diff --git a/dynamic_programming/rod_cut/readme b/dynamic_programming/rod_cut/readme
new file mode 100644
index 000000000..076d6ea3b
--- /dev/null
+++ b/dynamic_programming/rod_cut/readme
@@ -0,0 +1 @@
+Dynamic programming solution to the rod-cut problem from Cormen's Introduction to Algorithms.
diff --git a/dynamic_programming/rod_cut/rod-cut-main.c b/dynamic_programming/rod_cut/rod-cut-main.c
new file mode 100644
index 000000000..1f56d48c1
--- /dev/null
+++ b/dynamic_programming/rod_cut/rod-cut-main.c
@@ -0,0 +1,18 @@
+#include <stdlib.h>
+#include <stdio.h>
+
+#include <dynamic-programming/rod-cut/rod-cut.h>
+
+
+int main(int argc, char *argv[]) {
+
+  int p[] = {0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30};
+  size_t len_of_p = sizeof p / sizeof *p;
+
+  for(size_t i = 0; i < len_of_p; i++) {
+    printf("itr %zd\n", i);
+    printf("memoized: %d\n", memoized_cut_rod(p, i));
+    printf("bottomup: %d\n", bottom_up_cut_rod(p, i));
+    putchar('\n');
+  }
+}
diff --git a/dynamic_programming/rod_cut/rod-cut.c b/dynamic_programming/rod_cut/rod-cut.c
new file mode 100644
index 000000000..b9da14c3c
--- /dev/null
+++ b/dynamic_programming/rod_cut/rod-cut.c
@@ -0,0 +1,56 @@
+#include <stdlib.h>
+
+#include <dynamic-programming/rod-cut/rod-cut.h>
+#include <util/debug.h>
+
+
+int max(int a, int b) {
+  return a > b ? a : b;
+}
+
+int memoized_cut_rod_aux(int *p, size_t n, int *r) {
+  if(r[n] >= 0)
+    return r[n];
+
+  int q;
+  if(n == 0)
+    q = 0;
+
+  else {
+    q = -1;
+    for(size_t i = 1; i <= n; i++)
+      q = max(q, p[i] + memoized_cut_rod_aux(p, n - i, r));
+  }
+
+  r[n] = q;
+  return q;
+}
+
+int memoized_cut_rod(int *p, size_t n) {
+  int *r;
+  check_mem(r = malloc((n + 1) * sizeof *r));
+  for(size_t i = 0; i <= n; i++)
+    r[i] = -1;
+
+  return memoized_cut_rod_aux(p, n, r);
+
+  error:
+  return -1;
+}
+
+int bottom_up_cut_rod(int *p, size_t n) {
+  int *r;
+  check_mem(r = malloc((n + 1) * sizeof *r));
+  r[0] = 0;
+
+  for(size_t j = 1; j <= n; j++) {
+    int q = -1;
+    for (size_t i = 1; i <= j; i++)
+      q = max(q, p[i] + r[j - i]);
+    r[j] = q;
+  }
+  return r[n];
+
+  error:
+  return -1;
+}