fixing relative file path for hint figures

Author: kcpearce

Project#01/hints/hint7-1.md

@@ -4,8 +4,8 @@ Here are two approaches for the [diagonalization](http://en.wikipedia.org/wiki/D
 Since the moment of inertia tensor is only a 3x3 matrix, a brute-force approach via the secular determinant is feasible:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/determinant.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/determinant.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/determinant.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/determinant.png">
   <img src="../figures/determinant.png" height="60">
 </picture>
 

Project#03/hints/hint3-1.md

@@ -2,24 +2,24 @@
 How do we conveniently store the elements of the four-dimensional two-electron integral array in a one-dimensional array?  Consider the lower triangle of an *n x n* symmetric matrix:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/n-by-n-symmetric-matrix.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/n-by-n-symmetric-matrix.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/n-by-n-symmetric-matrix.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/n-by-n-symmetric-matrix.png">
   <img src="../figures/n-by-n-symmetric-matrix.png" height="125">
 </picture>
 
 The total number of elements in the lower triangle is *M = n(n+1)/2*.  We could store these in a one-dimensional array by ordering them from top-to-bottom, left-to-right:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/lower-triang-numbered-matrix.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/lower-triang-numbered-matrix.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/lower-triang-numbered-matrix.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/lower-triang-numbered-matrix.png">
   <img src="../figures/lower-triang-numbered-matrix.png" height="125">
 </picture>
 
 How do we translate row (*i*) and column (*j*) indices of the matrix to the position in the linear array (*ij*)?
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/ioff-compound-index.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/ioff-compound-index.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/ioff-compound-index.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/ioff-compound-index.png">
   <img src="../figures/ioff-compound-index.png" height="70">
 </picture>
 
@@ -41,15 +41,15 @@ Here's an equivalent piece of code that requires fewer keystrokes:
 The eight-fold permutational symmetry of the two-electron repulsion integrals can be viewed similarly.  The Mulliken-notation integrals are symmetric to permutations of the bra indices or of the ket indices.  Hence, we can view the integral list as a symmetric "super-matrix" of the form:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/symmetric-integral-matrix.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/symmetric-integral-matrix.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/symmetric-integral-matrix.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/symmetric-integral-matrix.png">
   <img src="../figures/symmetric-integral-matrix.png" height="125">
 </picture>
 
 Thus, just as for the two-dimensional case above, we only need to store the lower triangle of this matrix, and a one-dimensional array of length *M(M+1)/2* will do the trick.  Given four AO indices, *i*, *j*, *k*, and *l* corresponding to the integral, (ij|kl), we can translate these into compound row (*ij*) and column (*kl*) indices using the expression above, as well as the final compound index:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/ioff-final-compound-index.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/ioff-final-compound-index.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/ioff-final-compound-index.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/ioff-final-compound-index.png">
   <img src="../figures/ioff-final-compound-index.png" height="70">
 </picture>

Project#03/hints/hint3-2.md

@@ -1,16 +1,16 @@
 If you've written your program so that the two-electron integrals are stored in a one-dimensional array, you may have recognized that your program spends quite a bit of its time calculating compound indices.  Is there a way to speed up this process?  Sure!  Pre-compute the indices instead.  Look again at the structure of a symmetric matrix, with the elements of the lower triangle numbered top-to-bottom/left-to-right:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/lower-triang-numbered-matrix2.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/lower-triang-numbered-matrix2.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/lower-triang-numbered-matrix2.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/lower-triang-numbered-matrix2.png">
   <img src="../figures/lower-triang-numbered-matrix2.png" height="150">
 </picture>
 
 We calculate these above values from the raw row and column indices, *i* and *j*, respectively using:
 
 <picture>
-  <source media="(prefers-color-scheme: dark)" srcset="./figures/dark/ioff-compound-index2.png">
-  <source media="(prefers-color-scheme: light)" srcset="./figures/ioff-compound-index2.png">
+  <source media="(prefers-color-scheme: dark)" srcset="../figures/dark/ioff-compound-index2.png">
+  <source media="(prefers-color-scheme: light)" srcset="../figures/ioff-compound-index2.png">
   <img src="../figures/ioff-compound-index2.png" height="70">
 </picture>