Added Kahan summation algorithm to question 3.

simmod/ass1/.gitignore

@@ -9,3 +9,4 @@ report.pdf
 fd*
 pr
 floating_point.tex
+kahan

simmod/ass1/Makefile

@@ -4,7 +4,7 @@ SPEED_TYPES=float double LD
 SUM_TYPES=float double
 OPS=ADD DIV MULT SQRT
 
-all: fp speed highlight report.pdf sum pr
+all: fp speed highlight report.pdf sum kahan pr
 
 highlight: floating_point.tex
 
@@ -34,11 +34,14 @@ fp: floating_point.o
 sum: sum.c
 	for t in $(SUM_TYPES); do \
 		sed "s#{TYPE}#$$t#" $^ > sum.$$t.c; \
-		gcc $(FLAGS) -o sum.$$t sum.$$t.c; \
+		$(CC) $(FLAGS) -o sum.$$t sum.$$t.c; \
 		rm sum.$$t.c; \
 	done;
 	touch $@
 
+kahan: kahan_sum.o
+	$(CC) $(FLAGS) -o $@ $^
+
 %.o: %.c
 	$(CC) $(FLAGS) -o $@ -c $^
 
@@ -47,4 +50,4 @@ sum: sum.c
 
 clean:
 	rm -vf *.o *.i *.s fp pr fd* speed speed.*.* floating_point \
-		report.pdf *.aux *.log *.toc sum sum.float sum.double
+		report.pdf *.aux *.log *.toc sum sum.float sum.double kahan

simmod/ass1/kahan_sum.c

+#include <stdlib.h>
+#include <stdio.h>
+
+float kahan_sum(int N) {
+    float sum = 0.0, c = 0.0, t, y;
+
+    for( int i = 1; i <= N; i++ ) {
+        y = 1.0/i - c;
+        t = sum + y;
+        c = (t - sum) - y;
+        sum = t;
+    }
+
+    return sum;
+}
+
+int main(void) {
+    printf("N = 1e8: %f\n", kahan_sum(1e8));
+    printf("N = 2e8: %f\n", kahan_sum(2e8));
+
+    return 0;
+}

simmod/ass1/report.tex

@@ -112,13 +112,13 @@ We've calculated $\sum_{i=1}^{N}\frac{1}{i}$ for $N = 10^8$ and $N = 2 \cdot
 \texttt{float} and \texttt{double}. The results of this are in the table below.
 
 \begin{table}[H]
-\begin{tabular}{l|lll}
-Type            & N              & Forward    & Backward \\
+\begin{tabular}{l|llll}
+Type                   & N              & Forward    & Backward   & Kahan\\
 \hline
-\texttt{float}  & $10^8$         & $15.40368$ & $18.80792$ \\
-\texttt{float}  & $2 \cdot 10^8$ & $15.40368$ & $18.80792$ \\
-\texttt{double} & $10^8$         & $18.99790$ & $18.99790$ \\
-\texttt{double} & $2 \cdot 10^8$ & $19.69104$ & $19.69104$ \\
+\texttt{float}         & $10^8$         & $15.40368$ & $18.80792$ & $18.99790$ \\
+\texttt{float}         & $2 \cdot 10^8$ & $15.40368$ & $18.80792$ & $19.69104$ \\
+\texttt{double}        & $10^8$         & $18.99790$ & $18.99790$ & \\
+\texttt{double}        & $2 \cdot 10^8$ & $19.69104$ & $19.69104$ & \\
 \end{tabular}
 \caption{Results of various summation approaches on floats and doubles.}
 \end{table}
@@ -149,11 +149,13 @@ Type            & N              & Forward    & Backward \\
     to the same problem as described above: all numbers in $[\frac{1}{10^8},
     \frac{1}{2 \cdot 10^8}]$ are also represented as zero and therefore not added
     to the result.
+    \item To improve the precision of the \texttt{float} data type summation, we
+    implemented the Kahan summation algorithm.
 \end{itemize}
 
 % }}}
 
-\appendix{}
+\appendix
 
 \section{floating\_point.c} % {{{
 \label{sec:floating_point.c}

simmod/ass1/sum.c

@@ -23,12 +23,12 @@ int main(void) {
     puts("Using type {TYPE}.");
 
     puts("Forward summation:");
-    printf("N = 1e8: %e\n", sum_forward(1e8));
-    printf("N = 2e8: %e\n", sum_forward(2e8));
+    printf("N = 1e8: %f\n", sum_forward(1e8));
+    printf("N = 2e8: %f\n", sum_forward(2e8));
 
     puts("Backward summation:");
-    printf("N = 1e8: %e\n", sum_backward(1e8));
-    printf("N = 2e8: %e\n", sum_backward(2e8));
+    printf("N = 1e8: %f\n", sum_backward(1e8));
+    printf("N = 2e8: %f\n", sum_backward(2e8));
 
     return 0;
 }