... diagrams For the above λ, σ(λ) gives the 2- modular diagram 2222222222 the electronic journal of combinatorics 10 (20 03) , #R34 Namely, σ(λ) = (19, 13, 6, 5, 3) Clearly, the number of rows in ... diagram As an example, let λ = (10, 9, 7, 4, 4, 4, 3, 2, 2, 1) Then, λ gives the 2- modular diagram 2222222222 Since no odd part of λ is repeated, the 1’s can only occur at the bottom of ... at most 2m and with no repeated odd parts Theorem 2. 1 was established by Chapman [5] in his proof of the q-identity ∞ −q 2n−1 + q 2n n − q 2n n=1 n−1 j=1 ∞ − q 2j−1 − q 2j−1 = − q 2j − q 2j j=1...
... data[current_position] = DataIn count = count + return success 13 End InsertHeap Delete minimum element from min-heap 31 31 31 31 The element in the last position is put to the position of the ... subtree, seem to hurt balance of the tree 33 Implementations of Priority Queue Use min-heap: • Insertion requires O(log2 n) • DeleteMin requires O(log2 n) 34 Insert and Remove element into/from ... number of nodes n is between 2h-1 and (2h -1) • Complete tree: n = 2h -1 when last level is full • Nearly complete: All nodes in the last level are on the left • h = |log2 n| + • Can be represented...
... generated by z T1 + wT2 + xT3 , y T1 + xT2 + wT4 , xw2 T1 + 2 z T2 + y T3 , x2 wT1 + y T2 + z T4 , xwT1 + T2 + T3 T4 From this it follows that F (I) = K[T1 , T2 , T3 , T4 ]/(T2 + T3 T4 ) Symmetric ... xs xμ− α1 xt xμ+ 2 xt xν− 2 xs xν+ β1 2 xr xμ− xν− α1 2 xp xμ+ xν+ 2 we obtain 22 xν+ xν− fu − α1 β1 xμ+ xμ− fv − fu+v fu−v = Hence, it follows ν+ ν− μ+ μ− that Q := 22 x x Tu − α1 β1 ... = xt L2 + xr L3 − xp L4 , xs Q = Tu−v L2 − xμ+ Tv L3 − xν− Tu L4 , then Ap = (L2 , L3 , L4 )p , and it is easy to verify that (K[x, T ]/(L2 , L3 , L4 ))(xs ) ∼ (K[x, Tu , Tv , Tu−v ]/(L3 , L4...