Fix some more things related to control key weirdness

dotfiles/Xresources

@@ -127,6 +127,10 @@ URxvt.tabbed.title: no
 ! URxvt.keysym.Control-Left:     \033[1;5D
 ! URxvt.keysym.Control-Right:    \033[1;5C
 
+! URxvt doesn't nicely send C-S- to the underlying app so they have to be
+! added here as they become necessary.
+! URxvt.keysym.C-S-0:
+
 Xft.dpi: 96
 Xft.antialias: 1
 Xft.hinting:    1

dotfiles/emacs.d/init.el

@@ -216,7 +216,6 @@
 
 (require 'fill-column-indicator)
 
-(setq fci-rule-width 1)
 (setq fci-rule-color "black")
 (setq fci-rule-column 120)
 (add-hook 'after-change-major-mode-hook 'fci-mode)
@@ -306,14 +305,6 @@
 (projectile-global-mode)
 
 ;;-------------------------------------------------------------------------------------
-;; Rainbow Delimiters (on most programming modes)
-;;-------------------------------------------------------------------------------------
-
-(require 'rainbow-delimiters)
-
-(add-hook 'prog-mode-hook 'rainbow-delimiters-mode)
-
-;;-------------------------------------------------------------------------------------
 ;; Neotree Configuration
 ;;-------------------------------------------------------------------------------------
 

dotfiles/tmux.conf

@@ -87,5 +87,5 @@ bind p paste-buffer
 # Enable vi-like keybindings
 set-window-option -g mode-keys vi
 
-# Pass-through C-
-#set-window-option -g xterm-keys on
+# Pass-through C-<arrow>
+set-window-option -g xterm-keys on

sicp-2.5-notes

@@ -0,0 +1,41 @@
+(car (cons x y))
+
+(car (lambda (m) (m x y)))
+
+((lambda (m) (m x y)) (lambda (p q) p)))
+
+(lambda (x y) x)))
+
+x
+
+(define (cdr z) (z (lambda (x y) y)))
+
+
+
+2.5 - Given that we know the bases - 2, and 3, we only need to find the exponents.
+
+For example, 6 can only be represented one way - 2^1 3^1, meaning our pair of numbers a and b are 1
+
+So, cons is simple:
+
+(define (cons x y) (* (expt 2 x) (expt 3 y)))
+
+Now we're left solving for a and b
+
+log(2^a * 3^b) = alog(2) + blog(3)
+
+call our "cons" value x
+
+Because integer factorizations are unique we only need to find the biggest power of two and three that 
+go into the number.
+
+(define (power-of-x value base) 
+  (define (iter current-power)
+    (if (< ((expt base current-power) value)
+	(iter (+ current-power 1))
+	(current-power))))
+
+Then with this we can write car and cdr as
+
+(define (car int) (power-of-x int 2))
+(define (cdr int) (power-of-x int 3))