Understanding "median of medians" algorithm

The problem is in the step where you say to find the true median of the medians. In your example, you had these medians:

50 45 40 35 30 25 20 15 10

The true median of this data set is 30, not 15. You don't find this median by splitting the groups into blocks of five and taking the median of those medians, but instead by recursively calling the selection algorithm on this smaller group. The error in your logic is assuming that median of this group is found by splitting the above sequence into two blocks

50 45 40 35 30

and

25 20 15 10

then finding the median of each block. Instead, the median-of-medians algorithm will recursively call itself on the complete data set 50 45 40 35 30 25 20 15 10. Internally, this will split the group into blocks of five and sort them, etc., but it does so to determine the partition point for the partitioning step, and it's in this partitioning step that the recursive call will find the true median of the medians, which in this case will be 30. If you use 30 as the median as the partitioning step in the original algorithm, you do indeed get a very good split as required.

Hope this helps!

Here is the pseudocode for median of medians algorithm (slightly modified to suit your example). The pseudocode in wikipedia fails to portray the inner workings of the selectIdx function call.

I've added comments to the code for explanation.

// L is the array on which median of medians needs to be found.
// k is the expected median position. E.g. first select call might look like:
// select (array, N/2), where 'array' is an array of numbers of length N

select(L,k)
{

    if (L has 5 or fewer elements) {
        sort L
        return the element in the kth position
    }

    partition L into subsets S[i] of five elements each
        (there will be n/5 subsets total).

    for (i = 1 to n/5) do
        x[i] = select(S[i],3)

    M = select({x[i]}, n/10)

    // The code to follow ensures that even if M turns out to be the
    // smallest/largest value in the array, we'll get the kth smallest
    // element in the array

    // Partition array into three groups based on their value as
    // compared to median M

    partition L into L1<M, L2=M, L3>M

    // Compare the expected median position k with length of first array L1
    // Run recursive select over the array L1 if k is less than length
    // of array L1
    if (k <= length(L1))
        return select(L1,k)

    // Check if k falls in L3 array. Recurse accordingly
    else if (k > length(L1)+length(L2))
        return select(L3,k-length(L1)-length(L2))

    // Simply return M since k falls in L2
    else return M

}

Taking your example:

The median of medians function will be called over the entire array of 45 elements like (with k = 45/2 = 22):

median = select({48 49 50 51 52 43 44 45 46 47 38 39 40 41 42 33 34 35 36 37 28 29 30 31 32 23 24 25 26 27 18 19 20 21 22 13 14 15 16 17 8 9 10 53 54}, 45/2)

The first time M = select({x[i]}, n/10) is called, array {x[i]} will contain the following numbers: 50 45 40 35 30 20 15 10. In this call, n = 45, and hence the select function call will be M = select({50 45 40 35 30 20 15 10}, 4)
The second time M = select({x[i]}, n/10) is called, array {x[i]} will contain the following numbers: 40 20. In this call, n = 9 and hence the call will be M = select({40 20}, 0). This select call will return and assign the value M = 20.

Now, coming to the point where you had a doubt, we now partition the array L around M = 20 with k = 4.

Remember array L here is: 50 45 40 35 30 20 15 10.

The array will be partitioned into L1, L2 and L3 according to the rules L1 < M, L2 = M and L3 > M. Hence:
L1: 10 15
L2: 20
L3: 30 35 40 45 50

Since k = 4, it's greater than length(L1) + length(L2) = 3. Hence, the search will be continued with the following recursive call now:
return select(L3,k-length(L1)-length(L2))

which translates to:
return select({30 35 40 45 50}, 1)

which will return 30 as a result. (since L has 5 or fewer elements, hence it'll return the element in kth i.e. 1st position in the sorted array, which is 30).

Now, M = 30 will be received in the first select function call over the entire array of 45 elements, and the same partitioning logic which separates the array L around M = 30 will apply to finally get the median of medians.

Phew! I hope I was verbose and clear enough to explain median of medians algorithm.

Understanding "median of medians" algorithm

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