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+# 27 Nonstrictness
+
+## 27.5 Exercises: Evaluate
+
+Excercise 1
+
+```haskell
+const 1 undefined
+(\a -> \b -> a) 1 undefined
+(\b -> 1) undefined
+1
+```
+
+Exercise 2
+
+```haskell
+const undefined 1
+(\a -> \b -> a) undefined 1
+(\b -> undefined) 1
+undefined
+```
+
+Exercise 3
+
+```haskell
+flip const undefined 1
+(\f -> \a -> \b -> f b a) const undefined 1
+(\a -> \b -> const b a) undefined 1
+(\b -> const b undefined) 1
+const 1 undefined -- same as Exercise 1 now
+```
+
+Exercise 4 follows same principle as Exercise 3 but leads to Exercise 2.
+
+Exercise 5 follows same pattern as before but where 1 is undefined.
+
+Exercise 6
+
+```haskell
+foldr const 'z' ['a'..'e']
+const 'a' (foldr const 'z' ['b'..'e'])
+'a'
+```
+
+Exercise 7
+
+```haskell
+foldr (flip const) 'z' ['a'..'e']
+(flip const) 'a' (foldr (flip const) 'z' ['b'..'e'])
+const (foldr (flip const) 'z' ['b'..'e']) 'a'
+
+foldr (flip const) 'z' ['b'..'e']
+(flip const) 'b' (foldr (flip const) 'z' ['c'..'e'])
+const (foldr (flip const) 'z' ['c'..'e']) 'b'
+
+foldr (flip const) 'z' ['c'..'e']
+(flip const) 'c' (foldr (flip const) 'z' ['d','e'])
+const (foldr (flip const) 'z' ['d','e']) 'c'
+
+foldr (flip const) 'z' ['d','e']
+(flip const) 'd' (foldr (flip const) 'z' ['e'])
+const (foldr (flip const) 'z' ['e']) 'd'
+
+foldr (flip const) 'z' ['e']
+(flip const) 'e' (foldr (flip const) 'z' [])
+const (foldr (flip const) 'z' []) 'e'
+
+foldr (flip const) 'z' []
+'z'
+```
+
+## 27.14 Chapter Exercises
+
+### What will :sprint output
+
+1. `x = _`
+2. `x = "1"`
+3. `x = _`
+4. `x = 1`
+5. `x = _`
+6. `x = _`
+
+### Will printing this expression result in bottom
+
+1. No
+2. Yes
+3. Yes
+4. No
+5. No
+6. No
+7. Yes
+
+### Make the expression bottom
+
+There are different ways of doing this. This is one of them:
+
+```haskell
+x = undefined
+y = "blah"
+main = do
+    print (snd (x `seq` (x,y)))
+```