Greatest-Common-Divisor-Algorithms

A Comparison between different greatest common divisor algorithms:

• Euclidian GCD.
• Extended Euclidian GCD.
• Binary GCD recursive.
• Binary GCD non-recursive.

The Intuition Behind the Binary GCD Algorithm:

• $Since,$ $d = c.gcd(a, b)$ $Given$ $that:$ $d = gcd(ac, bc)$
• $This\ intuitively\ shows\ you\ that\ if\ two\ numbers\ share\ a\ common\ divisor,\ then\ the\ difference\ between\ the\ two\ numbers\ also\ shares\ the\ same\ divisor$

# if u and v are even, then
    return 2*binaryGCD_rec(u >> 1, v >> 1)  # 2*gcd(u/2, v/2) (From 1)

• $Since,$ $gcd(a, bc) = gcd(a, b)$ $if$ $gcd(a, c) = 1$
• $Since$ $gcd(odd\ number, 2) = 1$
• $This\ intuitively\ shows\ you\ that\ if\ one\ of\ the\ two\ numbers\ is\ even\ then,\ we\ can\ divide\ it\ by\ 2\ and\ its\ gcd\ won't\ change$

# if either u or v is even, then
    return binaryGCD_rec(u >> 1, v)  # 2*gcd(u/2, v) (From 2)
    return binaryGCD_rec(u, v >> 1)  # 2*gcd(u, v/2) (From 2)

• $Since,$ $gcd(a, b) = gcd(a−b, b)$ $assuming$ $a > b$
• $This\ intuitively\ shows\ you\ that\ we\ can\ reduce\ numbers\ using\ their\ difference$

# if either u and v are odd, then
# their difference is even, then
    return binaryGCD_rec((u-v) >> 1, v)  # 2*gcd((u-v)/2, v) (From 2, 3)
# or
    return binaryGCD_rec((v-u) >> 1, u)  # 2*gcd((v-u)/2, u) (From 2, 3)

Here, are the proofs of some of the used theroms and corollaries,

I. Prove that: if d = gcd(ac, bc), then d = c.gcd(a, b)

• $If$ $d = gcd(ac, bc)$
• $Then,$ $d|ac$ and $d|bc$
• $Then,$ $ac = dk_1$ and $bc = dk_2$
• $Then,$ $a = (d/c)k_1$ and $b = (d/c)k_2$
• $Then,$ $(d/c)|a$ and $(d/c)|b$
• $Then,$ $(d/c) = gcd(a, b)$
• $Then,$ $d = c.gcd(a, b)$

II. Prove that: if gcd(a, c) = 1 then, gcd(a, bc) = gcd(a, b)

• $Let,$ $d = gcd(a, bc)$
• $Then,$ $d|a$ and $d|bc$
• $Assume,$ $gcd(a, c) = 1$
• $Then,$ $d$ | $c$
• $But,$ $d|bc$
• $Then,$ $d|b$
• $Then,$ $d|gcd(a, b)$
• $Assume,$ $e = gcd(a, b)$
• $By\ definition,$ $e|a$ and $e|b$
• $Then,$ $e|bc$ $Theorem$
• $Then,$ $e|d$
• $But,$ $d|e$
• $Then,$ $d = e$
• $Then,$ $gcd(a, b) = gcd(a, bc)$

III. Prove that: gcd(a, b) = gcd(a−b, b) assuming a > b

We should prove it in the two directions:

• $Every\ common\ divisor\ of\ a\ and\ b\ is\ also\ a\ common\ divisor\ of\ a − b\ and\ b$
  • $If$ $a = dp$ and $b = dq$
  • $Then$ $a − b = d(p − q)$
• $Every\ common\ divisor\ of\ a − b\ and\ b \ is\ also\ a\ common\ divisor\ of\ a\ and\ b $
  • $If$ $a − b = dp$ and $b = dq$
  • $Then$ $a = d(p + q)$
• $Therefore,$ $gcd(a, b) = gcd(a − b, b)$