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Bicubic interpolation
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<h2 id="computation">Computation</h2> <p>Suppose the function values f {\displaystyle f} and the derivatives f x {\displaystyle f_{x}} , f y {\displaystyle f_{y}} and f x y {\displaystyle f_{xy}} are known at the four corners ( 0 , 0 ) {\displaystyle (0,0)} , ( 1 , 0 ) {\displaystyle (1,0)} , ( 0 , 1 ) {\displaystyle (0,1)} , and ( 1 , 1 ) {\displaystyle (1,1)} of the unit square. The interpolated surface can then be written as p ( x , y ) = ∑ i = 0 3 ∑ j = 0 3 a i j x i y j . {\displaystyle p(x,y)=\sum \limits _{i=0}^{3}\sum _{j=0}^{3}a_{ij}x^{i}y^{j}.} </p><p>The interpolation problem consists of determining the 16 coefficients a i j {\displaystyle a_{ij}} . Matching p ( x , y ) {\displaystyle p(x,y)} with the function values yields four equations: </p> <ol><li> f ( 0 , 0 ) = p ( 0 , 0 ) = a 00 , {\displaystyle f(0,0)=p(0,0)=a_{00},} </li> <li> f ( 1 , 0 ) = p ( 1 , 0 ) = a 00 + a 10 + a 20 + a 30 , {\displaystyle f(1,0)=p(1,0)=a_{00}+a_{10}+a_{20}+a_{30},} </li> <li> f ( 0 , 1 ) = p ( 0 , 1 ) = a 00 + a 01 + a 02 + a 03 , {\displaystyle f(0,1)=p(0,1)=a_{00}+a_{01}+a_{02}+a_{03},} </li> <li> f ( 1 , 1 ) = p ( 1 , 1 ) = ∑ i = 0 3 ∑ j = 0 3 a i j . {\displaystyle f(1,1)=p(1,1)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=0}^{3}a_{ij}.} </li></ol> <p>Likewise, eight equations for the derivatives in the x {\displaystyle x} and the y {\displaystyle y} directions: </p> <ol><li> f x ( 0 , 0 ) = p x ( 0 , 0 ) = a 10 , {\displaystyle f_{x}(0,0)=p_{x}(0,0)=a_{10},} </li> <li> f x ( 1 , 0 ) = p x ( 1 , 0 ) = a 10 + 2 a 20 + 3 a 30 , {\displaystyle f_{x}(1,0)=p_{x}(1,0)=a_{10}+2a_{20}+3a_{30},} </li> <li> f x ( 0 , 1 ) = p x ( 0 , 1 ) = a 10 + a 11 + a 12 + a 13 , {\displaystyle f_{x}(0,1)=p_{x}(0,1)=a_{10}+a_{11}+a_{12}+a_{13},} </li> <li> f x ( 1 , 1 ) = p x ( 1 , 1 ) = ∑ i = 1 3 ∑ j = 0 3 a i j i , {\displaystyle f_{x}(1,1)=p_{x}(1,1)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}a_{ij}i,} </li> <li> f y ( 0 , 0 ) = p y ( 0 , 0 ) = a 01 , {\displaystyle f_{y}(0,0)=p_{y}(0,0)=a_{01},} </li> <li> f y ( 1 , 0 ) = p y ( 1 , 0 ) = a 01 + a 11 + a 21 + a 31 , {\displaystyle f_{y}(1,0)=p_{y}(1,0)=a_{01}+a_{11}+a_{21}+a_{31},} </li> <li> f y ( 0 , 1 ) = p y ( 0 , 1 ) = a 01 + 2 a 02 + 3 a 03 , {\displaystyle f_{y}(0,1)=p_{y}(0,1)=a_{01}+2a_{02}+3a_{03},} </li> <li> f y ( 1 , 1 ) = p y ( 1 , 1 ) = ∑ i = 0 3 ∑ j = 1 3 a i j j . {\displaystyle f_{y}(1,1)=p_{y}(1,1)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}a_{ij}j.} </li></ol> <p>And four equations for the x y {\displaystyle xy} <a href="/facts/Mixed_partial_derivative/JWHsBkAu">mixed partial derivative</a>: </p> <ol><li> f x y ( 0 , 0 ) = p x y ( 0 , 0 ) = a 11 , {\displaystyle f_{xy}(0,0)=p_{xy}(0,0)=a_{11},} </li> <li> f x y ( 1 , 0 ) = p x y ( 1 , 0 ) = a 11 + 2 a 21 + 3 a 31 , {\displaystyle f_{xy}(1,0)=p_{xy}(1,0)=a_{11}+2a_{21}+3a_{31},} </li> <li> f x y ( 0 , 1 ) = p x y ( 0 , 1 ) = a 11 + 2 a 12 + 3 a 13 , {\displaystyle f_{xy}(0,1)=p_{xy}(0,1)=a_{11}+2a_{12}+3a_{13},} </li> <li> f x y ( 1 , 1 ) = p x y ( 1 , 1 ) = ∑ i = 1 3 ∑ j = 1 3 a i j i j . {\displaystyle f_{xy}(1,1)=p_{xy}(1,1)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}a_{ij}ij.} </li></ol> <p>The expressions above have used the following identities: p x ( x , y ) = ∑ i = 1 3 ∑ j = 0 3 a i j i x i − 1 y j , {\displaystyle p_{x}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}a_{ij}ix^{i-1}y^{j},} p y ( x , y ) = ∑ i = 0 3 ∑ j = 1 3 a i j x i j y j − 1 , {\displaystyle p_{y}(x,y)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}a_{ij}x^{i}jy^{j-1},} p x y ( x , y ) = ∑ i = 1 3 ∑ j = 1 3 a i j i x i − 1 j y j − 1 . {\displaystyle p_{xy}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}a_{ij}ix^{i-1}jy^{j-1}.} </p><p>This procedure yields a surface p ( x , y ) {\displaystyle p(x,y)} on the <a href="/facts/Unit_square/6aJUASbv">unit square</a> [ 0 , 1 ] × [ 0 , 1 ] {\displaystyle [0,1]\times [0,1]} that is continuous and has continuous derivatives. Bicubic interpolation on an arbitrarily sized <a href="/facts/Regular_grid/8vDdL4UB">regular grid</a> can then be accomplished by patching together such bicubic surfaces, ensuring that the derivatives match on the boundaries. </p><p>Grouping the unknown parameters a i j {\displaystyle a_{ij}} in a vector α = [ a 00 a 10 a 20 a 30 a 01 a 11 a 21 a 31 a 02 a 12 a 22 a 32 a 03 a 13 a 23 a 33 ] T {\displaystyle \alpha =\left[{\begin{smallmatrix}a_{00}&a_{10}&a_{20}&a_{30}&a_{01}&a_{11}&a_{21}&a_{31}&a_{02}&a_{12}&a_{22}&a_{32}&a_{03}&a_{13}&a_{23}&a_{33}\end{smallmatrix}}\right]^{T}} and letting x = [ f ( 0 , 0 ) f ( 1 , 0 ) f ( 0 , 1 ) f ( 1 , 1 ) f x ( 0 , 0 ) f x ( 1 , 0 ) f x ( 0 , 1 ) f x ( 1 , 1 ) f y ( 0 , 0 ) f y ( 1 , 0 ) f y ( 0 , 1 ) f y ( 1 , 1 ) f x y ( 0 , 0 ) f x y ( 1 , 0 ) f x y ( 0 , 1 ) f x y ( 1 , 1 ) ] T , {\displaystyle x=\left[{\begin{smallmatrix}f(0,0)&f(1,0)&f(0,1)&f(1,1)&f_{x}(0,0)&f_{x}(1,0)&f_{x}(0,1)&f_{x}(1,1)&f_{y}(0,0)&f_{y}(1,0)&f_{y}(0,1)&f_{y}(1,1)&f_{xy}(0,0)&f_{xy}(1,0)&f_{xy}(0,1)&f_{xy}(1,1)\end{smallmatrix}}\right]^{T},} the above system of equations can be reformulated into a matrix for the linear equation A α = x {\displaystyle A\alpha =x} . </p><p>Inverting the matrix gives the more useful linear equation A − 1 x = α {\displaystyle A^{-1}x=\alpha } , where A − 1 = [ 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 − 3 3 0 0 − 2 − 1 0 0 0 0 0 0 0 0 0 0 2 − 2 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 − 3 3 0 0 − 2 − 1 0 0 0 0 0 0 0 0 0 0 2 − 2 0 0 1 1 0 0 − 3 0 3 0 0 0 0 0 − 2 0 − 1 0 0 0 0 0 0 0 0 0 − 3 0 3 0 0 0 0 0 − 2 0 − 1 0 9 − 9 − 9 9 6 3 − 6 − 3 6 − 6 3 − 3 4 2 2 1 − 6 6 6 − 6 − 3 − 3 3 3 − 4 4 − 2 2 − 2 − 2 − 1 − 1 2 0 − 2 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 2 0 − 2 0 0 0 0 0 1 0 1 0 − 6 6 6 − 6 − 4 − 2 4 2 − 3 3 − 3 3 − 2 − 1 − 2 − 1 4 − 4 − 4 4 2 2 − 2 − 2 2 − 2 2 − 2 1 1 1 1 ] , {\displaystyle A^{-1}=\left[{\begin{smallmatrix}{\begin{array}{rrrrrrrrrrrrrrrr}1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\-3&3&0&0&-2&-1&0&0&0&0&0&0&0&0&0&0\\2&-2&0&0&1&1&0&0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&-3&3&0&0&-2&-1&0&0\\0&0&0&0&0&0&0&0&2&-2&0&0&1&1&0&0\\-3&0&3&0&0&0&0&0&-2&0&-1&0&0&0&0&0\\0&0&0&0&-3&0&3&0&0&0&0&0&-2&0&-1&0\\9&-9&-9&9&6&3&-6&-3&6&-6&3&-3&4&2&2&1\\-6&6&6&-6&-3&-3&3&3&-4&4&-2&2&-2&-2&-1&-1\\2&0&-2&0&0&0&0&0&1&0&1&0&0&0&0&0\\0&0&0&0&2&0&-2&0&0&0&0&0&1&0&1&0\\-6&6&6&-6&-4&-2&4&2&-3&3&-3&3&-2&-1&-2&-1\\4&-4&-4&4&2&2&-2&-2&2&-2&2&-2&1&1&1&1\end{array}}\end{smallmatrix}}\right],} which allows α {\displaystyle \alpha } to be calculated quickly and easily. </p><p>There can be another concise matrix form for 16 coefficients: [ f ( 0 , 0 ) f ( 0 , 1 ) f y ( 0 , 0 ) f y ( 0 , 1 ) f ( 1 , 0 ) f ( 1 , 1 ) f y ( 1 , 0 ) f y ( 1 , 1 ) f x ( 0 , 0 ) f x ( 0 , 1 ) f x y ( 0 , 0 ) f x y ( 0 , 1 ) f x ( 1 , 0 ) f x ( 1 , 1 ) f x y ( 1 , 0 ) f x y ( 1 , 1 ) ] = [ 1 0 0 0 1 1 1 1 0 1 0 0 0 1 2 3 ] [ a 00 a 01 a 02 a 03 a 10 a 11 a 12 a 13 a 20 a 21 a 22 a 23 a 30 a 31 a 32 a 33 ] [ 1 1 0 0 0 1 1 1 0 1 0 2 0 1 0 3 ] , {\displaystyle {\begin{bmatrix}f(0,0)&f(0,1)&f_{y}(0,0)&f_{y}(0,1)\\f(1,0)&f(1,1)&f_{y}(1,0)&f_{y}(1,1)\\f_{x}(0,0)&f_{x}(0,1)&f_{xy}(0,0)&f_{xy}(0,1)\\f_{x}(1,0)&f_{x}(1,1)&f_{xy}(1,0)&f_{xy}(1,1)\end{bmatrix}}={\begin{bmatrix}1&0&0&0\\1&1&1&1\\0&1&0&0\\0&1&2&3\end{bmatrix}}{\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix}}{\begin{bmatrix}1&1&0&0\\0&1&1&1\\0&1&0&2\\0&1&0&3\end{bmatrix}},} or [ a 00 a 01 a 02 a 03 a 10 a 11 a 12 a 13 a 20 a 21 a 22 a 23 a 30 a 31 a 32 a 33 ] = [ 1 0 0 0 0 0 1 0 − 3 3 − 2 − 1 2 − 2 1 1 ] [ f ( 0 , 0 ) f ( 0 , 1 ) f y ( 0 , 0 ) f y ( 0 , 1 ) f ( 1 , 0 ) f ( 1 , 1 ) f y ( 1 , 0 ) f y ( 1 , 1 ) f x ( 0 , 0 ) f x ( 0 , 1 ) f x y ( 0 , 0 ) f x y ( 0 , 1 ) f x ( 1 , 0 ) f x ( 1 , 1 ) f x y ( 1 , 0 ) f x y ( 1 , 1 ) ] [ 1 0 − 3 2 0 0 3 − 2 0 1 − 2 1 0 0 − 1 1 ] , {\displaystyle {\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix}}={\begin{bmatrix}1&0&0&0\\0&0&1&0\\-3&3&-2&-1\\2&-2&1&1\end{bmatrix}}{\begin{bmatrix}f(0,0)&f(0,1)&f_{y}(0,0)&f_{y}(0,1)\\f(1,0)&f(1,1)&f_{y}(1,0)&f_{y}(1,1)\\f_{x}(0,0)&f_{x}(0,1)&f_{xy}(0,0)&f_{xy}(0,1)\\f_{x}(1,0)&f_{x}(1,1)&f_{xy}(1,0)&f_{xy}(1,1)\end{bmatrix}}{\begin{bmatrix}1&0&-3&2\\0&0&3&-2\\0&1&-2&1\\0&0&-1&1\end{bmatrix}},} where p ( x , y ) = [ 1 x x 2 x 3 ] [ a 00 a 01 a 02 a 03 a 10 a 11 a 12 a 13 a 20 a 21 a 22 a 23 a 30 a 31 a 32 a 33 ] [ 1 y y 2 y 3 ] . {\displaystyle p(x,y)={\begin{bmatrix}1&x&x^{2}&x^{3}\end{bmatrix}}{\begin{bmatrix}a_{00}&a_{01}&a_{02}&a_{03}\\a_{10}&a_{11}&a_{12}&a_{13}\\a_{20}&a_{21}&a_{22}&a_{23}\\a_{30}&a_{31}&a_{32}&a_{33}\end{bmatrix}}{\begin{bmatrix}1\\y\\y^{2}\\y^{3}\end{bmatrix}}.} </p> <h2 id="extension-to-rectilinear-grids">Extension to rectilinear grids</h2> <p>Often, applications call for bicubic interpolation using data on a rectilinear grid, rather than the unit square. In this case, the identities for p x , p y , {\displaystyle p_{x},p_{y},} and p x y {\displaystyle p_{xy}} become p x ( x , y ) = ∑ i = 1 3 ∑ j = 0 3 a i j i x i − 1 y j Δ x , {\displaystyle p_{x}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=0}^{3}{\frac {a_{ij}ix^{i-1}y^{j}}{\Delta x}},} p y ( x , y ) = ∑ i = 0 3 ∑ j = 1 3 a i j x i j y j − 1 Δ y , {\displaystyle p_{y}(x,y)=\textstyle \sum \limits _{i=0}^{3}\sum \limits _{j=1}^{3}{\frac {a_{ij}x^{i}jy^{j-1}}{\Delta y}},} p x y ( x , y ) = ∑ i = 1 3 ∑ j = 1 3 a i j i x i − 1 j y j − 1 Δ x Δ y , {\displaystyle p_{xy}(x,y)=\textstyle \sum \limits _{i=1}^{3}\sum \limits _{j=1}^{3}{\frac {a_{ij}ix^{i-1}jy^{j-1}}{\Delta x\Delta y}},} where Δ x {\displaystyle \Delta x} is the x {\displaystyle x} spacing of the cell containing the point ( x , y ) {\displaystyle (x,y)} and similar for Δ y {\displaystyle \Delta y} . In this case, the most practical approach to computing the coefficients α {\displaystyle \alpha } is to let x = [ f ( 0 , 0 ) f ( 1 , 0 ) f ( 0 , 1 ) f ( 1 , 1 ) Δ x f x ( 0 , 0 ) Δ x f x ( 1 , 0 ) Δ x f x ( 0 , 1 ) Δ x f x ( 1 , 1 ) Δ y f y ( 0 , 0 ) Δ y f y ( 1 , 0 ) Δ y f y ( 0 , 1 ) Δ y f y ( 1 , 1 ) Δ x Δ y f x y ( 0 , 0 ) Δ x Δ y f x y ( 1 , 0 ) Δ x Δ y f x y ( 0 , 1 ) Δ x Δ y f x y ( 1 , 1 ) ] T , {\displaystyle x=\left[{\begin{smallmatrix}f(0,0)&f(1,0)&f(0,1)&f(1,1)&\Delta xf_{x}(0,0)&\Delta xf_{x}(1,0)&\Delta xf_{x}(0,1)&\Delta xf_{x}(1,1)&\Delta yf_{y}(0,0)&\Delta yf_{y}(1,0)&\Delta yf_{y}(0,1)&\Delta yf_{y}(1,1)&\Delta x\Delta yf_{xy}(0,0)&\Delta x\Delta yf_{xy}(1,0)&\Delta x\Delta yf_{xy}(0,1)&\Delta x\Delta yf_{xy}(1,1)\end{smallmatrix}}\right]^{T},} then to solve α = A − 1 x {\displaystyle \alpha =A^{-1}x} with A {\displaystyle A} as before. Next, the normalized interpolating variables are computed as x ¯ = x − x 0 x 1 − x 0 , y ¯ = y − y 0 y 1 − y 0 {\displaystyle {\begin{aligned}{\overline {x}}&={\frac {x-x_{0}}{x_{1}-x_{0}}},\\{\overline {y}}&={\frac {y-y_{0}}{y_{1}-y_{0}}}\end{aligned}}} where x 0 , x 1 , y 0 , {\displaystyle x_{0},x_{1},y_{0},} and y 1 {\displaystyle y_{1}} are the x {\displaystyle x} and y {\displaystyle y} coordinates of the grid points surrounding the point ( x , y ) {\displaystyle (x,y)} . Then, the interpolating surface becomes p ( x , y ) = ∑ i = 0 3 ∑ j = 0 3 a i j x ¯ i y ¯ j . {\displaystyle p(x,y)=\sum \limits _{i=0}^{3}\sum _{j=0}^{3}a_{ij}{\overline {x}}^{i}{\overline {y}}^{j}.} </p> <h2 id="finding-derivatives-from-function-values">Finding derivatives from function values</h2> <p>If the derivatives are unknown, they are typically approximated from the function values at points neighbouring the corners of the unit square, e.g. using <a href="/facts/Finite_differences/aY1txFhj">finite differences</a>. </p><p>To find either of the single derivatives, f x {\displaystyle f_{x}} or f y {\displaystyle f_{y}} , using that method, find the slope between the two <i>surrounding</i> points in the appropriate axis. For example, to calculate f x {\displaystyle f_{x}} for one of the points, find f ( x , y ) {\displaystyle f(x,y)} for the points to the left and right of the target point and calculate their slope, and similarly for f y {\displaystyle f_{y}} . </p><p>To find the cross derivative f x y {\displaystyle f_{xy}} , take the derivative in both axes, one at a time. For example, one can first use the f x {\displaystyle f_{x}} procedure to find the x {\displaystyle x} derivatives of the points above and below the target point, then use the f y {\displaystyle f_{y}} procedure on those values (rather than, as usual, the values of f {\displaystyle f} for those points) to obtain the value of f x y ( x , y ) {\displaystyle f_{xy}(x,y)} for the target point. (Or one can do it in the opposite direction, first calculating f y {\displaystyle f_{y}} and then f x {\displaystyle f_{x}} from those. The two give equivalent results.) </p><p>At the edges of the dataset, when one is missing some of the surrounding points, the missing points can be approximated by a number of methods. A simple and common method is to assume that the slope from the existing point to the target point continues without further change, and using this to calculate a hypothetical value for the missing point. </p> <h2 id="bicubic-convolution-algorithm">Bicubic convolution algorithm</h2> <p>Bicubic spline interpolation requires the solution of the linear system described above for each grid cell. An interpolator with similar properties can be obtained by applying a <a href="/facts/Convolution/PVPrdz9J">convolution</a> with the following kernel in both dimensions: W ( x ) = { ( a + 2 ) | x | 3 − ( a + 3 ) | x | 2 + 1 for | x | ≤ 1 , a | x | 3 − 5 a | x | 2 + 8 a | x | − 4 a for 1 < | x | < 2 , 0 otherwise , {\displaystyle W(x)={\begin{cases}(a+2)|x|^{3}-(a+3)|x|^{2}+1&{\text{for }}|x|\leq 1,\\a|x|^{3}-5a|x|^{2}+8a|x|-4a&{\text{for }}1<|x|<2,\\0&{\text{otherwise}},\end{cases}}} where a {\displaystyle a} is usually set to −0.5 or −0.75. Note that W ( 0 ) = 1 {\displaystyle W(0)=1} and W ( n ) = 0 {\displaystyle W(n)=0} for all nonzero integers n {\displaystyle n} . </p><p>This approach was proposed by Keys, who showed that a = − 0.5 {\displaystyle a=-0.5} produces third-order convergence with respect to the sampling interval of the original function.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p><p>If we use the matrix notation for the common case a = − 0.5 {\displaystyle a=-0.5} , we can express the equation in a more friendly manner: p ( t ) = 1 2 [ 1 t t 2 t 3 ] [ 0 2 0 0 − 1 0 1 0 2 − 5 4 − 1 − 1 3 − 3 1 ] [ f − 1 f 0 f 1 f 2 ] {\displaystyle p(t)={\tfrac {1}{2}}{\begin{bmatrix}1&t&t^{2}&t^{3}\end{bmatrix}}{\begin{bmatrix}0&2&0&0\\-1&0&1&0\\2&-5&4&-1\\-1&3&-3&1\end{bmatrix}}{\begin{bmatrix}f_{-1}\\f_{0}\\f_{1}\\f_{2}\end{bmatrix}}} for t {\displaystyle t} between 0 and 1 for one dimension. Note that for 1-dimensional cubic convolution interpolation 4 sample points are required. For each inquiry two samples are located on its left and two samples on the right. These points are indexed from −1 to 2 in this text. The distance from the point indexed with 0 to the inquiry point is denoted by t {\displaystyle t} here. </p><p>For two dimensions first applied once in x {\displaystyle x} and again in y {\displaystyle y} : b − 1 = p ( t x , f ( − 1 , − 1 ) , f ( 0 , − 1 ) , f ( 1 , − 1 ) , f ( 2 , − 1 ) ) , b 0 = p ( t x , f ( − 1 , 0 ) , f ( 0 , 0 ) , f ( 1 , 0 ) , f ( 2 , 0 ) ) , b 1 = p ( t x , f ( − 1 , 1 ) , f ( 0 , 1 ) , f ( 1 , 1 ) , f ( 2 , 1 ) ) , b 2 = p ( t x , f ( − 1 , 2 ) , f ( 0 , 2 ) , f ( 1 , 2 ) , f ( 2 , 2 ) ) , {\displaystyle {\begin{aligned}b_{-1}&=p(t_{x},f_{(-1,-1)},f_{(0,-1)},f_{(1,-1)},f_{(2,-1)}),\\[1ex]b_{0}&=p(t_{x},f_{(-1,0)},f_{(0,0)},f_{(1,0)},f_{(2,0)}),\\[1ex]b_{1}&=p(t_{x},f_{(-1,1)},f_{(0,1)},f_{(1,1)},f_{(2,1)}),\\[1ex]b_{2}&=p(t_{x},f_{(-1,2)},f_{(0,2)},f_{(1,2)},f_{(2,2)}),\end{aligned}}} p ( x , y ) = p ( t y , b − 1 , b 0 , b 1 , b 2 ) . {\displaystyle p(x,y)=p(t_{y},b_{-1},b_{0},b_{1},b_{2}).} </p> <h2 id="use-in-computer-graphics">Use in computer graphics</h2> <p>The bicubic algorithm is frequently used for scaling images and video for display (see <a href="/facts/Resampling_(bitmap)/qNoeujzm">bitmap resampling</a>). It preserves fine detail better than the common <a href="/facts/Bilinear_filtering/yoke6SzF">bilinear</a> algorithm. </p><p>However, due to the negative lobes on the kernel, it causes <a href="/facts/Overshoot_(signal)/l450BHuI">overshoot</a> (haloing). This can cause <a href="/facts/Clipping_(signal_processing)/jEekS8S4">clipping</a>, and is an artifact (see also <a href="/facts/Ringing_artifacts/JC2W86cf">ringing artifacts</a>), but it increases <a href="/facts/Acutance/xgE5cm2N">acutance</a> (apparent sharpness), and can be desirable. </p> <h2 id="see-also">See also</h2> <ul> <li>Mathematics portal</li></ul> <ul><li><a href="/facts/Spatial_anti-aliasing/od0SgvLy">Spatial anti-aliasing</a></li> <li><a href="/facts/B%C3%A9zier_surface/F1FlWnId">Bézier surface</a></li> <li><a href="/facts/Bilinear_interpolation/yoke6SzF">Bilinear interpolation</a></li> <li><a href="/facts/Cubic_Hermite_spline/S29nwRUF">Cubic Hermite spline</a>, the one-dimensional analogue of bicubic spline</li> <li><a href="/facts/Lanczos_resampling/uup50eym">Lanczos resampling</a></li> <li><a href="/facts/Natural_neighbor_interpolation/QSxn2NaR">Natural neighbor interpolation</a></li> <li><a href="/facts/Sinc_filter/1H6q0F5t">Sinc filter</a></li> <li><a href="/facts/Spline_interpolation/AowdpQwm">Spline interpolation</a></li> <li><a href="/facts/Tricubic_interpolation/U8EwEvLl">Tricubic interpolation</a></li> <li><a href="/facts/Directional_Cubic_Convolution_Interpolation/rmF2BgAy">Directional Cubic Convolution Interpolation</a></li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="https://web.archive.org/web/20051024202307/http://www.geovista.psu.edu/sites/geocomp99/Gc99/082/gc_082.htm">Application of interpolation to elevation samples</a></li> <li><a href="https://sepwww.stanford.edu/data/media/public/docs/sep107/paper_html/node20.html">Interpolation theory</a></li> <li><a href="http://www.paulinternet.nl/?page=bicubic">Explanation and Java/C++ implementation of (bi)cubic interpolation</a></li> <li><a href="http://mathformeremortals.wordpress.com/2013/01/15/bicubic-interpolation-excel-worksheet-function/">Excel Worksheet Function for Bicubic Lagrange Interpolation</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>R. Keys (1981). "Cubic convolution interpolation for digital image processing". IEEE Transactions on Acoustics, Speech, and Signal Processing. 29 (6): 1153–1160. Bibcode:1981ITASS..29.1153K. CiteSeerX 10.1.1.320.776. doi:10.1109/TASSP.1981.1163711. <a href="/wiki/Bibcode_(identifier)" target="_blank">/wiki/Bibcode_(identifier)</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>