diff --git a/homework/chapter2/65.c b/homework/chapter2/65.c
new file mode 100644
index 0000000..1e737f5
--- /dev/null
+++ b/homework/chapter2/65.c
@@ -0,0 +1,16 @@
+#include <stdio.h>
+
+/* Return 1 when x contains an odd number of 1s; 0 otherwise.
+   Assume w=32 */
+int odd_ones(unsigned x) {
+  x = x ^ (x >> 16);
+  x = x ^ (x >> 8);
+  x = x ^ (x >> 4);
+  x = x ^ (x >> 2);
+  x = x ^ (x >> 1);
+  return x & 1;
+}
+
+int main() {
+  printf("%d %d\n", odd_ones(0xF1F1F0F1), odd_ones(0xFF00AAAA));
+}
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diff --git a/homework/chapter2/66.c b/homework/chapter2/66.c
new file mode 100644
index 0000000..37518dd
--- /dev/null
+++ b/homework/chapter2/66.c
@@ -0,0 +1,26 @@
+#include <stdio.h>
+
+/*
+ * Generate mask indicating leftmost 1 in x.  Assume w=32.
+ * For example, 0xFF00 -> 0x8000, and 0x6600 --> 0x4000.
+ * If x = 0, then return 0.
+ */
+int leftmost_one(unsigned x) {
+  unsigned t = x;
+  t = t >> 1;
+
+  t = t | (t >> 1);
+  t = t | (t >> 2);
+  t = t | (t >> 4);
+  t = t | (t >> 8);
+  t = t | (t >> 16);
+
+  return (t + 1) & x;
+}
+
+int main() {
+  printf("%.8X %.8X\n", 0x06050403, leftmost_one(0x06050403));
+  printf("%.8X %.8X\n", 0xF0F0F0F0, leftmost_one(0xF0F0F0F0));
+  printf("%.8X %.8X\n", 0x00000000, leftmost_one(0x00000000));
+  printf("%.8X %.8X\n", 0x00000001, leftmost_one(0x00000001));
+}
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diff --git a/homework/chapter2/75.c b/homework/chapter2/75.c
new file mode 100644
index 0000000..75b68ee
--- /dev/null
+++ b/homework/chapter2/75.c
@@ -0,0 +1,23 @@
+#include <limits.h>
+#include <assert.h>
+
+int signed_high_prod(int x, int y) {
+  long long p = (long long) x * y;
+  return p >> 32;
+}
+
+unsigned unsigned_high_prod(unsigned x, unsigned y) {
+  int w = sizeof(int) << 3;
+  return signed_high_prod(x, y) + (((int) y >> (w - 1)) & x) + (((int) x >> (w - 1)) & y);
+}
+
+int main() {
+  unsigned x = 0x71827364, y = 0xF0E1D2C3;
+  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+  x = 0x91827364, y = 0xF0E1D2C3;
+  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+  x = 0x71827364, y = 0x60E1D2C3;
+  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+  x = 0x91827364, y = 0x60E1D2C3;
+  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+}
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diff --git a/homework/chapter2/80.c b/homework/chapter2/80.c
new file mode 100644
index 0000000..8d0950c
--- /dev/null
+++ b/homework/chapter2/80.c
@@ -0,0 +1,20 @@
+#include <assert.h>
+#include <stdio.h>
+
+int threefourths(int x) {
+  int p = x >> 2, q = x & 3;
+  return p + (p << 1) + q + ((!q | (x >> 31)) & 1) - 1;
+}
+
+int main() {
+  int x = 0x88659612;
+  for (int i = 0; i < 100; i++) {
+    x++;
+    assert(threefourths(x) == (long long) x * 3 / 4);
+  }
+  x = 0x79642695;
+  for (int i = 0; i < 100; i++) {
+    x++;
+    assert(threefourths(x) == (long long) x * 3 / 4);
+  }
+}
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diff --git a/homework/chapter2/README.md b/homework/chapter2/README.md
new file mode 100644
index 0000000..d16939b
--- /dev/null
+++ b/homework/chapter2/README.md
@@ -0,0 +1,21 @@
+## 2.65
+定义一次操作为将 $w$ 位的整数 $x$ 从中间分成两个 $\frac{w}{2}$ 位的整数 $y$ 和 $z$，接着令 `x=y^z`。进行该操作 5 次后得到答案。
+
+我们将这样的思路称为“折半递归法”。
+## 2.66
+第一次操作，我们令 `x = x | (x >> 1)`，这样 $x$ 的最高位所在 `1` 连续段长度必然不小于 2。
+
+第二次操作，令 `x = x | (x >> 2)`，依次类推。通过 5 次操作即可实现提示中的转换。
+## 2.75
+$x,y$ 是无符号整数，$x'=U2T_w(x),y'=U2T_w(y)$。
+
+$$
+\begin{aligned}
+x'\times y'&=(x+x_{w-1}2^w)\times(y+y_{w-1}2^w)\\
+&=x\times y+(x_{w-1}\times y+y_{w-1}\times x)2^w+x_{w-1}y_{w-1}2^{2w}
+\end{aligned}$$
+
+因此有 $U2B_{2w}(x'\times y')=T2B_{2w}(x\times y+(x_{w-1}\times y+y_{w-1}\times x)2^w)$。即令 `a = signed_high_prod(x, y), b = unsigned_high_prod(x', y')`，$T2B_w(a+x_{w-1}\times y+y_{w-1}\times x)=U2B_w(b)$。
+
+## 2.80
+先得到粗糙的答案 `(x >> 2) + ((x >> 2) << 1)`，再根据 `x & 3` 进行修正。
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