# Kinnim Mishnah (3:2)

The recent issue of Hakirah published my article on the above Mishnah. It will be available online in a a few months after the next issue of Hakirah is published.

Here is my latest version, updated and simplified, at least slightly.

__Kinnim ____(3:2) and the term merubah__

**:**

*Kinnim* (3:2) reads as follows:^{[1]}

*אַחַת לָזוֹ, וּשְׁתַּיִם לָזוֹ, וְשָׁלשׁ לָזוֹ, וְעֶשֶׂר לָזוֹ, וּמֵאָה לָזוֹ, עָשָׂה כֻלָּן לְמַעְלָה, מֶחֱצָה כָשֵׁר וּמֶחֱצָה פָסוּל. כֻּלָּן לְמַטָּן, מֶחֱצָה כָשֵׁר וּמֶחֱצָה פָסוּל. חֶצְיָן לְמַעְלָן וְחֶצְיָן לְמַטָּן, הַמְרֻבֶּה כָשֵׁר. זֶה הַכְּלָל, כָּל מָקוֹם שֶׁאַתָּה יָכוֹל לַחֲלֹק אֶת הַקִּנִּין וְלֹא יְהוּ מִשֶּׁל אִשָּׁה אַחַת, בֵּין מִלְמַעְלָן בֵּין מִלְּמַטָּן, מֶחֱצָה כָשֵׁר וּמֶחֱצָה פָסוּל. כָּל מָקוֹם שֶׁאֵין אַתָּה יָכוֹל לַחֲלֹק אֶת הַקִּנִּין עַד שֶׁיְּהוּ מִשֶּׁל אִשָּׁה אַחַת, בֵּין מִלְמַעְלָן בֵּין מִלְּמַטָּן, הַמְרֻבֶּה כָשֵׁר*

*If one [pair] belonged to one woman and two [pairs] to another, three [pairs] to another, ten pairs to another and a hundred to another, and he offered all of them above, then half are valid and half are invalid. [Similarly] if he offered all of them below, half are valid, and half are invalid. [If he offered] half of them above and* *half below, then the [number of birds as there is in the] larger part are valid. This is the general principle: whenever you can divide the pairs [of birds] so that those belonging to one woman need not have part of them [offered] above and part [offered] below, then half of them are valid and half are invalid; but whenever you cannot divide the pairs [of birds] without some of those belonging to one woman being [offered] above and some below, then [the number as there is in] the larger part are valid.*

When birds aresacrificed without consultation, we determine the worst possible scenario and then determine how many birds are nonetheless valid even under those conditions. When equal sized *kinnim* *stumot **are* intermingled, the worst possible case invalidates half of each nest, because the worst possible outcome results when each *ken* is sacrificed entirely as either* olot* or *ḥatta’ot*. When unequal sized nests are intermingled, in all cases, however, more than half of the birds are sacrificed correctly. Mishnah (3:2), conceptually perhaps the hardest Mishnah in *Kinnim*, deals with a case where (significantly) **more than half** of the intermingled *kinnim stumot* are valid; in the case in the Mishnah, 200 of 232 birds are valid. The difference between what happens with and without consultation can be illustrated in a remarkably simple case when two unspecified nests of unequal size are intermingled. As demonstrated in the introduction, with consultation, only the number of birds in the smaller nest are permitted to be sacrificed. When sacrificed without consultation, half as *hatta’ot *and half as *olot*, the number of birds in the larger nest are valid.[2]

The second half of the Mishnah following ** זֶה הַכְּלָל** distinguishes two situations. In the first situation, it is possible to arrange all the birds in all the

*kinnim*as either

*olot*or

*ḥatta’ot*without any

*ken*having birds

*sacrificed*as both

*olot*and

*ḥatta’ot*. The ability to arrange the nests exactly that way represents the worst possible outcome

*.*[3] This example is representative of how disqualification is maximized.

*Kinnim a*re meant to be sacrificed half as

*ḥatta’ot*and half as

*olot*. By sacrificing all as one type, disqualification is maximized.

When you cannot divide the nests without one nest[4] having to sacrifice some of its birds as* olot* and some as *ḥatta’ot*, more than half of the birds are valid. The Mishnah uses the phrase ** הַמְרֻבֶּה כָשֵׁר**, whose precise meaning will be defined going forward.[5]

An article by Dr. Phillip Reiss provides an elegant formal proof of the second part of Mishnah (3:2), which specifies that the number of valid sacrifices is the larger amount, ** הַמְרֻבֶּה כָשֵׁר**. Dr. Reiss defines this expression precisely, corresponding exactly to the

*halakhic*rule with which the Mishnah operates. That rule, as articulated by many classic commentators, is thatthe way to determine the minimum number of birds correctly sacrificed is to construct a scenario that

**maximizes**the number of birds incorrectly sacrificed. A formula that represents this concept is: (The minimum number of birds correctly sacrificed) = (the total number of birds.) – (the maximum number of birds potentially incorrectly sacrificed).

When we cannot divide the *kinnim* into two equal sized groups, more than half of the birds will be valid. That larger group, referred to by the phrase ** הַמְרֻבֶּה כָשֵׁר** however, must be the smallest possible larger group. Dr. Reiss formalized what exactly is meant by “the smallest majority” and I will reframe Dr. Reiss’s approach to make the Mishnah more intuitive and the proof more concise. It serves as a model for thinking about all the

*mishnayot*of the third chapter, which are examples of this general case as well as also addressing a situation where

*kinnim mefurashot*

**and**

*kinnim*

*stumot*are intermingled. What follows provides a somewhat less rigorous argument that may capture how the Mishnah may have been conceived by its authors and more traditional commentators.

To formalize the Mishnah and derive the minimum number of birds sacrificed correctly requires construction of a scenario that **provably** maximizes the number of birds incorrectly sacrificed.

Without consultation, the *Kohen* sacrifices half of the combined nest as *ḥatta’ot* and half as *olot*. To develop our model and without loss of generality, assume the *Kohen* places each of the birds into one of two equal sized storage containers labeled “O” for *olot* and “Ḥ” for *ḥatta’ot*. Together the two containers precisely hold the total number of birds originally in all the intermingled nests. The birds in container O are subsequently sacrificed as *olot *while the birds in container Ḥ are sacrificed as *ḥatta’ot*. Since the two containers are of equal size, the *Kohen* assumes, incorrectly of course,[6] that he has fulfilled his obligation to sacrifice half the nest as *ḥatta’ot* and half as *olot*.

To fully explain and prove the Mishnah in its most general case requires that we look at the situation prior to nests being combined and examine the original set of nests. There could have been any number of individual nests, but obviously the number of birds in each of the original nests, (all being *ḥ**ovot* and therefore containing an equal number of *ḥatta’ot* and *olot*,) must be an even number.^{[7]}

Prior to combining the nests, think of each of the birds in each of the individual *kinnim* being stored individually in identical cages; the cages of each *ken* are then stacked vertically one on top of another and not yet mixed with any other *ken*. That vertical stack of cages from a single *ken* forms a single package. The height of each package is proportional to the (even) number of bird cages in the package; a package holding a *ken* of 4 birds is ½ of the height of package holding a *ken* of 8 birds.

The individual cages are identical, all nests/packages[8] have the same length and width. The containers are constructed to allow the cages to only be stacked vertically in the O and Ḥ containers. Those containers can hold bird cages stacked vertically, but those cages must not exceed the height of the container, as will become clear later.

Return now to the two storage containers of equal size, (the *ḥatta’ot *and *olot* containers respectively,) into which each of the individual cages can be stacked. To match the case in the Mishnah, the two identically sized containers are assumed to have the exact capacity required to store all the bird cages in the N packages (i.e., the *kinnim)*, where N denotes the original number of (now intermingled) *kinnim*.

Were the birds not intermingled, we can assume that each woman’s *ken* is divided in half, with each half temporarily stored in one of the respective containers, before being correctly sacrificed. In the desired scenario, each *ken* is split equally across the two containers; the total number of bird cages fit precisely into the two equally sized containers.

To construct a worst-case scenario, we attempt to do exactly the opposite. Instead of dividing each package into an equal number of *ḥatta’ot *and o*lot,* we try to leave all the packages unsplit. To the greatest extent possible, something that will be defined formally below, all the birds in each *ken* are placed in **only** one container, i.e., all to be sacrificed as **either** *ḥatta’ot ***or** o*lot*. In terms of the two containers, we will try to fill both containers without **splitting **any individual *ken*, effectively sacrificing all birds from each unsplit package identically, thereby disqualifying exactly half of the birds in each *ken*. When attempting to do that, we may or may not be completely successful in dividing them evenly without splitting any given *ken*. Consider a case of 2, 4 and 6 birds; the worse we can do is put the two smaller *kinnim* with 2 and 4 birds in one container and the larger *ken* of 6 birds in the other, a split of *kinnim* without dividing any single *ken*. This case illustrates achieving maximal disqualification, where half (6) of the 12 birds are disqualified, but still allows for half (6) of the sacrificed birds in each nest to be valid. However, if we had 3 *kinnim* containing 2, 4, and 8 birds in each of the original *kinnim*, there is no way to divide the (2+4+8=) 14 birds into 2 groups of 7, without splitting a *ken*. Note that each *ken* has an even number of birds and the number of birds in any number of unsplit nests will have an even number of birds as well. In the worst-case scenario, we can put the two smaller *kinnim* totaling 6 birdsin one container and split the larger *ken* of 8 birds, putting 7 birds (half the total number of birds) in the other container and 1 bird in the container together with the two smaller unsplit nests, each container thereby holding 7 bird cages. As a result, 2 of the 8 birds in the larger nest are sacrificed correctly, one as an *olah* and one as a *ḥattat***,** while half of the remaining 12 birds in the two containers are also sacrificed correctly, for a total of 8, correctly sacrificed birds more than half of the 14 intermingled birds.

If we are trying to place unsplit packages into each container, it should be clear that it is always the case, that at most one package needs to be split across the two containers. That only one packet ever needs to be split is fundamental to understanding the proof of the Mishnah and is proven formally in the footnote below.^{[9]}

There are 2 ways to formalize the Mishnah using the paradigm of storing the maximum number of unsplit packages using either **one or both** containers. The first method, a 2-container storage solution, maximizes the total number of birds from unsplit packages that are stored entirely in either **one** of the two containers. However, optimizing the number of unsplit *kinnim* across both storage containers does not maximize the number of incorrectly sacrificed birds. Rather,^{[10]} the worst-case arises if we try to maximize the number of birds from unsplit packages that can be stored in only a single container (as opposed to both.) Therefore, to maximize the number of invalidly sacrificed birds out of the total group of *kinnim*, we place the largest number of unsplit packages in only one of the two containers. The other container will hold the remaining unsplit packages whose height is, by definition, less than or equal than the height of unsplit packages in the other container. If both containers are not full, the one remaining package is split, and it will fill both containers. Of course, even when unlike the case in the Mishnah where the size of one *ken* is not greater than the sum of all the rest, the package that must be split may not be the smallest package.^{[11]}

What we will now proceed to prove is that maximizing the number of birds from unsplit packages in only one container creates maximal disqualification. A case that illustrates why optimizing one container and not two container storage creates maximal disqualification, and a more intuitive discussion of this issue follows the proof.

If 2*K is the total number of birds in all the *kinnim*, then each container will have capacity for K birds.^{[12]} Let J be the largest number of birds that can be placed in one container without splitting a package. To simplify the presentation, we will assume that the container with J birds (cages) is the O container, as opposed to the Ḥ container.[13] Of course, the Ḥ container must therefore contain either as many as or fewer than J birds from unsplit containers; it cannot contain more than J birds. Note, if a container can be totally filled without splitting up any package, then J = K.[14] Whether or not a container can be filled with unsplit packages**, the minimal number of correctly sacrificed birds will be proven to equal J + 2*(K-J).**^{[15]} ^{[16]}** Once we have determined J, **the formula provides the (minimal) number of correctly sacrificed birds. The entire proof hinges on proving that determining J allows the identification of the minimal number of valid sacrifices, i.e., the “worst-case scenario.”[17]

Before providing an outline of the proof, some examples will help to facilitate understanding of the proof.

- For 2, 4, 6, 8, and 10 birds (1, 2, 3, 4, and 5
*kinnim*): In this example, K = 15, J = 14, and a minimum of 16 sacrifices are valid, since J+2*(K-J) = 14 + 2*(15-14) = 14 + 2 = 16. The O container holds either the*kinnim*with 10 and 4 birds, 6 and 8 birds, or 2, 4 and 8 birds, each group totaling to 14 birds. - For 2, 4, 6, and 8 birds J = K =10: One container holds the groups of 4 and 6 birds and the other holds the groups of 2 and 8 birds. Since every group of
*kinnim*can be stored unsplit, the number of correctly sacrificed bids is exactly half (this is the worst result that occurs only when both containers can be filled with unsplit nests.) - For the case in the Mishnah of 2, 4, 6, 20 and 200 birds, K = 116 and J = 32 and at least 200 birds were correctly sacrificed. In this example, the O container initially holds (2+4+6+20=) 32 birds in unsplit packages and the Ḥ container holds no birds
**in unsplit packages.**In this case where one of the*kinnim*contains more than half of the total number of birds, this large*ken*must contain K+ (K-J) birds. K is the complete capacity of the Ḥ container, (116 birds in the Mishnah’s case,). K must be added to the remaining capacity of the O container, i.e., (K-J), (84 birds in the Mishnah’s case,). (K-J) represents O’s original capacity of 116 birds minus the space already occupied by the 32 birds from unsplit packages, already present in the O container. This the total size of this large nest (K + (K-J)) = J + 2*(K-J). - In Dr. Reiss’s example of 8, 12, and 14 birds, K= 17 and J = 14, and even in the worst-case scenario, at least (14 + 2*(17-14) =) 20 birds were correctly sacrificed. In a 2-container storage optimization, unsplit nests with a total of 14 birds go in one container, unsplit nests with a total of 12 birds go into the other container, while the remaining nest of 8 birds is split across the two containers – 3 in the container with 14 birds and 5 in the container with 12 birds. In the
*kinnim*case of 1-container optimization however, once 14 birds are placed in the O container, how the remaining three slots in that container are filled (i.e., from the group of 8 or 12 or some combination) is irrelevant in determining the number of birds correctly sacrificed.[18]

An outline of a formal proof follows.

Remember that J is the largest number of birds that can be placed in the O container without splitting any package.^{[19]} Since those birds are all sacrificed identically as *olot*, exactly J/2 were correctly sacrificed, as the other J/2 birds in the O container should have been sacrificed as *ḥatta’ot*. No remaining *ken* has fewer than (K-J) birds. Otherwise, it could have been added to the O container that contains J birds from unsplit packages, thereby increasing J by the size of that *ken*. **After the O container is maximally filled with unsplit nests and the remainder of the unsplit nests are placed into the ****Ḥ ****container, then (K-J) is less than or equal to half the number of birds contained in the single remaining nest that now must be split. [20] This is the key point in the proof. The **

**Ḥ**

**container may also hold unsplit packages, but with less than or equal to J birds. Hence, after filling the O container with the split nest, there must be room for at least (K-J) additional birds from the remaining split package in the H container.**Thus, for each of the remaining (K-J) birds which are the birds from the split

*ken*placed in the O container, both it and some other member of its

*ken*which were placed in the

**Ḥ**container, were correctly sacrificed, one as an

*olah*and one as a

*ḥattat.*This adds 2*(K-J) correctly sacrificed birds to the J/2 correctly sacrificed birds from the unsplit nests in the O container. Beyond the K-J birds in the H container that have a mate in the O container and are therefore correctly sacrificed,

**of the remaining J birds in the Ḥ container, have a mate in the O container. Those J birds in the Ḥ container cannot be paired with any of the J birds from unsplit nests in the O container. This leaves J/2 additional valid sacrifices from, the Ḥ container, the J/2**

__none__*ḥatta’ot*that are validly sacrificed, (versus the J/2 that should have sacrificed as

*olot*and are invalid.) The J/2 correctly sacrificed birds in container H are added to the prior 2 correctly sacrificed groups of J/2 birds from container O and the 2*(K-J) correctly sacrificed birds from a single nest half in each of the O and Ḥ containers. The total of correctly sacrificed birds is now J/2 + 2*(K-J) + J/2, which equals J + 2*(K-J) or 2*K – J birds, Dr. Reiss equivalent expression for minimally valid sacrifices.[21]

It is critical to appreciate the difference between the improper two container optimization versus the one container optimization, in cases with no prior consultation. Only one container optimization correctly identifies the smallest number of validly sacrificed birds, by disqualifying the maximum number of birds. In examples a) through d) above, the two container and one container optimizations happen to yield equivalent solutions, but this is not always the case. Consider the case of *kinnim *consisting of 8, 10, 10, 10, 14, and 14 birds, respectively, totaling 66 birds. Note that the O and the Ḥ container are each of size K = 33 and one package (i.e., *ken*) must be split in the 2-container optimization. It is easy to see that the package of size 8 can be split, resulting in one container holding the three packages of size 10 and the other container holding the other two packages, both of size 14. In total, 58 birds in unsplit packages are stored across both containers. If we set J = 30 corresponding to the largest number of birds from unsplit nests in either container, the formula J + 2(K-J) would incorrectly yield (30 + 2(33-30) = (30 +6 =) 36 valid sacrifices. However, optimizing the number of birds from unsplit packages that can be placed in only one container allows you to put 32 = (8 + 10 + 14) items into one container, and hence J = 32. Note that this would **not** yield an optimization of unsplit packages across both containers since only 24 items would go into the second container for a total of 56 unsplit birds achieved in the one container optimization, versus the 58 birds that can be stored in unsplit packages with a 2-container optimization. In the 1 container optimization, a nest of 10 birds must be split, 9 birds in the container with 24 birds and only 1 bird in the container with 32 birds.^{[22]}

Since constructing the above example in 2001, I have tried to find the smallest example that illustrates the difference between the one and two container optimizations. Five *kinnim* of sizes 6, 6, 6, 10 and 10 birds, with 38 birds, is apparently the smallest example. When maximally filling two containers, each with a capacity to store 19 birds, we would put *kinnim *with 10 and 6 birds into both containers, leaving one of the three smaller *kinnim *with 6 birds to be split across both containers. Ostensibly, (J+2(K-J) = (16 + 2(19-16) =) 22 birds are correctly sacrificed. But that result does not comport to universally accepted interpretation of Mishnah that requires that the worst possible scenario be constructed to determine the smallest number of birds validly sacrificed. Using a one container optimization, all 3 *kinnim* that contain 6 birds are placed in the O container unsplit, leaving one unsplit nest of 10 birds in the Ḥ container. One of the 10 bird *kinnim* are then split with 1 bird placed in the O container and 9 in the Ḥ container. This results in (J+2(K-J) = (18 + 2(19-18) =) 20 birds, correctly sacrificed, which are valid even in the worst-case scenario, comporting with the universally accepted interpretation of the *Mishnah*.^{[23]}

**Conclusion:**

The only approach that comports with the Mishnah’s conclusion about intermingled nests when a *Kohen* sacrifices without prior consultation as if was dealing with a single nest was demonstrated to be the one container optimization. This approach renders more intuitive why the maximization of the number of invalid sacrifices (hence determining the smallest amount valid sacrifices) requires that one maximize the total size of those *kinnim *that can be placed unsplit in a container that can hold exactly half of the total number of birds in all the intermingled *kinnim*. One can imagine that mathematical reasoning like that contained in the proof can have been used in the formulation of the Mishnah by *tannaim *2000 years ago.

[1] The text of the Mishnah and its translation is largely taken from www.Sefaria.com.

[2] This is not the ruling with a *ken mefureshet*, in which case, even when sacrificed without consultation **no birds are valid, **because in the worst-case, it is possible that every bird designated as an *olah* was sacrificed as a *ḥattat *and every bird designated as a *ḥattat* was sacrificed as an *olah.*

[3] For example, if 4* kinnim *have 2, 4, 8 and 10 birds respectively. Combined there are a total of 24 birds, of which 12 should be sacrificed as* olot *and 12 as* ḥatta’ot. *Clearly one can sacrifice the 2 *kinnim *with 2 and 10 birds entirely as *olot *and the 2* kinnim *with 4 and 8 birds entirely as* hatta’ot.*

[4] It is shown later in this chapter, that in every case only one nest must be divided.

[5] For example, consider *kinnim *with 2, 4, 6 and 10 birds respectively. If the 22 birds had remained in their individual nests, the birds would be sacrificed as 11 *olot* and 11 *ḥatta’ot* in total. When combined they are sacrificed as 11 *olot* and 11 *ḥatta’ot*, without the ability to know how any nest was sacrificed. We could have sacrificed the nest of 10 birds all as *olot* and the nests of 4 and 6 birds all as *hatta’ot*. The nest of 2 birds would then have had to be sacrificed correctly, 1 bird as an *olah* and 1 bird as a *ḥattat*. Adding the 2 birds in the smallest nest to only half of the birds in the other 3 nests that are validly sacrificed, results in 12 valid sacrifices. Note that in this specific case, splitting any of the 4 nests would give the same result. Similarly, the case in the Mishnah with 2, 4, 6, 20 and 200 birds, the largest nest of 200 birds must be divided. The 232 birds must be sacrificed as 116 *olot* and 116 *ḥatta’ot*. The worst-case sacrifices the nests of 2, 4, 6, and 20 birds all as *hatta’ot *or *olot.* The nest of 200 birds is then split, with 116 of the birds sacrificed opposite to how the smaller nests are sacrificed; and 84 the birds sacrificed the same way as the smaller nests. This results in 200 valid sacrifices.

[6] Presented with one nest, the *Kohen* was assumed unaware of the nest being the result of combining several individual nests. However, when he sacrifices half and half in the case of mixed nests, one full nest can easily be sacrificed entirely as either *hatta’ot* or *olot* thereby half of such a nest is incorrectly sacrificed.

[7] Any even number can be expressed as 2*X for some value of X.

[8] The terms *kinnim*, packages and nests are used interchangeably.

[9] Assume that while constructing a worst-case scenario a *ken* must be split because it is not possible to fit that ken / package into either container. We now demonstrate that if a package must be split, then one of the containers must be able to be filled completely by that split *ken*. Were that not the case, i.e., the container was not filled with birds from that split *ken,* move as many birds as necessary from that split package to completely fill that container. Since that package had to be split, given that the complete package would not fully fit into either container, there must always be enough birds to fill that container from the split *ken*. This situation may occur when the containers are still empty of other *kinnim* as in the case of the Mishnah where the *ken* of 200 birds must always be split or may occur when some or all packages are already stored in the containers, depending on the size and grouping of the *kinnim*. Were that not the case and there were not enough birds in that package to fill that container, that entire package would fit into that container. Thus, at that point, after splitting one package one container must be completely full and unable to hold any more birds. The other container must exactly hold the remainder of the birds from the one split package and any (unsplit) packages not yet stored in a container. Those unsplit packages cannot be split since the other container is already full and no more room for bird cages exists within it. Thus, none of the remaining packages can be split proving that only one package can ever need to be split.

[10] An example demonstrating this is given below. It is not easy to construct such examples.

[11] In both the case in the Mishnah and in the example above of 3 *Kinnin* with 2, 4 and 8 birds, one *ken* was larger than the combination of all the other *kinnin*. Consider, however, a case of 6 packages of sizes 6, 20, 28, 28, 28 and 30 where to maximize disqualification a package of size 20 must be split. Note that the total of 140 birds, require 70 birds in both the O and Ḥ containers. The maximum number of unsplit packages contain (6 + 28 +30 =) 64 birds. The second container will hold the two remaining complete nests (28+28=) 56 birds. This leaves the 20-bird nest to be split, 6 birds going to the nest with 64 birds, and 14 birds going to the nest with 56 birds. The reader can try unsuccessfully to fit more than 64 birds in a container, demonstrating the correctness of the result.

[12] K is the total number of pairs of one *ḥattat* and *olah* in the intermingled nest.

[13] This has no impact on the correctness of what follows; we can easily reverse O and Ḥ.

[14] Of course, if one container can be filled by unsplit packages, the other container can be filled by unsplit packages as well.

[15]Note that (J + 2*(K-J)) equals (J+ 2*K – 2*J), which equals (2*K – J), the expression for the smallest number of validly sacrificed birds used in Dr. Reiss’s paper.

[16] The remainder of the O container, that was not filled with unsplit packages, has the capacity to still contain (K-J) birds, that will come from the split nest. That split nest, which fills the remaining K-J places, must have a mate from its nest in the other container as well. This contributes 2*(K-J) valid sacrifices, half from the O container and half from the Ḥ container.

[17] The worst-case equals the total number of birds minus the maximal number of birds that could have been sacrificed incorrectly.

[18]Once 14 birds are placed in the O container, it does not matter if the 12 or the 8-bird nest is put in the Ḥ container unsplit, with the other nest split to fill both the O and Ḥ containers. Even if both nests with 12 and 8 birds are both split, and any combination of birds from these nests are used to complete the O and H containers the resulting minimum number of valid birds does not change.

[19] Note that J can always be computed, if necessary, by an exhaustive examination of all options.

[20] Remember the H container cannot hold more than J birds from unsplit packages, leaving K-J or more spaces to fill the rest of the Ḥ container.

[21] The Ḥ container also contains J birds beyond the (K-J) birds with a mate in the J container. K-J birds are matched up in the two containers, one group as *olot* and one group as *ḥatta’ot*, the remaining birds in the H container is a mixture of unsplit nests and excess birds from the split nest. Thus, there are J/2 valid sacrifices coming from the Ḥ container as well.

[22] The (8+10+14 =) 32 birds in the O container, contribute 16 valid sacrifices. 2 birds from the split nest one in from the O container and one from the Ḥ container are correctly sacrificed and contribute another 2 valid sacrifices. The remaining 32 birds in the Ḥ container have no mate in the O container, so only half are validly sacrificed as *hatta’o*t and contribute another 16 valid sacrifices, resulting in a total of 34 valid sacrifices vs. the 36 valid sacrifices achieved with the two-container optimization. This demonstrates the under certain combinations of intermingled *kinnim*, the two-container optimization will not result in the maximum invalid sacrifices, whereas the single container optimization does, resulting in the worst-case scenario.

[23] My nephew, Joshua Blumenkopf, proved that there is no such example with fewer than 5 *kinnim*. While I assume there is no example with fewer than 38 birds in total across 5 *kinnim*, I do not have a simple proof. One can exhaustively examine 10 through 36 birds to confirm this conclusion.

## Leave a Comment