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+{
+ "metadata": {
+  "name": "chapter11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h1>Chapter 11: Pointers<h1>"
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.1, Page 11.2<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# realtionship between two variables, their address and associated pointers\n\n\nu,v=[3],[]\npu=u\nv=pu\npv=v\n\nprint \"u=\",u[0],\nprint \"&u=\",id(u),\nprint \"pu=\",id(u),\nprint \"*pu\",pu[0]\n\nprint \"v=\",v[0],\nprint \"&v=\",id(v),\nprint \"pv=\",id(v),\nprint \"*pv\",pv[0]\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "u= 3 &u= 56575840 pu= 56575840 *pu 3\nv= 3 &v= 56575840 pv= 56575840 *pv 3\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.2, Page number: 11.3<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# a simple program using pointers\n\n\nv=[3]\nu1=2*(v[0]+5)\npv=v\nu2=2*(pv[0]+5)\n\nprint \"u1=%d u2=%d\" %(u1,u2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "u1=16 u2=16\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.3, Page munber: 11.4<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# a simple programme using pointers\n\nv=[3]\npv=v\nprint \"*pv=%d  v=%d\" %(pv[0],v[0])\n\npv[0]=0\nprint \"*pv=%d  v=%d\" %(pv[0],v[0])",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "*pv=3  v=3\n*pv=0  v=0\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.7, Page number: 11.6<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# difference between by pass by reference and pass by value\n\n\ndef funct1(u,v):\n    u=0\n    v=0\n    print \"Within funct1        :   u=%d    v=%d\" %(u,v)\n    return\n\ndef funct2(u,v):\n    u[0]=0\n    v[0]=0\n    print \"Within funct2        :  *pu=%d  *pv=%d\" %(u[0],v[0])\n    return\n\nu=[1]\nv=[3]\n\nprint \"Before calling funct1:   u=%d    v=%d\" %(u[0],v[0])\nfunct1(u[0],v[0])\nprint \"After calling funct1 :   u=%d    v=%d\" %(u[0],v[0])\n\nprint \"Before calling funct2:   u=%d    v=%d\" %(u[0],v[0])\nfunct2(u,v)\nprint \"After calling funct2 :   u=%d    v=%d\" %(u[0],v[0])",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Before calling funct1:   u=1    v=3\nWithin funct1        :   u=0    v=0\nAfter calling funct1 :   u=1    v=3\nBefore calling funct2:   u=1    v=3\nWithin funct2        :  *pu=0  *pv=0\nAfter calling funct2 :   u=0    v=0\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.8, Page number: 11.8<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Analyzing a line of text\n\n\n\ndef scan_line(line,pv,pc,pd,pw,po):\n\n    for c in line:\n\n        if c=='A' or c=='E' or c=='I' or c=='O' or c=='U':\n            pv[0]+=1\n        elif c>='A' and c<='Z':\n            pc[0]+=1\n        elif c>='0' and c<='9':\n            pd[0]+=1\n        elif c==' ' or c=='\\t':\n            pw[0]+=1\n        else:\n            po[0]+=1\n\n    return\n\n\nvowel,consonants,digits,whitespc,other=[0],[0],[0],[0],[0]\nline=\"Personal computers with memories in excess of 4096 KB are now quite common.\"\nline=line.upper()\nscan_line(line,vowel,consonants,digits,whitespc,other)\nprint \"\\n\\n\"\nprint \"No. of vowels                : \",vowel[0]\nprint \"No. of consonants            : \",consonants[0]\nprint \"No. of digits                : \",digits[0]\nprint \"No. of whitespace characters : \",whitespc[0]\nprint \"No. of other characters      : \",other[0]",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "\n\n\nNo. of vowels                :  23\nNo. of consonants            :  35\nNo. of digits                :  4\nNo. of whitespace characters :  12\nNo. of other characters      :  1\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.12, Page number: 11.15<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# unlike C in python the elements of a list are not arranged in memory uniformly\n\nx=[10,11,12,13,14,15,16,17,18,19]\n\nfor i in range(0,10):\n    print \"i=%d   x[i]=%d   *(x+1)=%d\" %(i,x[i],x[i]),\n    print \"   &x[i]=\",id(x[i]),\n    print \"   x+i=\",id(x[i])",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "i=0   x[i]=10   *(x+1)=10    &x[i]= 30696620    x+i= 30696620\ni=1   x[i]=11   *(x+1)=11    &x[i]= 30696608    x+i= 30696608\ni=2   x[i]=12   *(x+1)=12    &x[i]= 30696596    x+i= 30696596\ni=3   x[i]=13   *(x+1)=13    &x[i]= 30696584    x+i= 30696584\ni=4   x[i]=14   *(x+1)=14    &x[i]= 30696572    x+i= 30696572\ni=5   x[i]=15   *(x+1)=15    &x[i]= 30696560    x+i= 30696560\ni=6   x[i]=16   *(x+1)=16    &x[i]= 30696548    x+i= 30696548\ni=7   x[i]=17   *(x+1)=17    &x[i]= 30696536    x+i= 30696536\ni=8   x[i]=18   *(x+1)=18    &x[i]= 30696524    x+i= 30696524\ni=9   x[i]=19   *(x+1)=19    &x[i]= 30696512    x+i= 30696512\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.14, Page number: 11.17<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# string as one dimensional character array\n\ndef main():\n    global x\n    y=\"This string is declared within main\"\n    y=list(y)\n    x=list(x)\n    print x\n    print y\n\n    return\n\nx=\"This string is declared externally\"\nmain()\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "['T', 'h', 'i', 's', ' ', 's', 't', 'r', 'i', 'n', 'g', ' ', 'i', 's', ' ', 'd', 'e', 'c', 'l', 'a', 'r', 'e', 'd', ' ', 'e', 'x', 't', 'e', 'r', 'n', 'a', 'l', 'l', 'y']\n['T', 'h', 'i', 's', ' ', 's', 't', 'r', 'i', 'n', 'g', ' ', 'i', 's', ' ', 'd', 'e', 'c', 'l', 'a', 'r', 'e', 'd', ' ', 'w', 'i', 't', 'h', 'i', 'n', ' ', 'm', 'a', 'i', 'n']\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.16, Page number: 11.19<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Reoedering a list of numbers\n\n\ndef reorder(x):\n    n=len(x)\n    for item in range(0,n-1):\n        for i in range(item+1,n):\n            if x[i]<x[item]:\n                temp=x[item]\n                x[item]=x[i]\n                x[i]=temp\n\n    return\n\nn=10\nx=[]\n\nfor i in range(0,n):\n    inp=i+1\n    print \"\\ni=%d x=%d\" %(i+1,inp),\n    x.append(inp)\n\nreorder(x)\n\nprint \"\\n\"\nfor i in range(0,n):\n    print \"i=%d  x=%d\" %(i+1,x[i])\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "\ni=1 x=1 \ni=2 x=2 \ni=3 x=3 \ni=4 x=4 \ni=5 x=5 \ni=6 x=6 \ni=7 x=7 \ni=8 x=8 \ni=9 x=9 \ni=10 x=10 \n\ni=1  x=1\ni=2  x=2\ni=3  x=3\ni=4  x=4\ni=5  x=5\ni=6  x=6\ni=7  x=7\ni=8  x=8\ni=9  x=9\ni=10  x=10\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.17, Page number: 11.21<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Pointer arithmetic doesnt exixt in python\n\ni=[4]\ni[0]=1\nf=0.3\nd=0.005\nc='*'\n\npx=i\nprint 'Values: i=%d f=%f d=%f c=%c \\n' %(i[0],f,d,c)\nprint 'Addresses: &i=%X &f=%X &d=%X &c=%X \\n' %(id(i[0]),id(f),id(d),id(c))\nprint 'Pointer Values: px=%X  px+1=%X  px+2=%X  px+3=%X' %(id(px[0]),id(px[0]+1),id(px[0]+2),id(px[0]+3))\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Values: i=1 f=0.300000 d=0.005000 c=* \n\nAddresses: &i=1D46518 &f=32BD940 &d=32BD950 &c=1D9F158 \n\nPointer Values: px=1D46518  px+1=1D4650C  px+2=1D46500  px+3=1D464F4\n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.18, Page number: 11.22<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# two different pointer variable points to the first and last element of an integer array\n\na=[1,2,3,4,5,6]\n\npx=a[0]\npy=a[5]\n\nprint 'px=%X   py=%X \\n' %(id(px),id(py))\nprint 'py - px = %X' %(id(py)-id(px))",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "px=1D46518   py=1D464DC \n\npy - px = -3C\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.22, Page number: 11.26<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# adding two tables of numbers\n\ndef readinput(m,n,i=0):\n\n    at=[]\n    for row in range(0,m):\n        temp=[]\n        for col in range(0,n):\n            t=i\n            i+=2\n            temp.append(t)\n        at.append(temp)\n\n\n    return at\n        \ndef computesum(a,b,m,n):\n\n    c=[]\n\n    for row in range(0,m):\n        temp=[]\n        for col in range(0,n):\n            t=a[row][col]+b[row][col]\n            temp.append(t)\n        c.append(temp)\n\n    return c\n\ndef writeoutput(c,m,n):\n\n    for row in range(0,m):\n        for col in range(0,n):\n            print \"%4d\" %(c[row][col]),\n        print\n\n    return\n\n\n\nprint \"\\n FIRST TABLE : \\n\"\na=readinput(5,5,1)\nwriteoutput(a,5,5)\n\nprint \"\\n SECOND TABLE : \\n\"\nb=readinput(5,5,50)\nwriteoutput(b,5,5)\n\nc=computesum(a,b,5,5)\nprint \"Sums of the elements : \\n\"\nwriteoutput(c,5,5)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "\n FIRST TABLE : \n\n   1    3    5    7    9\n  11   13   15   17   19\n  21   23   25   27   29\n  31   33   35   37   39\n  41   43   45   47   49\n\n SECOND TABLE : \n\n  50   52   54   56   58\n  60   62   64   66   68\n  70   72   74   76   78\n  80   82   84   86   88\n  90   92   94   96   98\nSums of the elements : \n\n  51   55   59   63   67\n  71   75   79   83   87\n  91   95   99  103  107\n 111  115  119  123  127\n 131  135  139  143  147\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.24, Page number: 11.31<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# adding two tables of numbers\n\ndef readinput(m,n,i=0):\n\n    at=[]\n    for row in range(0,m):\n        temp=[]\n        for col in range(0,n):\n            t=i\n            i+=2\n            temp.append(t)\n        at.append(temp)\n\n\n    return at\n        \ndef computesum(a,b,m,n):\n\n    c=[]\n\n    for row in range(0,m):\n        temp=[]\n        for col in range(0,n):\n            t=a[row][col]+b[row][col]\n            temp.append(t)\n        c.append(temp)\n\n    return c\n\ndef writeoutput(c,m,n):\n\n    for row in range(0,m):\n        for col in range(0,n):\n            print \"%4d\" %(c[row][col]),\n        print\n\n    return\n\n\n\nprint \"\\n FIRST TABLE : \\n\"\na=readinput(5,5,1)\nwriteoutput(a,5,5)\n\nprint \"\\n SECOND TABLE : \\n\"\nb=readinput(5,5,50)\nwriteoutput(b,5,5)\n\nc=computesum(a,b,5,5)\nprint \"Sums of the elements : \\n\"\nwriteoutput(c,5,5)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "\n FIRST TABLE : \n\n   1    3    5    7    9\n  11   13   15   17   19\n  21   23   25   27   29\n  31   33   35   37   39\n  41   43   45   47   49\n\n SECOND TABLE : \n\n  50   52   54   56   58\n  60   62   64   66   68\n  70   72   74   76   78\n  80   82   84   86   88\n  90   92   94   96   98\nSums of the elements : \n\n  51   55   59   63   67\n  71   75   79   83   87\n  91   95   99  103  107\n 111  115  119  123  127\n 131  135  139  143  147\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.26, Page number: 11.34<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Reordering a list of strings\n\n\ndef reorder(x):\n\n    n=len(x)\n    for item in range(0,n-1):\n        for i in range(item+1,n):\n            if x[item]>x[i]:\n                temp=x[item]\n                x[item]=x[i]\n                x[i]=temp\n\n\n    return\n\nx=['PACIFIC','ATLANTIC','INDIAN','CARIBBEAN','BERING','BLACK','RED','NORTH','BALTIC','CASPIAN']\nprint 'Original list of strings :\\n\\n'\n\nfor i in x:\n    print \"String : \",i\n\nreorder(x)\n\nprint \"\\nReodered list of strings : \\n\\n\"\n\nfor i in x:\n    print \"String : \",i",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Original list of strings :\n\n\nString :  PACIFIC\nString :  ATLANTIC\nString :  INDIAN\nString :  CARIBBEAN\nString :  BERING\nString :  BLACK\nString :  RED\nString :  NORTH\nString :  BALTIC\nString :  CASPIAN\n\nReodered list of strings : \n\n\nString :  ATLANTIC\nString :  BALTIC\nString :  BERING\nString :  BLACK\nString :  CARIBBEAN\nString :  CASPIAN\nString :  INDIAN\nString :  NORTH\nString :  PACIFIC\nString :  RED\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.28, Page number: 11.37<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Display the day of the year\n\n\ndef convert(mm,dd,yy):\n    yy-=1900\n    ndays=long(30.42*(mm-1)+dd)\n    if mm==2:\n        ndays+=1\n    if mm>2 and mm<8:\n        ndays-=1\n    if yy%4==0 and mm<2:\n        ndays+=1\n\n    ncycles=yy/4\n    ndays+=ncycles*1461\n\n    nyears=yy%4\n    if nyears>0:\n        ndays+=365*nyears+1\n    if ndays>59:\n        ndays-=1\n\n    day=ndays%7\n\n    return day\n\ndef main(mm,dd,yy):\n    day_of_week=convert(mm,dd,yy)\n    print \"%s, %s %d %d\" %(weekday[day_of_week],month[mm-1],dd,yy)\n    return\n    \nweekday=['Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday']\nmonth=['January','February','March','April','May','June','July','August',\\\n       'September','October','November','December']\n\nmain(10,29,1929)\nprint\nmain(8,15,1945)\nprint\nmain(7,20,1969)\nprint\nmain(5,24,1997)\nprint\nmain(8,30,2010)\nprint\nmain(4,12,2069)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Tuesday, October 29 1929\n\nWednesday, August 15 1945\n\nSunday, July 20 1969\n\nSaturday, May 24 1997\n\nMonday, August 30 2010\n\nFriday, April 12 2069\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": "<h3>Example 11.30, Page number: 11.44<h3>"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "# Future value of monthly deposits (compound interest calculations)\n\n\ndef ratio_choose(freq,a,m,n):\n\n    if freq=='C':\n        ratio=md3(a,m,n)\n    elif freq=='D':\n        ratio=md2(a,m,n)\n    else:\n        ratio=md1(a,m,n)\n\n    return ratio\n\n\ndef table(freq,a,m,n):\n\n\n    print \"Interest Rate          Future Amount\\n\\n\"\n    for count in range(1,21):\n        i=0.01*count\n        f=a*ratio_choose(freq,i,m,n)\n        print \"    %2d                      %.2f\" %(count,f)\n\n    return\n\ndef md1(i,m,n):\n    \n    factor=1+i/m\n    ratio=12*(factor**(m*n)-1)/i\n    return ratio\n\ndef md2(i,m,n):\n\n    factor=1+i/m\n    ratio=(factor**(m*n)-1)/(factor**(m/12)-1)\n    return ratio\n\ndef md3(i,dummy,n):\n    ratio=(10**(i*n)-1)/(10**(i/12)-1)\n    return ratio\n\ndef main(freq):\n    m=0\n    freq=freq.upper()\n    if freq=='A':\n        m=1\n        print \"\\nAnual Compounding\\n\"\n        \n    elif freq=='S':\n        m=2\n        print \"\\nSemiannual Compounding\\n\"\n        \n    elif freq=='Q':\n        print \"\\nQuaterly Compounding\\n\"\n        m=4\n        \n    elif freq=='M':\n        m=12\n        print \"\\nMonthly Compounding\\n\"\n        \n    elif freq=='D':\n        m=360\n        print \"\\nDaily Compounding\\n\"\n\n    elif freq=='C':\n        m=0\n        print \"\\nContinuous Compounding\\n\"\n        \n    else:\n        print \"\\nERROR!!! Please Repeat\\n\\n\"\n        return\n        \n    return m\n\n\na,n,freq=100,3,'m'\nprint \"FUTURE VALUE OF A SERIES OF MONTHLY DEPOSITS\\n\\n\"\nprint \"Frequency of Compunding (A,S,Q,M,D,C): \",freq\nprint \"Amount of Each Monthly Payement : \",a\nprint \"Number of years: \",n\nm=main(freq)\n\ntable(freq,a,m,n)\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "FUTURE VALUE OF A SERIES OF MONTHLY DEPOSITS\n\n\nFrequency of Compunding (A,S,Q,M,D,C):  m\nAmount of Each Monthly Payement :  100\nNumber of years:  3\n\nMonthly Compounding\n\nInterest Rate          Future Amount\n\n\n     1                      3653.00\n     2                      3707.01\n     3                      3762.06\n     4                      3818.16\n     5                      3875.33\n     6                      3933.61\n     7                      3993.01\n     8                      4053.56\n     9                      4115.27\n    10                      4178.18\n    11                      4242.31\n    12                      4307.69\n    13                      4374.33\n    14                      4442.28\n    15                      4511.55\n    16                      4582.17\n    17                      4654.18\n    18                      4727.60\n    19                      4802.45\n    20                      4878.78\n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 25
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file