Displaying 6 results from an estimated 6 matches for "b_d".
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2006 Apr 29
2
[LLVMdev] Register allocation in LLVM
...the register
allocation phase, and adds one permutation of registers to each edge that
reaches the basic block where phi functions were destroyed. With three
xor, we can implement a swap without an extra register. For instance, if I
have (after RegAlloc), these PHI functions at the header of block B_d:
a(r1) <- (a1(r1), a2(r2))
b(r2) <- (b1(r2), b2(r1))
assuming that a2, b2 come from block B_s, at the end of block B_s I would
have to add the permutation: (r1, r2) <- perm (r2, r1)
which I can implement with six Xor instructions. Well, I would like to
get some comments about the best way...
2004 Oct 01
2
multiple dimensional diag()
...0 0 0 0 2 2
[5,] 0 0 0 0 2 2
I am trying to generalize this to two higher dimensional arrays.
If x <- adiag(a,b) then I want all(dim(x)==dim(a)+dim(b)); and if
dim(a)=c(a_1, a_2,...a_d) then x[1:a_1,1:a_2,...,1:a_d]=a, and
x[(a_1+1):(a_1+b_1),...,(a_d+1):(a_d+b_d)]=b. Other elements of x are
zero.
The fact that I'm having difficulty expressing this succinctly makes
me think I'm missing something basic.
If a and b have identical dimensions [ie all(dim(a)==dim(b)) ], the
following ghastly kludge (which is one of many) works:
adiag <- function(a...
2006 May 01
0
[LLVMdev] Register allocation in LLVM
...llocation phase, and adds one permutation of registers to each edge that
> reaches the basic block where phi functions were destroyed. With three
> xor, we can implement a swap without an extra register. For instance, if I
> have (after RegAlloc), these PHI functions at the header of block B_d:
> a(r1) <- (a1(r1), a2(r2))
> b(r2) <- (b1(r2), b2(r1))
> assuming that a2, b2 come from block B_s, at the end of block B_s I would
> have to add the permutation: (r1, r2) <- perm (r2, r1)
> which I can implement with six Xor instructions.
Ok, seems complicated, but potent...
2016 Apr 28
2
Combinaciones sin repetición...con restricciones
...-----
>
> Ejemplo:
>
> > set_examp <- c('a','b','c','d')
> > my_choice(set_examp, 'a', 'b')
> [1] "a" "b" "a_b" "a_c" "a_d" "b_c" "b_d"
> "a_b_c"
> [9] "a_b_d" "a_c_d" "b_c_d" "a_b_c_d"
>
> La función calcula todas las combinaciones sin repetición posibles y
> extrae los elementos donde están presentes "cual_1" y "cual_2". No está
>...
2006 May 01
2
[LLVMdev] Register allocation in LLVM
...o each
>> edge that
>> reaches the basic block where phi functions were destroyed. With
>> three
>> xor, we can implement a swap without an extra register. For
>> instance, if I
>> have (after RegAlloc), these PHI functions at the header of block
>> B_d:
>> a(r1) <- (a1(r1), a2(r2))
>> b(r2) <- (b1(r2), b2(r1))
>> assuming that a2, b2 come from block B_s, at the end of block B_s
>> I would
>> have to add the permutation: (r1, r2) <- perm (r2, r1)
>> which I can implement with six Xor instructions.
>...
2016 Apr 27
4
Combinaciones sin repetición...con restricciones
Hola, tengo que resolver un problema para el que normalmente utilizaría
excel, pero me gustaría intentar resolverlo con R. Se trata de lo siguiente:
Tengo tres elementos: a, b y c.
Dichos elementos están agrupados en siete objetos, producto de todas las
combinaciones sin repetición posibles:
Objeto 1: a
Objeto 2: b
Objeto 3: c
Objeto 4: a y b
Objeto 5: a y c
Objeto 6: b y c
Objeto 7: a, b y c