Have you ever tried to compare two large expressions but you couldn't easily see how they differed? For example:

e1 := 1/4*(2*T[x,j,g]*T[i,c,k]-2*T[x,j,c]*T[i,g,k]-2*d[j](T[x,c,i],[X])+2*d[x](T[c,j,i],[X]))*g_[a,b]+1/4*(T[a,x,b]+T[b,x,a])*T[c,j,i]-1/4*T[b,x,j]*T[c,a,i]-1/2*T[x,j,a]*T[i,b,c]+1/2*T[x,a,b]*T[j,c,i]-1/4*T[a,x,j]*T[c,b,i]+1/4*d[x](T[a,b,c],[X])+1/4*d[x](T[b,a,c],[X])-1/2*d[x](T[c,a,b],[X])-1/4*d[a](T[b,x,c],[X])+1/2*d[b](T[x,a,c],[X])-1/4*d[b](T[a,x,c],[X]);

$\mathrm{e1}≔\frac{\left(2T_{x,j,g}{T}_{i,c,k}-2T_{x,j,c}{T}_{i,g,k}-2d_j\left(T_{x,c,i}\,\left[X\right]\right)+2d_x\left(T_{c,j,i}\,\left[X\right]\right)\right){\mathrm{g\_}}_{a,b}}{4}+\frac{\left(T_{a,x,b}+T_{b,x,a}\right)T_{c,j,i}}{4}-\frac{T_{b,x,j}{T}_{c,a,i}}{4}-\frac{T_{x,j,a}{T}_{i,b,c}}{2}+\frac{T_{x,a,b}{T}_{j,c,i}}{2}-\frac{T_{a,x,j}{T}_{c,b,i}}{4}+\frac{d_x\left(T_{a,b,c}\,\left[X\right]\right)}{4}+\frac{d_x\left(T_{b,a,c}\,\left[X\right]\right)}{4}-\frac{d_x\left(T_{c,a,b}\,\left[X\right]\right)}{2}-\frac{d_a\left(T_{b,x,c}\,\left[X\right]\right)}{4}+\frac{d_b\left(T_{x,a,c}\,\left[X\right]\right)}{2}-\frac{d_b\left(T_{a,x,c}\,\left[X\right]\right)}{4}$

e2 := 1/4*(2*T[x,j,g]*T[i,c,k]-2*T[x,j,c]*T[i,g,k]+2*d[j](T[i,x,c],[X])-2*d[x](T[j,c,i],[X]))*g_[a,b]+1/4*(T[a,x,b]+T[b,x,a])*T[c,j,i]-1/4*T[b,x,j]*T[c,a,i]-1/2*T[x,j,a]*T[i,b,c]+1/2*T[x,a,b]*T[j,c,i]-1/4*T[a,x,j]*T[c,b,i]+1/4*d[x](T[a,b,c],[X])+1/4*d[x](T[b,a,c],[X])-1/2*d[x](T[c,a,b],[X])-1/4*d[a](T[b,x,c],[X])+1/2*d[b](T[x,a,c],[X])-1/4*d[b](T[a,x,c],[X]);

$\mathrm{e2}≔\frac{\left(2T_{x,j,g}{T}_{i,c,k}-2T_{x,j,c}{T}_{i,g,k}+2d_j\left(T_{i,x,c}\,\left[X\right]\right)-2d_x\left(T_{j,c,i}\,\left[X\right]\right)\right){\mathrm{g\_}}_{a,b}}{4}+\frac{\left(T_{a,x,b}+T_{b,x,a}\right)T_{c,j,i}}{4}-\frac{T_{b,x,j}{T}_{c,a,i}}{4}-\frac{T_{x,j,a}{T}_{i,b,c}}{2}+\frac{T_{x,a,b}{T}_{j,c,i}}{2}-\frac{T_{a,x,j}{T}_{c,b,i}}{4}+\frac{d_x\left(T_{a,b,c}\,\left[X\right]\right)}{4}+\frac{d_x\left(T_{b,a,c}\,\left[X\right]\right)}{4}-\frac{d_x\left(T_{c,a,b}\,\left[X\right]\right)}{2}-\frac{d_a\left(T_{b,x,c}\,\left[X\right]\right)}{4}+\frac{d_b\left(T_{x,a,c}\,\left[X\right]\right)}{2}-\frac{d_b\left(T_{a,x,c}\,\left[X\right]\right)}{4}$

evalb(e1=e2);

$\mathrm{false}$

The new package ExpressionTools permits you to perform a visual comparison of two expressions.

The main command in this package is Compare, which, given two expressions, compares them and highlights their differences:

with(ExpressionTools);

$\left[\mathrm{Compare}\,\mathrm{Options}\right]$

Compare(e1, e2);

$\frac{\left(2T_{x,j,g}{T}_{i,c,k}-2T_{x,j,c}{T}_{i,g,k}-2d_{\left(\begin{array}{c}j\\ x\end{array}\right)}\left(T_{\left(\begin{array}{c}x\\ j\end{array}\right),c,i}\,\left[X\right]\right)+2d_{\left(\begin{array}{c}x\\ j\end{array}\right)}\left(T_{\left(\begin{array}{c}c\\ i\end{array}\right),\left(\begin{array}{c}j\\ x\end{array}\right),\left(\begin{array}{c}i\\ c\end{array}\right)}\,\left[X\right]\right)\right){\mathrm{g\_}}_{a,b}}{4}+\frac{\left(T_{a,x,b}+T_{b,x,a}\right)T_{c,j,i}}{4}-\frac{T_{b,x,j}{T}_{c,a,i}}{4}-\frac{T_{x,j,a}{T}_{i,b,c}}{2}+\frac{T_{x,a,b}{T}_{j,c,i}}{2}-\frac{T_{a,x,j}{T}_{c,b,i}}{4}+\frac{d_x\left(T_{a,b,c}\,\left[X\right]\right)}{4}+\frac{d_x\left(T_{b,a,c}\,\left[X\right]\right)}{4}-\frac{d_x\left(T_{c,a,b}\,\left[X\right]\right)}{2}-\frac{d_a\left(T_{b,x,c}\,\left[X\right]\right)}{4}+\frac{d_b\left(T_{x,a,c}\,\left[X\right]\right)}{2}-\frac{d_b\left(T_{a,x,c}\,\left[X\right]\right)}{4}$

For smaller expressions (both having length less than 250, an adjustable parameter), the two expressions are printed one after the other, and the differences are simply highlighted by changing their colors:

x1 := (a-b)*(A-B)*c;

$\mathrm{x1}≔\left(a-b\right)\left(A-B\right)c$

x2 := (b-a)*(B-A)*C;

$\mathrm{x2}≔\left(b-a\right)\left(B-A\right)C$

Compare(x1, x2);

$\left(a+\mathrm{-1}b\right)\left(A+\mathrm{-1}B\right)c$

$\left(b+\mathrm{-1}a\right)\left(B+\mathrm{-1}A\right)C$

If a verification is provided as an option, the comparison ignores any differences it can find that satisfy that verification:

Compare(x1, x2, sign);

$\left(a-b\right)\left(A-B\right)c$

$\left(b-a\right)\left(B-A\right)C$

For sums, products, and sets, operands are compared according to the best match, even when none of the top-level subexpressions match exactly:

x3, x4 := a/2+b/3+c/4, c/5+b/6+d/7:

Compare(x3, x4, combine);

$\left(\begin{array}{c}\frac{a}{2}\\ \frac{d}{7}\end{array}\right)+\left(\begin{array}{c}\frac{1}{3}\\ \frac{1}{6}\end{array}\right)b+\left(\begin{array}{c}\frac{1}{4}\\ \frac{1}{5}\end{array}\right)c$

Here's a larger example, showing the use of some other options to control the display:

x3 := Vector(2,[w(x), z(x)]) = Vector(2,[1/15*exp(-x)*(-15*c__1+3)+1/15*(10*c__2+2)*exp(4*x)-1/3*exp(x), 1/5*exp(-x)*(-1+5*c__1)+1/5*exp(4*x)*(1+5*c__2)]);

$\mathrm{x3}≔\left[\begin{array}{c}w\left(x\right)\\ z\left(x\right)\end{array}\right]=\left[\begin{array}{c}\frac{{\ⅇ}^{-x}\left(-15\mathrm{c\_\_1}+3\right)}{15}+\frac{\left(10\mathrm{c\_\_2}+2\right){\ⅇ}^{4x}}{15}-\frac{{\ⅇ}^x}{3}\\ \frac{{\ⅇ}^{-x}\left(-1+5\mathrm{c\_\_1}\right)}{5}+\frac{{\ⅇ}^{4x}\left(1+5\mathrm{c\_\_2}\right)}{5}\end{array}\right]$

x4 := Vector(2,[w(x), z(x)]) = Vector(2, [1/15*exp(-x)*(-15*_C1+3)+1/15*(10*_C2+2)*exp(4*x)-1/3*exp(x),1/5*exp(-x)*((1+5*_C2)*exp(5*x)-1+5*_C1)]);

$\mathrm{x4}≔\left[\begin{array}{c}w\left(x\right)\\ z\left(x\right)\end{array}\right]=\left[\begin{array}{c}\frac{{\ⅇ}^{-x}\left(-15\mathrm{\_C1}+3\right)}{15}+\frac{\left(10\mathrm{\_C2}+2\right){\ⅇ}^{4x}}{15}-\frac{{\ⅇ}^x}{3}\\ \frac{{\ⅇ}^{-x}\left(\left(1+5\mathrm{\_C2}\right){\ⅇ}^{5x}-1+5\mathrm{\_C1}\right)}{5}\end{array}\right]$

Compare(x3, x4, Silver, Gold, showvectorbrackets=false, showNULLoperands);

$\left[\begin{array}{c}w\left(x\right)\\ z\left(x\right)\end{array}\right]=\left[\begin{array}{c}\frac{{\ⅇ}^{-x}\left(-15\begin{array}{c}\mathrm{c\_\_1}\\ \mathrm{\_C1}\end{array}+3\right)}{15}+\frac{\left(10\begin{array}{c}\mathrm{c\_\_2}\\ \mathrm{\_C2}\end{array}+2\right){\ⅇ}^{4x}}{15}-\frac{{\ⅇ}^x}{3}\\ \begin{array}{c}\frac{{\ⅇ}^{-x}\left(-1+5\mathrm{c\_\_1}\right)}{5}\\ \end{array}+\frac{{\ⅇ}^{\begin{array}{c}4\\ \mathrm{-1}\end{array}x}\left(\left(1+\begin{array}{c}\\ 5\mathrm{\_C2}\end{array}\right)\begin{array}{c}\\ {\ⅇ}^{5x}\end{array}+5\begin{array}{c}\mathrm{c\_\_2}\\ \mathrm{\_C1}\end{array}+\begin{array}{c}\\ \mathrm{-1}\end{array}\right)}{5}\end{array}\right]$

Compare uses the uneval parameter modifier, which means it does not evaluate its arguments before comparing them:

x5 := 'hypergeom'([a,b],[c],x);

$\mathrm{x5}≔\mathrm{hypergeom}\left(\left[a\,b\right]\,\left[c\right]\,x\right)$

x6 := 'hypergeom'([b,a],[c],x);

$\mathrm{x6}≔\mathrm{hypergeom}\left(\left[b\,a\right]\,\left[c\right]\,x\right)$

evalb(x5=x6);

$\mathrm{true}$

Compare(x5, x6);

$\mathrm{hypergeom}\left(\left[a\,b\right]\,\left[c\right]\,x\right)$

$\mathrm{hypergeom}\left(\left[b\,a\right]\,\left[c\right]\,x\right)$

Note, however, that this could lead to unexpected results:

Compare(x1, subs(C=c, x2), combine, sign);

$\left(\begin{array}{c}\left(a-b\right)\left(A-B\right)c\\ \mathrm{subs}\left(C=c\,\mathrm{x2}\right)\end{array}\right)$

So be sure to evaluate results if desired before passing them to Compare:

x3 := subs(C=c, x2);

$\mathrm{x3}≔\left(b-a\right)\left(B-A\right)c$

Compare(x1, x3, combine, sign);

$\mathrm{The\; expressions\; satisfy\; the\; given\; verification}$

Compare(x5, x6, evaluate);

$\mathrm{The\; expressions\; are\; the\; same}$

Compare(x1, subs(C=c, x2), sign, evaluate);

$\mathrm{The\; expressions\; satisfy\; the\; given\; verification}$

A secondary ExpressionTools command called Options allows you to query and set the default values for each option available to the Compare command:

Options();

$\mathrm{backgroundcolor\; :\; [LightGreen,\; Pink]}$

$\mathrm{combine\; :\; default}$

$\mathrm{combinethreshold\; :\; 250}$

$\mathrm{evaluate\; :\; false}$

$\mathrm{foregroundcolor\; :\; [Green,\; Red]}$

$\mathrm{highlight\; :\; default}$

$\mathrm{showNULLoperands\; :\; false}$

$\mathrm{showvectorbrackets\; :\; true}$

$\mathrm{verification\; :\; boolean}$

x1 := (2*a-3*b)*(A-B)*c:

x2 := (3*b-2*a)*(B-A)*C:

Compare(x1, x2);

$\left(2a+\mathrm{-3}b\right)\left(A+\mathrm{-1}B\right)c$

$\left(3b+\mathrm{-2}a\right)\left(B+\mathrm{-1}A\right)C$

Options(combine, highlight, backgroundcolor);

$\mathrm{combine\; :\; default}$

$\mathrm{highlight\; :\; default}$

$\mathrm{backgroundcolor\; :\; [LightGreen,\; Pink]}$

Options(combine=true, highlight=background, backgroundcolor = [Yellow, Orange]);

$\mathrm{combine\; :\; true}$

$\mathrm{highlight\; :\; background}$

$\mathrm{backgroundcolor\; :\; [Yellow,\; Orange]}$

Compare(x1, x2);

$\left(\left(\begin{array}{c}2\\ \mathrm{-2}\end{array}\right)a+\left(\begin{array}{c}\mathrm{-3}\\ 3\end{array}\right)b\right)\left(\left(\mathrm{-1}\right)A+\left(\mathrm{-1}\right)B\right)\left(\begin{array}{c}c\\ C\end{array}\right)$