( τ k ( | λ | ) − | λ ≻ u | +1 2) + 2, so by Lemma 3.7, we have τ k ( | λ | ) − ≤ τ k ( | λ | − | λ ≻ u | − . Therefore, applying (4.3) we have q k ( | λ | ) ≥ τ k ( | λ u | ) + 1 = τ k ( | λ | − | λ ≻ u | − 1) + 1 ≥ ( τ k ( | λ | ) − 1) + 1 = τ k ( | λ | ) > q k ( | λ | ) , a contradiction. Therefore τ k ( | λ | ) ≤ q k ( | λ | ), as desired. This completes the induction step, andthe proof. (cid:3) With the following two lemmas, we prove that the bounds of Theorem 4.2 are tight withrespect to arbitrary posets. Lemma 4.5. Let λ be a poset with trivial partial order. Then q k ( λ ) = | λ | .Proof. If λ = ∅ or k = 1, the claim follows by Definition 2.4. Now let k > | λ | > 0, and assume q k ′ ( λ ′ ) = | λ ′ | for all k ′ < k , and trivial posets λ ′ with | λ ′ | < | λ | . Let u ∈ λ . Then we have that λ ≻ u = ∅ , and λ u = λ \{ u } is itself a trivial poset. Therefore by the induction assumption wehave q k ( λ ) = min { max { q k ( λ u ) , q k − ( λ ≻ u ) } | u ∈ λ } + 1= min { max {| λ | − , } | u ∈ λ } + 1 = ( | λ | − 1) + 1 = | λ | , as desired. (cid:3) Lemma 4.6. Let λ be a totally ordered set. Then q k ( λ ) = τ k ( | λ | ) .Proof. As usual, we note that the claim holds for k = 1 , n = 0 by Definition 2.4. We now let k > | λ | > 0, and make the induction assumption that q k ′ ( λ ′ ) = τ k ′ ( | λ ′ | ) for all k ′ < k andtotally ordered λ ′ with | λ ′ | < | λ | . EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 9 We may assume λ = [1 , n ], as any totally ordered set of cardinality n is equivalent to thisinterval. Note that we have 0 ≤ T k ( τ k ( n ) − < n by Definition 3.3, so v := T k ( τ k ( n ) − 1) + 1 ∈ [1 , n ]. Then, applying Lemma 3.4, we have q k ( λ v ) = τ k ( | [1 , v − | ) = τ k ( v − 1) = τ k ( T k ( τ k ( n ) − τ k ( n ) − . On the other hand, we have q k − ( λ ≻ v ) = τ k − ( | [ v + 1 , n ] | ) = τ k − ( n − v ) = τ k − ( n − T k ( τ k ( n ) − − . Then we have q k − ( λ ≻ v ) = τ k − ( n − T k ( τ k ( n ) − − ≤ τ k − ( T k ( τ k ( n )) − T k ( τ k ( n ) − − τ k − (( T k − ( τ k ( n ) − 1) + 1) − 1) = τ k − ( T k − ( τ k ( n ) − τ k ( n ) − , using Lemma 3.5 and the fact that n ≤ T k ( τ k ( n )) by Definition 3.3 for the first inequality,Lemma 3.6 for the second equality, and Lemma 3.4 for the last equality.Thus we have q k ( λ ) = min { max { q k ( λ u ) , q k − ( λ ≻ u ) } + 1 | u ∈ λ }≤ max { q k ( λ v ) , q k − ( λ ≻ v ) } + 1 = ( τ k ( n ) − 1) + 1 = τ k ( n ) = τ k ( | λ | ) . Since q k ( λ ) ≥ τ k ( | λ | ) by Theorem 4.2, we have q k ( λ ) = τ k ( | λ | ). This completes the inductionstep, and the proof. (cid:3) Remark 4.7. The proof of Lemma 4.6 contains a solution to the strategy question from Prob-lem 1 for totally ordered sets, defined recursively for any k ∈ N . Namely, one should query theelement v such that | λ ≺ v | = T k ( τ k ( | λ | ) − λ ≺ v . If the query is positive and k = 1, stop. Otherwise, repeat theprocess with the totally ordered set λ ≻ v and k := k − 1. The final positive query will identifythe element which generates the ideal µ . Remark 4.8. In view of Theorem 4.2 and Lemmas 4.5 and 4.6, one may be led to conjecturethat q k ( λ ′ ) ≤ q k ( λ ) when λ ′ is a refinement of the poset λ . This does not hold in general,however. For a counterexample, see Example 2.5, where the posets λ , λ , λ , λ are sequentialrefinements, but the corresponding sequence of q k values is not monotonic when k ≥ Strategy in the k = 2 case We will now narrow our focus to the k = 2 setting. We develop a combinatorial language fordescribing query strategies in response to Problem 1. We fix some nonempty poset λ throughoutthis section. Definition 5.1. Let r ∈ N , and u = ( u , . . . , u r ) be a sequence of elements of λ . For each t = 1 , . . . , r , define the subset: λ ( t ) u := λ (cid:23) u t \ λ (cid:23){ u ,...,u t − } = { v ∈ λ | v (cid:23) u t , v u i for all i = 1 , . . . , t − } . If λ (cid:23) u = λ and λ ( t ) u = ∅ for all t = 1 , . . . , r , we call u a λ -strategy .By definition, the sets λ (1) u , . . . , λ ( r ) u are mutually disjoint, so if u is a λ -strategy, we have: λ = λ (1) u ⊔ · · · ⊔ λ ( r ) u . (5.2) ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH The Q -function.Definition 5.3. For a sequence of elements u = ( u , . . . , u r ) in λ , we define: Q ( λ, u ) := max {| λ ( t ) u | + t − | t = 1 , . . . , r } . We will primarily be concerned with the value of Q ( λ, u ) when u is a λ -strategy. Example 5.4. Let λ = J , K , and define the λ -strategy u = ((2 , , (5 , , (1 , , (3 , , (2 , , (1 , , (1 , . Then we may visually represent u in the diagram below: (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) The poset λ = J , K with λ -strategy u The elements u , . . . , u are marked with circled numbers. For each i ∈ { , . . . , } , λ ( i ) u is the setof boxes in the same colored region as the box marked i (cid:13) . The cardinalities of these sets are8 , , , , , , Q ( λ, u ) = max { , , , , , , } = 13 . We consider now some special choices of λ -strategies. Lemma 5.5. For any nonempty poset λ , let u = ( u , . . . , u | λ | ) be any arrangement of theelements of λ which is non-increasing with respect to the partial order. Then u is a λ -strategyand Q ( λ, u ) = | λ | .Proof. By the condition on u we have | λ ( t ) u | = 1 for all t , so u is a λ -strategy and Q ( λ, u ) = max {| λ ( t ) u | + t − | t = 1 , . . . , | λ |} = max { t − | t = 1 , . . . , | λ |} = | λ | , as desired. (cid:3) Lemma 5.6. Let λ be a nonempty poset, and assume there exists a λ -strategy u = ( u ) of lengthone. Then we have Q ( λ, u ) = | λ | .Proof. By the definition of λ -strategies u , we must have λ = λ (cid:23) u = λ (cid:23) u = λ (1) u . Thus we have Q ( λ, u ) = | λ (1) u | = | λ | , as desired. (cid:3) For sequences of elements v = ( v , . . . , v s ) and w = ( w , . . . , w r ) in λ , we will write vw forthe concatenation ( v , . . . , v s , w , . . . , w r ), or just v w if v = ( v ). For u ∈ λ with λ (cid:23) u = λ , notethat u w is a λ -strategy if and only if w is a λ u -strategy. Lemma 5.7. Let λ be a nonempty poset. Let v = ( v , . . . , v s ) be a sequence of elements of λ ,and w = ( w , . . . , w r ) be a sequence of elements of λ v . Then, setting u = vw , we have Q ( λ, u ) = max { Q ( λ, v ) , Q ( λ v , w ) + s } . EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 11 Proof. Note that for t = 1 , . . . , s , we have λ ( t ) u = λ ( t ) v , and for t = s + 1 , . . . , s + r , we have λ ( t ) u = λ (cid:23) u t \ λ (cid:23){ u ,...,u t − } = ( λ u ,...,u s } ) (cid:23) u t \ ( λ u ,...,u s } ) (cid:23){ u s +1 ,...,u t − } = ( λ v ,...,v s } ) (cid:23) w t − s \ ( λ (cid:23){ v ,...,v s } ) (cid:23){ w ,...,w t − s − } = ( λ v ) ( t − s ) w . Thus we have Q ( λ, u ) = max {| λ ( t ) u | + t − | t = 1 , . . . , s + r } = max { max {| λ ( t ) u | + t − | t = 1 , . . . , s } , max {| λ ( t ) u | + t − | u = s + 1 , . . . , s + r }} = max { max {| λ ( t ) v | + t − | t = 1 , . . . , r } , max {| ( λ v ) ( t − s ) w | + t − | u = s + 1 , . . . , s + r }} = max { Q ( λ, v ) , max {| ( λ v ) ( t ) w | + t + s − | t = 1 , . . . , r }} = max { Q ( λ, v ) , max {| ( λ v ) ( t ) w | + t − | t = 1 , . . . , r } + s } = max { Q ( λ, v ) , Q ( λ v , w ) + s } , as desired. (cid:3) Connecting Q and q .Theorem 5.8. Let λ be a nonempty poset. We have q ( λ ) = min { Q ( λ, u ) | u a λ -strategy } . (5.9) Proof. We go by induction on | λ | . The base case | λ | = 1 follows immediately from Lemma 5.6.Now assume | λ | > | λ ′ | < | λ | . Note that by Lemmas 5.5 and 5.6, itsuffices to take the minimum on the right of (5.9) over λ -strategies of length greater than one.Thus we havemin { Q ( λ, u ) | u a λ -strategy } = min { Q ( λ, u ) | u a λ -strategy of length greater than one } = min { Q ( λ, u w ) | u ∈ λ, u w a λ -strategy } = min { Q ( λ, u w ) | u ∈ λ, w a λ u -strategy } = min { max { Q ( λ, ( u )) , Q ( λ u , w ) + 1 } | u ∈ λ, w a λ u -strategy } = min { max {| λ (cid:23) u | , Q ( λ u , w ) + 1 } | u ∈ λ, w a λ u -strategy } = min { max {| λ ≻ u | + 1 , Q ( λ u , w ) + 1 } | u ∈ λ, w a λ u -strategy } = min { max { q ( λ ≻ u ) + 1 , Q ( λ u , w ) + 1 } | u ∈ λ, w a λ u -strategy } = min { min { max { q ( λ ≻ u ) + 1 , Q ( λ u , w ) + 1 } | w a λ u -strategy } | u ∈ λ } = min { max { q ( λ ≻ u ) + 1 , min { Q ( λ u , w ) | w a λ u -strategy } + 1 } | u ∈ λ } = min { max { q ( λ ≻ u ) + 1 , q ( λ u ) + 1 } | u ∈ λ } = min { max { q ( λ u ) , q ( λ ≻ u ) } | u ∈ λ } + 1= q ( λ ) . The fourth equality above follows from Lemma 5.7, and the tenth equality follows from theinduction assumption. This completes the induction step, and the proof. (cid:3) Some examples. Combining Theorems 4.2 and 5.8 can be a useful method of computing q ( λ ), as shown in the examples below. ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH Example 5.10. Let λ = J , K , and consider the λ -strategy: u = ((4 , , (2 , , (1 , , (1 , , (4 , , (1 , , (2 , , (1 , . Then, as in Example 5.4, we visually represent u in the diagram below: (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) The poset λ = J , K with λ -strategy u This gives Q ( λ, u ) = max { , , , , , , , } = 8 . Thus by Theorem 5.8 we have q ( λ ) ≤ 8. But by Theorem 4.2 we also have q ( λ ) ≥ τ ( | λ | ) = τ (35) = 8 , so q ( λ ) = 8. Example 5.11. Let λ = J , K . As | λ | = 36, any λ -strategy u = ( u , . . . , u r ) which satisfies Q ( λ, u ) = τ ( | λ | ) = 8 must have r = 8 and | λ ( t ) u | = 9 − t for all t = 1 , . . . , 8. It is straightforwardto check that no such λ -strategy exists, so by Theorems 4.2 and 5.8, we have q ( λ, u ) > 8. Nowconsider the λ -strategy : v = ((5 , , (4 , , (2 , , (3 , , (1 , , (2 , , (1 , , (1 , . We visually represent v in the diagram: (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) The poset λ = J , K with λ -strategy v This gives Q ( λ, v ) = 9, so it follows from Theorem 5.8 that q ( λ ) = 9.5.4. Strategies for Problem 1 in the k = 2 case. We now relate these definitions and resultsback to Problem 1, in the case where only two positive query results are permitted. Recall asin § µ = ∅ or µ = λ (cid:22) x for some x ∈ λ . The λ -strategy u = ( u , . . . , u r ) defines a search strategy for µ as follows.We query the elements u , u , . . . in sequence, until we have a positive query. If all the queriesare negative, then, since λ (cid:23) u = λ , we have that µ = ∅ , and we are done after r ≤ | λ ( r ) u | + r − u t is positive. Then the element x is known to belong to λ (cid:23) u t ,and known to not belong to λ (cid:23){ u ,...,u t − } . Thus x may be any of the elements in λ ( t ) u . Withone positive query remaining, the elements in λ ( t ) u \{ u t } must be sequentially queried in anynon-increasing order, as in § u t query is positive, | λ ( t ) u | + t − µ . EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 13 Therefore, by Definition 5.3, the value Q ( λ, u ) represents the maximum number of queriesnecessary to identify µ via the search strategy defined by u . Thus, in view of Theorem 5.8, wemay reframe the k = 2 case of Problem 1 in this combinatorial language: Problem 1, k = 2 . Find the value q ( λ ), and identify a λ -strategy u such that Q ( λ, u ) = q ( λ ).6. Product posets of finite totally ordered sets If u = ( a, b ) ∈ N , we define the transpose element u T := ( b, a ). We extend this definition tosequences of elements u = ( u , . . . , u r ) in N and subsets S ⊂ N by setting: u T := ( u T , . . . , u Tr ) , S T := { s T | s ∈ S } The transpose map induces an isomorphism of posets J m, n K ∼ = J n, m K , for all m, n ∈ N .In this section it will be convenient to make use of a horizontally compressed visual shorthandfor sequences of elements v = ( v , . . . , v r ) in λ = J m, n K . Using the ‘box array’ representationof J m, n K , we will label the element v i with i (cid:13) as usual, and then label every row in λ ( i ) v withthe number of elements in that row. This visual information is sufficient to describe exactly allelements v i in v , and the related sets λ ( i ) u . Example 6.1. Let λ = J , K . If v = ((3 , , (2 , , (2 , , (1 , , (3 , , (1 , v (on the left) and the compressed shorthandrepresentation of v (on the right). (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ↔ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Theorem 6.2. Let λ = J m, n K , with m ≤ or n ≤ . Then we have: q ( λ ) = ( m = n = 6; τ ( mn ) otherwise . Moreover, Algorithm 6.3 below produces an explicit λ -strategy u such that Q ( λ, u ) = q ( λ ) . Algorithm 6.3. We assume λ = J m, n K , with one of m, n less than or equal to 6. This algo-rithm produces a λ -strategy u such that Q ( λ, u ) = q ( λ ).( Step 0) Let u = () be the empty sequence. Go to ( Step Step 1) If the number of columns of λ is greater than the number of rows, then redefine λ := λ T ,and set flip = 1. Otherwise set flip = 0. Redefine m, n if necessary such that λ = J m, n K . Goto ( Step m + 1).( Step λ = J , n K ). Define t := τ ( n ). Define v to be the one-element sequence in λ depictedbelow. Go to ( Step (cid:13) λ v t ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH ( Step λ = J , n K ). Define t := τ (2 n ). Define v to be the element sequence in λ depictedbelow which corresponds to the appropriate condition on t . Go to ( Step (cid:13) λ v t t (cid:13) (cid:13) (cid:13) λ v tt − ( t − ( t − t ≡ t ≡ Step λ = J , n K ). Define t := τ (3 n ). Define v to be the element sequence in λ depictedbelow which corresponds to the appropriate condition on t . Go to ( Step (cid:13) λ v n (cid:13) λ v ⌊ t ⌋⌊ t ⌋⌊ t ⌋ (cid:13) (cid:13) λ v t t ( t − ( t − (2 t − t = 5 t ≡ , t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t t − t − ( t − (2 t − ( t − ( t − ( t − ≤ t ≡ Step λ = J , n K ). Define t := τ (4 n ). Define v to be the element sequence in λ depictedbelow which corresponds to the appropriate condition on t . Go to ( Step (cid:13) λ v n (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v 114 5 10 65 874 t = 6 , t = 11 (cid:13) λ v ⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v ( t − ( t − ( t − ( t + 2) tt − t − t − t ≡ , ≤ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − t ( t − ( t − t − t − t − (2 t − ( t − ( t − ( t − ( t − ( t − t ≡ EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 151 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (2 t − λ v t ( t − ( t − ( t − ( t − ( t − t − t − t − ( t − ( t − ( t − ( t − ( t − 4) 19 ≤ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − (2 t − t − t − t − ( t − ( t − ( t − (3 t − ≤ t ≡ 11 (mod 12)( Step λ = J , n K ). Define t := τ (5 n ). Define v to be the element sequence in λ depictedbelow which corresponds to the appropriate condition on t . Go to ( Step (cid:13) λ v n (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v 22 9562221111 77 66 (cid:13) λ v ⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋ t = 9 t = 14 t ≡ , , (cid:13) (cid:13) (cid:13) λ v ( t − ( t − ( t − ( t − (3 t − ( t − ( t − ( t − ( t − (cid:13) (cid:13) (cid:13) λ v ( t − ( t − ( t + 2) ( t + 2) (3 t − t t ( t − ( t − t ≡ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − (2 t − (2 t − (2 t − (2 t − (3 t − (3 t − t − t − ( t − ( t − t − t − ≤ t ≡ ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − (2 t − (2 t − (2 t − (2 t − (3 t + 3) (3 t + 13) t − ( t − t − t − ( t − t − ≤ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − (2 t − (2 t − (2 t − (2 t − (3 t + 18) (3 t + 18) t − t − ( t − ( t − t − t − ≤ t ≡ 14 (mod 30) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − ( t − ( t − (2 t − (2 t − (2 t − (2 t − (7 t − t − t − ( t − ( t − ( t − t − t − t ≡ 19 (mod 30) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − t ( t − ( t − ( t − (2 t − (2 t − (2 t − (3 t + 8) (3 t + 8) t − t − ( t − ( t − t − t − t ≡ 24 (mod 30) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t ( t − ( t − ( t − ( t − (2 t − (2 t − (2 t − (2 t − (3 t + 3) (3 t + 3) t − t − ( t − ( t − t − t − t ≡ 29 (mod 30)( Step λ = J , n K ). Define t := τ (6 n ). Define v to be the element sequence in λ depictedbelow which corresponds to the appropriate conditions on n and t . Go to ( Step EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 171 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v n (cid:13) λ v ⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋⌊ t ⌋ (cid:13) (cid:13) λ v ( t − ( t − ( t − ( t − ( t − ( t − n = 6 7 ≤ n ≤ n ∈ { , , , , , } , ≤ t ≡ t ≡ , (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n = 20 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v t t t ( t − ( t − ( t − ( t − (2 t − t − t − t − ( t − ( t − ( t − ( t − ( t + 3) ( t + 9)21 ≤ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v tt − t − t − t − t − ( t − ( t − ( t − ( t + 2) ( t − ( t − ( t − ( t − ( t + 1) ( t + 4)26 ≤ t ≡ ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v tt − t − ( t − ( t − ( t − ( t − (2 t − ( t − ( t − ( t − ( t − ( t − ( t − ( t + 3) ( t − ( t − ( t − ( t − ( t − ( t − ≤ t ≡ , n ≥ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v tt − t − t − t − t − ( t − ( t − ( t − t ( t − ( t − ( t − ( t − ( t − ( t + 1)20 ≤ t ≡ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) λ v tt − t − t − t − t − ( t − ( t − ( t − ( t + 1) ( t − ( t − ( t − ( t − ( t − ( t + 4)23 ≤ t ≡ 11 (mod 12)( Step 8) Set λ ′ = λ v . If flip = 1, set v := v T and λ ′ := ( λ ′ ) T . Redefine u to be the concate-nation uv . Go to ( Step Step 9) If λ ′ = ∅ , END and return u . Otherwise, redefine λ := λ ′ and go to ( Step Proof. First, one must check that the algorithm is well-defined; this entails verifying that thediagrams depicted in ( Steps v (in particular, that therow labels are non-negative integers), and rests on the modular conditions for t below eachdiagram. This is a straightforward exercise, and is left to the reader.To begin, we consider the case λ = J , K . As discussed in Example 5.11, an exhaustive checkshows that Q ( λ, w ) > λ -strategies w , so q ( λ ) > 8. The λ -strategy defined in ( Step < q ( λ ) ≤ Q ( λ, u ) = 9, so we have q ( λ ) = Q ( λ, u ) = 9, as desired.With that special case out of the way, we now prove, for all other diagrams under considera-tion, that Algorithm 6.3 produces a λ -strategy u such that q ( λ ) = Q ( λ, u ) = τ ( | λ | ). We goby induction on | λ | . The base case λ = J , K is clear, as the algorithm produces u = ((1 , Q ( λ, u ) = q ( λ ) = 1.Now let λ = J m, n K , where m ≤ n ≤ 6, and m, n are not both 6. Make the inductionassumption that, if ν satisfies these conditions as well, with | ν | < | λ | , then Algorithm 6.3produces a ν -strategy w such that q ( ν ) = Q ( ν, w ) = τ ( | ν | ). EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 19 Via the transpose operations in ( Steps m ≤ n , so we make that additional assumption now. We insert λ = J m, n K intoAlgorithm 6.3, letting t = τ ( | λ | ) = τ ( mn ), and letting the element sequence v = ( v , . . . , v s )be as it stands at the end of ( Step 8) in the first loop of the algorithm. We begin by arguingthat λ, v satisfy the following three conditions:(C1) Q ( λ, v ) ≤ t .(C2) | λ (cid:23) v | + T ( t ) − | λ | ≥ st − T ( s − λ v = J , K .First we check that (C1) is satisfied by considering every diagram in ( Steps J , K diagram. The homogeneously-colored component of the diagram marked with the element i (cid:13) in the southwest corner is exactly the set λ ( i ) v . By adding up the elements in each row of λ ( i ) v ,it is straightforward to check that in all cases, we have | λ ( i ) v | ≤ t − i + 1. Then we have: Q ( λ, v ) = max {| λ ( i ) v | + i − | i = 1 , . . . , s } ≤ max { ( t − i + 1) + i − | i = 1 , . . . , s } = t. Now we check that λ, v satisfy (C2) by considering every diagram in ( Steps J , K diagram. We do so in the separate Cases 1–7 below.(Case 1) Consider the small cases of the form: • m = 3, n ∈ { , } (and so t = 5) • m = 4, n ∈ { , . . . , } (and so t ∈ { , } ) • m = 5, n ∈ { , } (and so t = 8) • m = 6, n ∈ { , . . . , } (and so t ∈ { , . . . , } )In all these cases, we have s = 1, and | λ (cid:23) v | = n . It is easily checked on a case-by-case basis that T ( t ) − | λ | = T ( t ) − mn ≥ t − n , so we have | λ (cid:23) v | + T ( t ) − | λ | ≥ n + ( t − n ) = t = t − T (0) , satisfying (C2).(Case 2) Consider the small cases of the form: • m = 6, n ∈ { , , , , , } (and so t ∈ { , , } )In all these cases, we have s = 1, and | λ ≥ v | = 6 · ⌊ t ⌋ = 12. It is easily checked on a case-by-casebasis that 12 + T ( t ) − n ≥ t , so we have | λ (cid:23) v | + T ( t ) − | λ | = 12 + T ( t ) − n ≥ t = t − T (0) , satisfying (C2).(Case 3) Consider the case λ = J m, n K , where 3 ≤ m ≤ 6, and t ≡ m ) , as in ( Steps s = 1, and T ( t ) m ) by Lemma 3.11. We also have | λ | ≡ m ) ,so T ( t ) − | λ | ≥ | λ (cid:23) v | + T ( t ) − | λ | ≥ | λ (cid:23) v | + 1 = m (cid:22) tm (cid:23) + 1 = m · t − m + 1 = t = t − T (0) , satisfying (C2).(Case 4) Consider the case λ = J , n K and t ≡ Step s = 1, and T ( t ) ≡ T (2) ≡ | λ | ≡ T ( t ) − | λ | ≥ | λ (cid:23) v | + T ( t ) − | λ | ≥ | λ (cid:23) v | + 3 = 5 (cid:22) t (cid:23) + 3 = 5 · t − 25 + 3 = t + 1 > t = t − T (0) , satisfying (C2). ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH (Case 5) Consider the case λ = J , n K and t ≡ t ≡ t ≡ Step s = 3, and T ( t ) ≡ T (3) ≡ | λ | ≡ T ( t ) − | λ | ≥ | λ (cid:23) v | + T ( t ) − | λ | ≥ | λ (cid:23) v | + 1 = (3 t − 4) + 1 = 3 t − t − T (2) , satisfying (C2).(Case 6) Consider the case λ = J , n K and 16 ≤ t ≡ Step s = 2,and T ( t ) | λ | ≡ T ( t ) − | λ | ≥ | λ (cid:23) v | + T ( t ) − | λ | ≥ | λ (cid:23) v | + 1 = (2 t − 2) + 1 = 2 t − t − T (1) , satisfying (C2).(Case 7) Now we may consider the remaining cases in one fell swoop. In all remaining cases,it may be checked that v is defined such that | λ ( i ) v | = t − i + 1 for t = 1 , . . . , s , and thus | λ (cid:23) v | = st − T ( s − | λ (cid:23) v | + T ( t ) − | λ | ≥ | λ (cid:23) v | = st − T ( s − , satisfying (C2).Now we check that λ, v satisfy (C3). The case λ = J m, n K for m < λ = J , n K . As with the last claim, we check (C3) in the separate Cases 1–10 below.(Case 1) If 7 ≤ n ≤ 11, then λ v is a 5-row diagram, so is not equal to J , K .(Case 2) If 12 ≤ n ≤ 15, then t ∈ { , } , so t ≡ , ⌊ t ⌋ = 2, and we have | λ v | = | λ | − · (cid:4) t (cid:5) ≥ · − 12 = 60 , so λ v = J , K .(Case 3) If n ∈ { , , , , , } , then t ∈ { , , } , so ⌊ t ⌋ = 2, and we have | λ v | = | λ | − · (cid:4) t (cid:5) ≥ · − 12 = 84 , so λ v = J , K .(Case 4) If n = 20, then λ v = ∅ = J , K .(Case 5) If n ∈ { , } , then t = 16. Then | λ v | = | λ | − (2 t − ≥ (21 · − 30 = 96, so λ v = J , K .(Case 6) If n = 25, then t = 17. Then | λ v | = | λ | − (8 t − 28) = (25 · − (8 · − 28) = 42,so λ v = J , K .In the remaining cases, we assume that n ≥ 26. Then we have t ≥ 18. Note that by thedefinition of t = τ ( | λ | ), we have | λ | > T ( t − t ≡ , | λ v | = | λ | − (cid:22) t (cid:23) > T ( t − − t = T ( t − − ( t − − T ( t − − ≥ T (16) − , so λ v = J , K .(Case 8) Say t ≡ | λ v | = | λ | − (2 t − > T ( t − − (2 t − T ( t − − ( t − − ( t − − T ( t − − ≥ T (15) − , so λ v = J , K .(Case 9) Say t ≡ | λ v | = | λ | − (7 t − > T ( t − − ( t − − · · · − ( t − − T ( t − − ≥ T (10) − , so λ v = J , K . EARCHING FOR QUICKSAND IDEALS IN PARTIALLY ORDERED SETS 21 (Case 10) Say t ≡ , | λ v | = | λ | − (8 t − > T ( t − − ( t − − · · · − ( t − − T ( t − − ≥ T (9) − , so λ v = J , K .Thus, in every case we have λ v = J , K , and so (C3) holds.Therefore (C1), (C2), (C3) hold for λ, v . By (C2), we have | λ v | = | λ | − | λ (cid:23) v | ≤ T ( τ ( | λ | )) − sτ ( | λ | ) + T ( s − , so by Lemmas 3.5 and 3.10 we have τ ( | λ v | ) ≤ τ ( T ( τ ( | λ | )) − sτ ( | λ | ) + T ( s − ≤ τ ( | λ | ) − s. (6.4)By (C3), the induction assumption holds for λ v , so inserting λ v into the algorithm yields a λ v -strategy w such that Q ( λ v , w ) = q ( λ v ) = τ ( | λ v | ). By the inductive nature of thealgorithm, inserting λ into the algorithm yields the λ -strategy u = vw . Then we have Q ( λ, u ) = max { Q ( λ, v ) , Q ( λ v , w ) + s } by Lemma 5.7 ≤ max { τ ( | λ | ) , Q ( λ v , w ) + s } by (C1)= max { τ ( | λ | ) , τ ( | λ v | ) + s } by induction assumption ≤ max { τ ( | λ | ) , ( τ ( | λ | ) − s ) + s } by (6.4)= τ ( | λ | ) . Then, by Theorems 4.2 and 5.8, we have τ ( | λ | ) ≤ q ( λ ) ≤ Q ( λ, u ) ≤ τ ( | λ | ) , so we have q ( λ ) = Q ( λ, u ) = τ ( | λ | ) as desired, completing the proof. (cid:3) As κ × ν ∼ = J m, n K when κ , ν are totally ordered sets of cardinality m, n respectively, we havethe immediate corollary thanks to Lemma 4.1: Corollary 6.5. Let κ, ν be finite totally ordered sets, with | κ | ≤ or | ν | ≤ . Then q ( κ × ν ) = ( | κ | = | ν | = 6; τ ( | κ || ν | ) otherwise . A conjecture. We end with a conjectural bound for product posets of totally ordered sets. Conjecture 6.6. Let m, n ∈ N . Then q ( J m, n K ) ≤ τ ( mn ) + 1 . This suggests q ( J m, n K ) ∈ { τ ( mn ) , τ ( mn ) + 1 } for all m, n ∈ N . By Theorem 6.2, theposets J m, n K obey this claim when m ≤ n ≤ 6. In fact, all but J , K have the minimalpossible value q ( J m, n K ) = τ ( mn ) allowed by Theorem 4.2. Moving beyond these results,computations show that exceptional cases like J , K , where no λ -strategy u can be found thatrealizes Q ( λ, u ) = τ ( | λ | ), seem to occur fairly rarely (the poset J , K is another). Butallowing for a λ -strategy that realizes Q ( λ, u ) = τ ( | λ | ) + 1 instead seems to afford so muchflexibility that we expect such a λ -strategy can always be found, even in these exceptional cases.For instance, while there are no J , K -strategies that realize Q ( J , K , u ) = 8, there are 53,688distinct J , K -strategies which realize Q ( J , K , u ) = 9. This is the authors’ line of reasoningbehind positing Conjecture 6.6. ALEXAS IAMS, HANNAH JOHNSTON, AND ROBERT MUTH References [1] B. A. Davey, H. A. Priestley, Introduction to Lattices and Order , Cambridge University Press, New York, 2002.[2] R. Denman, D. Hailey, and M. Rothenberg, The Tower and Glass Marbles Problem, College Math. J. , Amer. J. Math. , (1941) 600–610.[4] J. D. E. Konhauser, D. Velleman, and S. Wagon, (1996). Which way did the Bicycle Go? Dolciani MathematicalExpositions, No. 18, The Mathematical Association of America, 1996.[5] G. L. McDowell, Cracking the Coding Interview, 5th ed. , CareerCup, 2011.[6] S.S. Skiena, The Algorithm Design Manual , Springer-Verlag, New York, 1997. Washington & Jefferson College, Washington, PA 15301, USA E-mail address : [email protected] Washington & Jefferson College, Washington, PA 15301, USA E-mail address : [email protected] Department of Mathematics, Washington & Jefferson College, Washington, PA 15301, USA E-mail address ::